Simple algebras

Abstract

In this chapter, we return to the characterization of quaternion algebras. We initially defined quaternion algebras in terms of generators and relations in Chapter 2; in the chapters that followed, we showed that quaternion algebras are equivalently noncommutative algebras with a nondegenerate standard involution. Here, we pursue another approach, and we characterize quaternion algebras in a different way, as central simple algebras of dimension 4.

In this chapter, we return to the characterization of quaternion algebras. We initially defined quaternion algebras in terms of generators and relations in Chapter 2; in the chapters that followed, we showed that quaternion algebras are equivalently noncommutative algebras with a nondegenerate standard involution. Here, we pursue another approach, and we characterize quaternion algebras in a different way, as central simple algebras of dimension 4.

1 $\triangleright$ Motivation and summary

Consider now the “simplest” sorts of algebras. Like the primes among the integers or the finite simple groups among finite groups, it is natural to seek algebras that cannot be “broken down” any further. Accordingly, we say that a ring A is simple if it has no nontrivial two-sided (bilateral) ideals, i.e., the only two-sided ideals are $\{0\}$ and A. To show the power of this notion, consider this: if $\phi :A \rightarrow A'$ is a ring homomorphism and A is simple, then $\phi$ is either injective or the zero map (since $\ker \phi \subseteq B$ is a two-sided ideal).

A division ring A is simple, since every nonzero element is a unit and therefore every nonzero ideal (left, right, or two-sided) contains 1 so is equal to A. In particular, a field is a simple ring, and a commutative ring is simple if and only if it is a field. The matrix ring ${{\,\mathrm{M}\,}}_n(F)$ over a field F is also simple, something that can be checked directly by multiplying by matrix units (Exercise 7.5).

Moreover, quaternion algebras are simple. The shortest proof of this statement, given what we have done so far, is to employ Main Theorem 5.4.4 (and Theorem 6.4.11 in characteristic 2): a quaternion algebra B over F is either isomorphic to ${{\,\mathrm{M}\,}}_2(F)$ or is a division ring, and in either case is simple. One can also prove this directly (Exercise 7.1).

Although the primes are quite mysterious and the classification of finite simple groups is a monumental achievement in group theory, the situation for algebras is quite simple, indeed! Our first main result is as follows (Main Theorem 7.3.10).

Theorem 7.1.1

(Wedderburn–Artin). Let F be a field and B be a finite-dimensional F-algebra. Then B is simple if and only if $B \simeq {{\,\mathrm{M}\,}}_n(D)$ where $n \ge 1$ and D is a finite-dimensional division F-algebra.

As a corollary of Theorem 7.1.1, we give another characterization of quaternion algebras.

Corollary 7.1.2

Let B be an F-algebra. Then the following are equivalent:

• (i) B is a quaternion algebra;

• (ii) $B \otimes _F F{}^{al }\simeq {{\,\mathrm{M}\,}}_2(F{}^{al })$, where $F{}^{al }$ is an algebraic closure of F; and

• (iii) B is a central simple algebra of dimension $\dim _F B=4$.

Moreover, a central simple algebra B of dimension $\dim _F B=4$ is either a division algebra or has $B \simeq {{\,\mathrm{M}\,}}_2(F)$.

This corollary has the neat consequence that a division algebra B over F is a quaternion algebra over F if and only if it is central of dimension $\dim _F B=4$.

For the reader in a hurry, we now give a proof of this corollary without invoking the Wedderburn–Artin theorem; this proof also serves as a preview of some of the ideas that go into the theorem.

Proof of Corollary 7.1.2. The statement (i) $\Rightarrow$ (ii) was proven in Exercise 2.4(d).

To prove (ii) $\Rightarrow$ (iii), suppose B is an algebra with $B{}^{al }:=B \otimes _F F{}^{al }\simeq {{\,\mathrm{M}\,}}_2(F{}^{al })$. The $F{}^{al }$-algebra $B{}^{al }$ is central simple, from above. Thus $Z(B) = Z(B{}^{al }) \cap B = F$. And if I is a two-sided ideal of B then $I{}^{al }:=I \otimes _F F{}^{al }$ is a two-sided ideal of $B{}^{al }$, so $I{}^{al }=\{0\}$ or $I{}^{al }= B{}^{al }$ is trivial, whence $I = I{}^{al }\cap F$ is trivial. Finally, $\dim _F B = \dim _{F{}^{al }} B{}^{al }= 4$.

Finally, we prove (iii) $\Rightarrow$ (i). Let B a central simple F-algebra of dimension 4. If B is a division algebra we are done; so suppose not. Then B has a nontrivial left ideal (e.g., one generated by a nonunit); let $\{0\} \subsetneq I \subsetneq B$ be a nontrivial left ideal with $0<m=\dim _F I$ minimal. Then there is a nonzero homomorphism $B \rightarrow {{\,\mathrm{End}\,}}_F(I) \simeq {{\,\mathrm{M}\,}}_m(F)$ which is injective, since B is simple. By dimension, we cannot have $m=1$; if $m=2$, then $B \simeq {{\,\mathrm{M}\,}}_2(F)$ and we are done. So suppose $m=3$. Then by minimality, every nontrivial left ideal of B has dimension 3. But for any $\alpha \in B$, we have that $I\alpha$ is a left ideal, so the left ideal $I \cap I\alpha$ is either $\{0\}$ or I; in either case, $I\alpha \subseteq I$ and I is a right ideal as well. But this contradicts the fact that B is simple. $\square$

The Wedderburn–Artin theorem is an important structural result used throughout mathematics, so we give in this chapter a self-contained account of its proof. More generally, it will be convenient to work withsemisimple algebras, finite direct products of simple algebras. When treating ideals of an algebra we would be remiss if we did not discuss more generally modules over the algebra, and the notions of simple and semisimple module are natural concepts in linear algebra and representation theory: a semisimple module is one that is a direct sum of simple modules (“completely reducible”), analogous to a semisimple operator where every invariant subspace has an invariant complement (e.g., a diagonalizable matrix).

The second important result in this chapter is a theorem that concerns the simple subalgebras of a simple algebra, as follows (Main Theorem 7.7.1).

Theorem 7.1.3

(Skolem–Noether).  Let A, B be simple F-algebras and suppose that B is central. Suppose that $f,g :A \rightarrow B$ are homomorphisms. Then there exists $\beta \in B$ such that $f(\alpha )=\beta ^{-1}g(\alpha )\beta$ for all $\alpha \in A$.

Corollary 7.1.4

Every F-algebra automorphism of a simple F-algebra B is inner, i.e., ${{\,\mathrm{Aut}\,}}(B) \simeq B^\times /F^\times$.

Just as above, for our quaternionic purposes, we can give a direct proof.

Corollary 7.1.5

Let B be a quaternion algebra over F and let $K_1,K_2 \subset B$ be quadratic subfields. Suppose that is an isomorphism of F-algebras. Then $\phi$ lifts to an inner automorphism of B, i.e., there exists $\beta \in B$ such that $\alpha _2=\phi (\alpha _1)=\beta ^{-1} \alpha _1 \beta$ for all $\alpha _1 \in K_1$. In particular, $K_2=\beta ^{-1} K_1 \beta$.

Proof. Write $K_1=F(\alpha _1)$ with $\alpha _1 \in B$ and let $\alpha _2 =\phi (\alpha _1) \in K_2 \subset B$, so $K_2=F(\alpha _2)$. We want to find $\beta \in B^\times$ such that $\alpha _2=\beta ^{-1}\alpha _1 \beta$. In the special case $B \simeq {{\,\mathrm{M}\,}}_2(F)$, then $\alpha _1,\alpha _2 \in {{\,\mathrm{M}\,}}_2(F)$ satisfy the same irreducible characteristic polynomial, so by the theory of rational canonical forms, $\alpha _2=\beta ^{-1}\alpha _1\beta$ where $\beta \in B^\times \simeq {{\,\mathrm{GL}\,}}_2(F)$ as desired.

Suppose then that B is a division ring. Then the set

$$\begin{aligned} W=\{\beta \in B : \beta \alpha _2=\alpha _1\beta \} \end{aligned}$$

(7.1.6)

is an F-vector subspace of B. Let $F{}^{sep }$ be a separable closure of F. (Or, apply Exercise 6.5 and work over a splitting field K linearly disjoint from $K_1 \simeq K_2$.) Then we have $B \otimes _F F{}^{sep }\simeq {{\,\mathrm{M}\,}}_2(F{}^{sep })$, and the common characteristic polynomial of $\alpha _1,\alpha _2$ either remains irreducible over $F{}^{sep }$ (if $K \supset F$ is inseparable) or splits as the product of two linear factors with distinct roots. In either case, the theory of rational canonical forms again applies, and there exists $\beta \in B \otimes _F F{}^{sep }\simeq {{\,\mathrm{GL}\,}}_2(F{}^{sep })$ that will do; but then by linear algebra $\dim _{F{}^{sep }} W \otimes _F F{}^{sep }= \dim _F W>0$, so there exists $\beta \in B \setminus \{0\}=B^\times$ with the desired property. $\square$

As shown in the above proof, Corollary 7.1.5 can be seen as a general reformulation of the rational canonical form from linear algebra.

2 Simple modules

Basic references for this section include Drozd–Kirichenko [DK94, §1–4], Curtis–Reiner [CR81, §3], Lam [Lam2001, §2–3], and Farb–Dennis [FD93, Part I]. An elementary approach to the Weddernburn–Artin theorem is given by Brešar [Bre2010]. An overview of the subject of associative algebras is given by Pierce [Pie82] and Jacobson [Jacn2009].

Throughout this chapter, let B be a finite-dimensional F-algebra.

To understand the algebra B, we look at its representations. Arepresentation of B (over F) is a vector space V over F together with an F-algebra homomorphism $B \rightarrow {{\,\mathrm{End}\,}}_F(V)$. Equivalently, a representation is given by a left (or right) B-module V: this is almost a tautology. Although one can define infinite-dimensional representations, they will not interest us here, and we suppose throughout that $\dim _F V<\infty$, or equivalently that V is a finitely generated (left or right) B-module. If we choose a basis for V, we obtain an isomorphism ${{\,\mathrm{End}\,}}_F(V) \simeq {{\,\mathrm{M}\,}}_n(F)$ where $n=\dim _F V$, so a representation is just a homomorphic way of thinking of the algebra B as an algebra of matrices.

Example 7.2.1

The space of column vectors $F^n$ is a left ${{\,\mathrm{M}\,}}_n(F)$-module; the space of row vectors is a right ${{\,\mathrm{M}\,}}_n(F)$-module.

Example 7.2.2

B is itself a left B-module, giving rise to the left regular representation $B \rightarrow {{\,\mathrm{End}\,}}_F(B)$ over F (cf. Remark 3.3.8).

Example 7.2.3

Let G be a finite group. Then a representation of F[G] (is the same as an F[G]-module which) is the same as a homomorphism $G \rightarrow {{\,\mathrm{GL}\,}}(V)$, where V is an F-vector space (Exercise 3.8).

Definition 7.2.4

Let V be a left B-module. We say V issimple (or irreducible) if $V \ne \{0\}$ and the only B-submodules of V are $\{0\}$ and V.

We say V isindecomposable if V cannot be written as $V=V_1 \oplus V_2$ with $V_1,V_2 \ne \{0\}$ left B-modules.

A simple module is indecomposable, but the converse need not hold, and this is a central point of difficulty in understanding representations.

Example 7.2.5

If $B=\left\{ \begin{pmatrix} a &{} b \\ 0 &{} c \end{pmatrix} : a,b,c \in F\right\} \subseteq {{\,\mathrm{M}\,}}_2(F)$, then the space $V=F^2$ of column vectors is not simple, since the subspace spanned by $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ is a B-submodule; nevertheless, V is indecomposable (Exercise 7.4).

The importance of simple modules is analogous to that of simple groups. Arguing by induction on the dimension of V, we have the following lemma analogous to the Jordan–Hölder theorem on composition series.

Lemma 7.2.6

A (finite-dimensional) left B-module V admits a filtration

$$\begin{aligned} V=V_0 \supsetneq V_1 \supsetneq V_2 \supsetneq \dots \supsetneq V_r = \{0\} \end{aligned}$$

such that $V_i/V_{i+1}$ is simple for each i.

This filtration is not unique, but up to isomorphism and permutation, the quotients $V_i/V_{i+1}$ are unique.

Lemma 7.2.7

If I is a maximal left ideal of B, then B/I is a simple B-module. Conversely, if V is a simple B-module, then $V \simeq B/I$ for a maximal left ideal I: more precisely, for any $x \in V \smallsetminus \{0\}$, we may take

$$\begin{aligned} I={{\,\mathrm{ann}\,}}(x) :=\{\alpha \in B : \alpha x = 0\}. \end{aligned}$$

Proof. For the first statement, a submodule of B/I corresponds to a left ideal containing I, so B/I is simple if and only if I is maximal. Conversely, letting $x \in V \smallsetminus \{0\}$ we have $\{0\} \ne Bx \subseteq V$ a B-submodule and so $Bx=V$; and consequently $V \simeq B/I$ where $I={{\,\mathrm{ann}\,}}(x)$ and again I is a maximal left ideal. $\square$

Having defined the notion of simplicity for modules, we now consider simplicity of the algebra B.

Definition 7.2.8

An F-algebra B issimple if the only two-sided ideals of B are $\{0\}$ and B.

Equivalently, B is simple if and only if any F-algebra (or even ring) homomorphism $B \rightarrow A$ is either injective or the zero map.

Example 7.2.9

A division F-algebra D is simple. In fact, the F-algebra ${{\,\mathrm{M}\,}}_n(D)$ is simple for any division F-algebra D (Exercise 7.5), and in particular ${{\,\mathrm{M}\,}}_n(F)$ is simple.

Example 7.2.10

Let $F{}^{al }$ be an algebraic closure of F. If $B \otimes _F F{}^{al }$ is simple, then B is simple. The association $I \mapsto I \otimes _F F{}^{al }$ is an injective map from the set of two-sided ideals of B to the set of two-sided ideals of $B \otimes _F F{}^{al }$.

7.2.11

If B is a quaternion algebra over F, then B is simple. We have $B \otimes _F F{}^{al }\simeq {{\,\mathrm{M}\,}}_2(F{}^{al })$, which is simple by Example 7.2.9, and B is simple by Example 7.2.10.

Example 7.2.9 shows that algebras of the form ${{\,\mathrm{M}\,}}_n(D)$ with D a division F-algebra yield a large class of simple F-algebras. In fact, these are all such algebras, a fact we will now prove. First, a few preliminary results.

Lemma 7.2.12

(Schur). Let B be an F-algebra. Let $V_1,V_2$ be simple B-modules. Then any homomorphism $\phi :V_1 \rightarrow V_2$ of B-modules is either zero or an isomorphism.

Proof. We have that $\ker \phi$ and ${{\,\mathrm{img}\,}}\phi$ are B-submodules of $V_1$ and $V_2$, respectively, so either $\phi =0$ or $\ker \phi =\{0\}$ and ${{\,\mathrm{img}\,}}\phi =V_2$, hence $V_1 \simeq V_2$. $\square$

Corollary 7.2.13

If V is a simple B-module, then ${{\,\mathrm{End}\,}}_B(V)$ is a division ring.

7.2.14

Let B be an F-algebra and consider B as a left B-module. Then there is a map

$$\begin{aligned} \rho :B^{op }&\rightarrow {{\,\mathrm{End}\,}}_B(B) \\ \alpha&\mapsto (\rho _\alpha :\beta \mapsto \beta \alpha ), \end{aligned}$$

where $B^{op }$ is the opposite algebra of B defined in 3.2.2. The map $\rho$ is injective since $\rho _\alpha =0$ implies $\rho _\alpha (1)=\alpha =0$; it is also surjective, since if $\phi \in {{\,\mathrm{End}\,}}_B(B)$ then letting $\alpha =\phi (1)$ we have $\phi (\beta )=\beta \phi (1)=\beta \alpha$ for all $\beta \in B$. Finally, it is an F-algebra homomorphism, since

$$\begin{aligned} \rho _{\alpha \beta }(\mu )=\mu (\alpha \beta )=(\mu \alpha )\beta =(\rho _\beta \circ \rho _\alpha )(\mu ), \end{aligned}$$

and therefore $\rho$ is an isomorphism of F-algebras.

One lesson here is that a left module has endomorphisms that act naturally on the right; but the more common convention is that endomorphisms also act on the left. In order to make this compatible, the opposite algebra intervenes.

7.2.15

More generally, the decomposition of modules is determined by idempotent endomorphisms as follows. Let V be a left B-module. Then V is indecomposable if and only if ${{\,\mathrm{End}\,}}_B(V)$ has no nontrivialidempotents: that is to say, if $e \in {{\,\mathrm{End}\,}}_B(V)$ satisfies $e^2=e$, then $e \in \{ 0,1\}$. Given a nontrivial idempotent, we can write $V=eV \oplus (1-e)V$, and conversely if $V=V_1 \oplus V_2$ then the projection $V \rightarrow V_1 \subseteq V$ gives an idempotent.

7.2.16

Many theorems of linear algebra hold equally well over division rings as they do over fields, as long as one is careful about the direction of scalar multiplication. For example, let D be a division F-algebra and let V be a left D-module. Then $V \simeq D^n$ is free, and choice of basis for V gives an isomorphism ${{\,\mathrm{End}\,}}_D(V) \simeq {{\,\mathrm{M}\,}}_n(D^{op })$. When $n=1$, this becomes ${{\,\mathrm{End}\,}}_D(D) \simeq D^{op }$, as in 7.2.14.

Lemma 7.2.17

Let B be a (finite-dimensional) simple F-algebra. Then there exists a simple left B-module which is unique up to isomorphism.

Proof. Since B is finite-dimensional over F, there is a nonzero left ideal I of B of minimal dimension, and such an ideal I is necessarily simple. Moreover, if $\nu \in I$ is nonzero then $B\nu = I$, since $B\nu \subseteq I$ is nonzero and I is simple. Let $I=B\nu$ with $\nu \in I$.

Now let V be any simple B-module; we will show $I \simeq V$ as B-modules. Since B is simple, the natural map $B \rightarrow {{\,\mathrm{End}\,}}_F(V)$ is injective (since it is nonzero). Therefore, there exists $x \in V$ such that $\nu x \ne 0$, so $Ix \ne \{0\}$. Thus, the map $I \rightarrow V$ by $\beta \mapsto \beta x$ is a nonzero B-module homomorphism, so it is an isomorphism by Schur’s lemma. $\square$

Example 7.2.18

The unique simple left ${{\,\mathrm{M}\,}}_n(F)$-module (up to isomorphism) is the space $F^n$ of column vectors (Example 7.2.1).

7.2.19

Every algebra can be decomposed according to its idempotents 7.2.15. Let B be a finite-dimensional F-algebra. Then we can write $B=I_1 \oplus \dots \oplus I_r$ as a direct sum of indecomposable left B-modules: this follows by induction, as the decomposing procedure must stop because each factor is a finite-dimensional F-vector space. This means we may write

$$\begin{aligned} 1=e_1+\dots +e_r \end{aligned}$$

with $e_i \in I_i$. For each $\alpha \in I_i$ we have $\alpha =\sum _i \alpha e_i$ whence $\alpha e_i=\alpha$ and $\alpha e_j=0$ for $j \ne i$, which implies that

$$\begin{aligned} e_i^2=e_i, \quad e_ie_j=0 \quad \text {for }j \ne i, \quad \text { and }I_i=Be_i. \end{aligned}$$

Thus each $e_i$ is idempotent; we call $\{e_1,\dots ,e_r\}$ a complete set ofprimitive orthogonal idempotents: the orthogonal is because $e_ie_j=0$ for $j \ne i$, and the primitive is because each $e_i$ is not the sum of two other orthogonal idempotents (by 7.2.15).

Remark 7.2.20. The tight connection between F and ${{\,\mathrm{M}\,}}_n(F)$ is encoded in the fact that the two rings areMorita equivalent: there is an equivalence of categories between F-vector spaces and left ${{\,\mathrm{M}\,}}_n(F)$-modules. For more on this rich subject, see Lam [Lam99, §18], Reiner [Rei2003, Chapter 4], and Curtis–Reiner [CR81, §35].

3 Semisimple modules and the Wedderburn–Artin theorem

We continue our assumptions that B is a finite-dimensional F-algebra and a B-module V is finite-dimensional.

Definition 7.3.1

A B-module V is semisimple (or  completely reducible) if V is isomorphic to a (finite) direct sum of simple B-modules ${V \simeq \bigoplus _i V_i}$.

B is a semisimple F-algebra if B is semisimple as a left B-module.

Remark 7.3.2. More precisely, we have defined the notion of left semisimple and could equally well define right semisimple; below we will see that these two notions are the same.

Example 7.3.3

If $B=F$, then simple F-modules are one-dimensional vector spaces, and as F is simple these are the only ones. Every F-vector space has a basis and so is the direct sum of one-dimensional subspaces, thus every F-module is semisimple.

Example 7.3.4

A finite-dimensional commutative F-algebra B is semisimple if and only if B is the product of field extensions of F, i.e., $B \simeq K_1 \times \dots \times K_r$ with $K_i \supseteq F$ a finite extension of fields.

Lemma 7.3.5

The following statements hold.

• (a) A B-module V is semisimple if and only if it is the sum of simple B-modules.

• (b) A submodule or a quotient module of a semisimple B-module is semisimple.

• (c) If B is a semisimple F-algebra, then every B-module is semisimple.

Proof. For (a), let $V=\sum _i V_i$ be the sum of simple B-modules. Since V is finite-dimensional, we can rewrite it as an irredundant finite sum; and then since each $V_i$ is simple, the intersection of any two distinct summands is $\{0\}$, so the sum is direct.

For (b), let $W \subseteq V$ be a submodule of the semisimple B-module V. Every $x \in W$ with $x \ne 0$ is contained in a simple B-submodule of W by minimality, so $W = \sum _i W_i$ is a sum of simple B-modules. The result now follows from (a) for submodules. For quotient modules, suppose $\phi :V \rightarrow Z$ is a surjective B-module homomorphism; then $\phi ^{-1}(Z) \subseteq V$ is a B-submodule, and $\phi ^{-1}(Z)=\sum _i W_i$ is a sum of simple B-modules, and hence by Schur’s lemma $Z=\sum _i \phi (W_i)$ is semisimple.

For (c), let V be a B-module. Since V is finitely generated as a B-module, there is a surjective B-module homomorphism $B^r \rightarrow V$ for some $r \ge 1$. Since $B^r$ is semisimple, so too is V by (b). $\square$

Remark 7.3.6. Doing linear algebra with semisimple modules mirrors very closely linear algebra over a field. We have already seen that every submodule and quotient module of a semisimple module is again semisimple. Moreover, every module homomorphism $V \rightarrow W$ with V semisimple splits, and every submodule of a semisimple module is a direct summand. The extent to which this fails over other rings concerns the structure of projective modules; we take this up in Chapter 20.

Lemma 7.3.7

If B is a simple F-algebra, then B is a semisimple F-algebra.

Proof. Let $I \subseteq B$ be a minimal nonzero left ideal, the unique simple left B-module up to isomorphism as in Lemma 7.2.17. For all $\alpha \in B$, the left ideal $I\alpha$ is a homomorphic image of I, so by Schur’s lemma, either $I\alpha =\{0\}$ or $I\alpha$ is simple. Let $A :=\sum _{\alpha \in B} \,I\, \alpha$. Then A is a nonzero two-sided ideal of B, so since B is simple, we conclude $A=B$. Thus B is the sum of simple B-modules, and the result follows from Lemma 7.3.5(a). $\square$

Corollary 7.3.8

A (finite) direct product of simple F-algebras is a semisimple F-algebra.

Proof. If $B \simeq B_1 \times \dots \times B_r$ with each $B_i$ simple, then by Lemma 7.3.7, each $B_i$ is semisimple so $B_i = \bigoplus _j I_{ij}$ is the direct sum of simple $B_i$-modules $I_{ij}$. Each $I_{ij}$ has the natural structure of a B-module (extending by zero), and with this structure it is simple, and $B=\bigoplus _{i,j} I_{ij}$ is semisimple. $\square$

The converse of Corollary 7.3.8 is true and is proven as Corollary 7.3.14, a consequence of the Wedderburn–Artin theorem.

In analogy to 7.2.16, we have the following corollary.

Corollary 7.3.9

Let B be a simple F-algebra and let V be a left B-module. Then $V \simeq I^{\oplus n}$ for some $n \ge 1$, where I is a simple left B-module. In particular, two left B-modules $V_1,V_2$ are isomorphic if and only if $\dim _F V_1=\dim _F V_2$.

Proof. Since B is simple, B is semisimple by Lemma 7.3.7, and V is semisimple by Lemma 7.3.5. But by Lemma 7.2.17, there is a unique simple left B-module I, and the result follows. $\square$

In other words, this corollary says that if B is simple then every left B-module V is free over B, so has a left basis over B; if we define the rank of a left B-module V to be cardinality of this basis (the integer n such that $V \simeq I^{\oplus n}$ as in Corollary 7.3.9), then two such modules are isomorphic if and only if they have the same rank.

We now come to one of the main results of this chapter.

MainTheorem 7.3.10

(Wedderburn–Artin).  Let B be a finite-dimensional F-algebra. Then B is semisimple if and only if there exist integers $n_1,\dots ,n_r$ and division algebras $D_1,\dots ,D_r$ such that

$$\begin{aligned} B \simeq {{\,\mathrm{M}\,}}_{n_1}(D_1) \times \dots \times {{\,\mathrm{M}\,}}_{n_r}(D_r). \end{aligned}$$

Such a decomposition is unique up to permuting the integers $n_1,\dots ,n_r$ and applying an isomorphism to the division rings $D_1,\dots ,D_r$.

Proof. If $B \simeq \prod _i M_{n_i}(D_i)$, then each factor $M_{n_i}(D_i)$ is a simple F-algebra by Example 7.2.9, so by Corollary 7.3.8, B is semisimple.

So suppose B is semisimple. Then we can write B as a left B-module as the direct sum $B \simeq I_1^{\oplus n_1} \oplus \dots \oplus I_r^{\oplus n_r}$ of simple B-modules $I_1,\dots ,I_r$, grouped up to isomorphism. We have ${{\,\mathrm{End}\,}}_B(B) \simeq B^{op }$ by 7.2.14. By Schur’s lemma,

$$\begin{aligned} {{\,\mathrm{End}\,}}_B(B) \simeq \bigoplus _i {{\,\mathrm{End}\,}}_B\bigl (I_i^{\oplus n_i}\bigr ); \end{aligned}$$

by 7.2.16,

$$\begin{aligned} {{\,\mathrm{End}\,}}_B\left( I_i^{\oplus n_i}\right) \simeq {{\,\mathrm{M}\,}}_{n_i}(D_i) \end{aligned}$$

where $D_i={{\,\mathrm{End}\,}}_B(I_i)$ is a division ring. So

$$\begin{aligned} B \simeq {{\,\mathrm{End}\,}}_B(B)^{op } \simeq {{\,\mathrm{M}\,}}_{n_1}(D_1^{op }) \times \dots \times {{\,\mathrm{M}\,}}_{n_r}(D_r^{op }). \end{aligned}$$

The statements about uniqueness are then clear. $\square$

Remark 7.3.11. Main Theorem 7.3.10 as it is stated was originally proven by Wedderburn [Wed08], and so is sometimes called Wedderburn’s theorem. However, this term may also apply to the theorem of Wedderburn that a finite division ring is a field; and Artin generalized Main Theorem 7.3.10 to rings where the ascending and descending chain condition holds for left ideals [Art26]. We follow the common convention by referring to Main Theorem 7.3.10 as the Wedderburn–Artin theorem.

Corollary 7.3.12

Let B be a simple F-algebra. Then $B \simeq {{\,\mathrm{M}\,}}_n(D)$ for a unique $n \in \mathbb Z _{\ge 1}$ and a division algebra D unique up to isomorphism.

Example 7.3.13

Let B be a division F-algebra. Then $V=B$ is a simple B-module, and in Corollary 7.3.12 we have $D={{\,\mathrm{End}\,}}_B(B)=B^{op }$, and the Wedderburn–Artin isomorphism is just $B \simeq {{\,\mathrm{M}\,}}_1((B^{op })^{op })$.

Corollary 7.3.14

An F-algebra B is semisimple if and only if B is the direct product of simple F-algebras.

Proof. Immediate from the Wedderburn–Artin theorem, as each factor ${{\,\mathrm{M}\,}}_{n_i}(D_i)$ is simple. $\square$

4 Jacobson radical

We now consider an important criterion for establishing the semisimplicity of an F-algebra. Let B be a finite-dimensional F-algebra.

Definition 7.4.1

TheJacobson radical ${{\,\mathrm{rad}\,}}B$ of B is the intersection of all maximal left ideals of B.

We will in Corollary 7.4.6 see that this definition has left-right symmetry. Before doing so, we see right away the importance of the Jacobson radical in the following lemma.

Lemma 7.4.2

B is semisimple if and only if ${{\,\mathrm{rad}\,}}B=\{0\}$.

Proof. First, suppose B is semisimple. Then B as a left B-module is isomorphic to the direct sum of simple left ideals of B. Suppose ${{\,\mathrm{rad}\,}}B \ne \{0\}$; then ${{\,\mathrm{rad}\,}}B$ contains a minimal, hence simple, nonzero left ideal $I \subseteq B$. Then $B=I \oplus I'$ for some B-submodule $I'$ and $B/I' \simeq I$ so $I'$ is a maximal left ideal. Therefore ${{\,\mathrm{rad}\,}}B \subseteq I'$, but then ${{\,\mathrm{rad}\,}}B \cap I = \{0\}$, a contradiction.

Conversely, suppose ${{\,\mathrm{rad}\,}}B=\{0\}$. Suppose B is not semisimple. Let $I_1$ be a minimal left ideal of B. Since $I_1 \ne \{0\}={{\,\mathrm{rad}\,}}B$, there exists a maximal left ideal $J_1$ not containing $I_1$, so $I_1 \cap J_1 = \{0\}$ and $B=I_1 \oplus J_1$. Since B is not semisimple, $J_1 \ne \{0\}$, and there exists a minimal left ideal $I_2 \subsetneq J_1 \subseteq B$. Continuing in this fashion, we obtain a descending chain $J_1 \supsetneq J_2 \supsetneq \dots$, a contradiction. $\square$

Corollary 7.4.3

$B/{{\,\mathrm{rad}\,}}B$ is semisimple.

Proof. Let $J={{\,\mathrm{rad}\,}}B$. Under the natural map $B \rightarrow B/J$, the intersection of all maximal left ideals of $B/{{\,\mathrm{rad}\,}}B$ corresponds to the intersection of all maximal left ideals of B containing J; but ${{\,\mathrm{rad}\,}}B$ is the intersection thereof, so ${{\,\mathrm{rad}\,}}(B/J)=\{0\}$ and by Lemma 7.4.2, B/J is semisimple. $\square$

We now characterize the Jacobson radical in several ways.

7.4.4

For a left B-module V, define its annihilator  by

$$\begin{aligned} {{\,\mathrm{ann}\,}}V :=\{\alpha \in B : \alpha V = 0\}. \end{aligned}$$

Every annihilator ${{\,\mathrm{ann}\,}}V$ is a two-sided ideal of B: if $\alpha \in {{\,\mathrm{ann}\,}}(V)$ and $\beta \in B$, then $\alpha \beta V \subseteq \alpha V = \{0\}$ so $\alpha \beta \in {{\,\mathrm{ann}\,}}(V)$.

Lemma 7.4.5

The Jacobson radical is equal to the intersection of the annihilators of all simple left B-modules: i.e., we have ${{\,\mathrm{rad}\,}}B= \bigcap _V {{\,\mathrm{ann}\,}}V$, the intersection taken over all simple left B-modules. Moreover, if $\alpha \in {{\,\mathrm{rad}\,}}B$, then $1-\alpha \in B^\times$.

Proof. We begin with the containment $(\supseteq )$. Let $\alpha \in \bigcap _V {{\,\mathrm{ann}\,}}V$ and let I be a maximal left ideal. Then $V=B/I$ is a simple left B-module, so $\alpha \in {{\,\mathrm{ann}\,}}(B/I)$ whence $\alpha B \subseteq I$ and $\alpha \in I$.

The containment $(\subseteq )$ follows with a bit more work. Let $\alpha \in {{\,\mathrm{rad}\,}}B$, and let V be a simple left B-module. Assume for purposes of contradiction that $x \in V$ has $\alpha x \ne 0$. Then as in Lemma 7.2.7, $V=B(\alpha x)$ so $x=\beta \alpha x$ for some $\beta \in B$ and $(1-\beta \alpha )x=0$. Let I be a maximal left ideal containing $1-\beta \alpha$. Since $\alpha \in {{\,\mathrm{rad}\,}}B$, we have $\alpha \in I$, and thus $1=(1-\beta \alpha )+\beta \alpha \in I$, a contradiction. Thus $\alpha V = \{0\}$ and $\alpha \in {{\,\mathrm{ann}\,}}V$.

The final statement follows along similar lines as the previous paragraph (Exercise 7.10). $\square$

Corollary 7.4.6

The Jacobson radical ${{\,\mathrm{rad}\,}}B$ is a two-sided ideal of B.

Proof. The statement follows by combining 7.4.4 and Lemma 7.4.5: ${{\,\mathrm{rad}\,}}B$ is the intersection of two-sided ideals and so is itself a two-sided ideal. $\square$

Example 7.4.7

If B is commutative (and still a finite-dimensional F-algebra), then ${{\,\mathrm{rad}\,}}B=\sqrt{(0)}$ is the nilradical of B, the set of all nilpotent elements of B.

A two-sided ideal $J \subseteq B$ isnilpotent if $J^n=\{0\}$ for some $n \ge 1$, i.e., every product of n elements from J is zero. Every element of a nilpotent ideal is itself nilpotent.

Lemma 7.4.8

$J={{\,\mathrm{rad}\,}}B$ contains every nilpotent two-sided ideal, and J itself is nilpotent.

Proof. If $I \subseteq B$ is a nilpotent two-sided ideal, then $I + J$ is a nilpotent two-sided ideal of B/J; but ${{\,\mathrm{rad}\,}}(B/J) = \{0\}$ by Corollary 7.4.3, so B/J is the direct product of simple algebras (Corollary 7.3.14) and therefore has no nonzero nilpotent two-sided ideals. Therefore $I \subseteq I+J \subseteq J$.

Now we prove that J is nilpotent. Consider the descending chain

$$\begin{aligned} B \supset J \supseteq J^2 \supseteq \dots . \end{aligned}$$

There exists $n \in \mathbb Z _{\ge 1}$ such that $J^n=J^{2n}$. We claim that $J^n=\{0\}$. Assume for the purposes of contradiction that $I \subseteq J^n$ is a minimal left ideal such that $J^n I \ne \{0\}$. Let $\alpha \in I$ be such that $J^n \alpha \ne \{0\}$; by minimality $J^n \alpha = I$, so $\alpha =\eta \alpha$ for some $\eta \in J^n$, thus $(1-\eta )\alpha =0$. But $\eta \in J^n \subseteq J={{\,\mathrm{rad}\,}}B$. By Lemma 7.4.5, $1-\eta \in B^\times$ is a unit hence $\alpha =0$, a contradiction. $\square$

Example 7.4.9

Suppose B has a standard involution. Then by Lemma 7.4.8 and the fact that B has degree 2, we conclude that ${{\,\mathrm{rad}\,}}B \subseteq \{\epsilon \in B : \epsilon ^2 = 0 \}$. If ${{\,\mathrm{char}\,}}F \ne 2$ and we define ${{\,\mathrm{rad}\,}}({{\,\mathrm{nrd}\,}})$ as in 4.3.9 for the quadratic form ${{\,\mathrm{nrd}\,}}$, then ${{\,\mathrm{rad}\,}}({{\,\mathrm{nrd}\,}})={{\,\mathrm{rad}\,}}B$ (Exercise 7.21).

Corollary 7.4.10

The Jacobson radical ${{\,\mathrm{rad}\,}}B$ is the intersection of all maximal right ideals of B.

Proof. Lemma 7.4.8 gives a left-right symmetric characterization of the Jacobson radical, so ${{\,\mathrm{rad}\,}}B={{\,\mathrm{rad}\,}}B^{op }$. There is a bijection between simple left B-modules and simple right $B^{op }$-modules, and the result follows. $\square$

5 Central simple algebras

For more on central simple algebras (and in particular division algebras), see e.g. Saltman [Sal99] or Draxl [Dra83].

Recall (2.1.1) that the center of B is defined as

$$\begin{aligned} Z(B) :=\{\alpha \in B : \alpha \beta =\beta \alpha \text { for all }\alpha \in B\}. \end{aligned}$$

Remark 7.5.1. An F-algebra B is a central Z(B)-algebra when Z(B) is a field. (Under a more general definition of algebra, every algebra is an algebra over its center.)

Example 7.5.2

The center Z(B) of a simple F-algebra is a field, since it is a simple commutative F-algebra. One reaches the same conclusion by applying Corollary 7.3.12 together with $Z({{\,\mathrm{M}\,}}_n(D))=Z(D)$ (Exercise 7.5).

The category of central simple algebras is closed under tensor product, as follows.

Proposition 7.5.3

Let A, B be F-algebras and suppose that B is central.

• (a) The center of $A \otimes _F B$ is the image of $Z(A) \hookrightarrow A \otimes _F B$ under $z \mapsto z \otimes 1$.

• (b) Suppose that A, B are simple. Then $A \otimes _F B$ is simple.

Proof. First, centrality in part (a). Suppose that $\gamma = \sum _i \alpha _i \otimes \beta _i \in Z(A \otimes B)$ (a finite sum). By rewriting the tensor, without loss of generality, we may suppose that $\alpha _i$ are linearly independent over F. Then by properties of tensor products, the elements $\beta _i \in B$ in the representation $\gamma = \sum _i \alpha _i \otimes \beta _i$ are unique. But then for all $\beta \in B$,

$$\begin{aligned} \sum _i (\alpha _i \otimes \beta \beta _i) = (1 \otimes \beta )\left( \sum _i \alpha _i \otimes \beta _i\right) = \left( \sum _i \alpha _i \otimes \beta _i\right) (1 \otimes \beta )=\sum _i (\alpha _i \otimes \beta _i\beta ) \end{aligned}$$

so $\beta \beta _i=\beta _i\beta$ for each i; thus $\beta _i=b_i \in Z(B)=F$. Hence

$$\begin{aligned} \gamma = \sum _i \alpha _i \otimes b_i = \sum _i \alpha _i b_i \otimes 1 = \left( \sum _i \alpha _i b_i\right) \otimes 1; \end{aligned}$$

since $\alpha \otimes 1$ also commutes with $\gamma$ for all $\alpha \in A$, we have $\sum _i \alpha _i b_i \in Z(A)$. Thus $\gamma \in Z(A) \otimes F = Z(A)$.

Next, simplicity in part (b). Let I be a nontrivial two-sided ideal in $A \otimes B$, and let $\gamma = \sum _{i=1}^m \alpha _i \otimes \beta _i \in I \smallsetminus \{0\}$. Without loss of generality, we may suppose $\beta _1 \ne 0$. Then $B \beta _1 B = B$ since B is simple; multiplying on the left and right by elements of $B \subseteq A \otimes B$, we may suppose further that $\beta _1=1$. Let $\gamma \in I \smallsetminus \{0\}$ be such an element that is minimal with respect to m; then in particular the elements $\beta _i$ are linearly independent over F. Now for each $\beta \in B$,

$$\begin{aligned} (1 \otimes \beta )\gamma - \gamma (1 \otimes \beta ) = \sum _{i=2}^m \left( \alpha _i \otimes (\beta \beta _i - \beta _i\beta )\right) \in I; \end{aligned}$$

but by minimality of m, the right-hand side is zero, so $\beta \beta _i=\beta _i\beta$ for all i. Hence $\beta _i \in Z(B)=F$ for all i and as above $\gamma =\alpha \otimes 1$ for some $0 \ne \alpha \in A$. But then

$$\begin{aligned} I \supseteq (A \otimes 1)(\alpha \otimes 1)(A \otimes 1)=(A\alpha A) \otimes 1 = A \otimes 1 \end{aligned}$$

since A is simple, so $I \supseteq (A \otimes 1)(1 \otimes B)=A \otimes B$, and thus $I=A \otimes B$ and $A \otimes B$ is simple. $\square$

Lemma 7.5.4

If B is a finite-dimensional algebra over F, then B is a central simple F-algebra if and only if the map

is an isomorphism.

Proof. First, the implication $(\Rightarrow )$. Just as in 7.2.14, $\phi$ is a nonzero F-algebra homomorphism. By Proposition 7.5.3, $B \otimes _F B^{op }$ is simple, so $\phi$ is injective. Since $\dim _F (B \otimes _F B^{op }) = \dim _F {{\,\mathrm{End}\,}}_F(B)=(\dim _F B)^2$, $\phi$ is an isomorphism.

Now the converse $(\Leftarrow )$; suppose $\phi$ is an isomorphism. If I is an ideal of B then $\phi (I \otimes B^{op }) \subseteq {{\,\mathrm{End}\,}}_F(B)$ is an ideal; but ${{\,\mathrm{End}\,}}_F(B)$ is simple over F, therefore I is trivial. And if $\alpha \in Z(B)$ then $\phi (\alpha \otimes 1) \in Z({{\,\mathrm{End}\,}}_F(B))=F$, so $\alpha \in F$. $\square$

7.5.5

Among central simple algebras over a field, quaternion algebras have an especially nice presentation because the quadratic norm form can be put into a standard form (indeed, diagonalized in characteristic not 2). More generally, one may look at algebras with a similarly nice presentation, as follows.

Let F be a field, let $K \supset F$ be a cyclic extension of F of degree $n=[K:F]$, let $\sigma \in {{\,\mathrm{Gal}\,}}(K\,|\,F)$ be a generator, and let $b \in F^\times$. For example, if F contains a primitive nth root of unity $\zeta \in F^\times$, and $a \in F^\times \smallsetminus F^{\times n}$, then we may take $K=F(\root n \of {a})$ and $\sigma (\root n \of {a})=\zeta \root n \of {a}$. We then define thecyclic algebra

$$\begin{aligned} \displaystyle {\biggl (\frac{K,\sigma ,b}{F}\biggr )} = K \oplus Kj \oplus \dots \oplus K j^{n-1} \end{aligned}$$

to be the left K-vector space with basis $1,j,\dots ,j^{n-1}$ and with multiplication $j^n=b$ and $j\alpha =\sigma (\alpha )j$ for $\alpha \in K$. The definition of a cyclic algebra generalizes that of 6.1.5, where there is only one choice for the generator $\sigma$. A cyclic algebra is a central simple algebra over F of dimension $n^2$, and indeed $({K,\sigma ,b} \mid {K}) \simeq {{\,\mathrm{M}\,}}_n(K)$. (See Exercise 7.12.) More generally, we may relax the condition that G be cyclic: there is an analogous construction for any finite Galois extension, yielding a central simple algebra called a crossed product algebra (and giving an interpretation to a second cohomology group): see Reiner [Rei2003, §29–30]. There are significant open problems relating cyclic algebras and crossed products to central simple algebras in general [ABGV2006].

It is a consequence of the main theorem of class field theory that if F is a global field then every (finite-dimensional) central simple algebra over F is isomorphic to a cyclic algebra.

Remark 7.5.6. The theory of central simple algebras and Brauer groups extends to one over commutative rings (or even schemes), and this becomes the theory of Azumaya algebras: see Saltman [Sal99, §2].

6 Quaternion algebras

Having set the stage, we are now ready to prove the following final characterizations of quaternion algebras.

Proposition 7.6.1

Let B be an F-algebra. Then the following are equivalent:

• (i) B is a quaternion algebra;

• (ii) B is a central simple F-algebra with $\dim _F B=4$;

• (iii) B is a central semisimple F-algebra with $\dim _F B=4$; and

• (iv) $B \otimes _F F{}^{al }\simeq {{\,\mathrm{M}\,}}_2(F{}^{al })$, where $F{}^{al }$ is an algebraic closure of F.

Proof. First, (i) $\Rightarrow$ (ii): if B is a quaternion algebra, then B is central simple (7.2.11).

The equivalence (ii) $\Leftrightarrow$ (iii) follows from the Wedderburn–Artin theorem:

$$\begin{aligned} 1=\dim Z(B) = \sum _{i=1}^r \dim _F Z(D_i) \ge r \end{aligned}$$

so $r=1$.

Next we prove (ii) $\Rightarrow$ (iv). If B is central simple, then $B \otimes _F F{}^{al }$ is a central simple $F{}^{al }$-algebra by Proposition 7.5.3. But by Exercise 2.9, the only division $F{}^{al }$-algebra is $F{}^{al }$, so by the Wedderburn–Artin theorem, $B \otimes _F F{}^{al }\simeq {{\,\mathrm{M}\,}}_n(F{}^{al })$; by dimensions, $n=2$.

It remains to prove (iv) $\Rightarrow$ (i). So suppose $B \otimes _F F{}^{al }\simeq {{\,\mathrm{M}\,}}_2(F{}^{al })$. Then B is simple by Example 7.2.10 and $\dim _F B=4$. By the Wedderburn–Artin theorem (Corollary 7.3.12), we have $B \simeq {{\,\mathrm{M}\,}}_n(D)$ with $n \in \mathbb Z _{\ge 1}$ and D a division ring. Since $4=\dim _F B=n^2 \dim _F D$, either $n=2$ and $B \simeq {{\,\mathrm{M}\,}}_2(F)$, or $n=1$ and B is a division ring.

In this latter case, the result will follow from Theorem 3.5.1 (and Theorem 6.2.8 for the case ${{\,\mathrm{char}\,}}F =2$) if we show that B has degree 2. But for any $\alpha \in B$ we have that $\alpha \in B \otimes _F F{}^{al }\simeq {{\,\mathrm{M}\,}}_2(F{}^{al })$ satisfies its characteristic polynomial of degree 2, so that $1,\alpha ,\alpha ^2$ are linearly dependent over $F{}^{al }$ and hence linearly dependent over F, by linear algebra. $\square$

Inspired by the proof of this result, we reconsider and reprove our splitting criterion for quaternion algebras.

Proposition 7.6.2

Let B be a quaternion algebra over F. Then the following are equivalent:

• (i) $B \simeq {{\,\mathrm{M}\,}}_2(F)$;

• (ii) B is not a division ring;

• (iii) There exists $0 \ne \epsilon \in B$ such that $\epsilon ^2=0$;

• (iv) B has a nontrivial left ideal $I \subseteq B$;

Proof. The equivalence (i) $\Leftrightarrow$ (ii) follows from the Wedderburn–Artin theorem (also proved in Main Theorem 5.4.4 and Theorem 6.4.11). The implications (i) $\Rightarrow$ (iii) $\Rightarrow$ (ii) and (i) $\Rightarrow$ (iv) $\Rightarrow$ (ii) are clear. $\square$

7.6.3

We showed in Lemma 7.2.17 that a simple algebra B has a unique simple left B-module I up to isomorphism, obtained as a minimal nonzero left ideal. If B is a quaternion algebra, this simple module I can be readily identified using the above proposition. If B is a division ring, then necessarily $I=B$. Otherwise, $B \simeq {{\,\mathrm{M}\,}}_2(F)$, and then $I \simeq F^2$, and the map $B \rightarrow {{\,\mathrm{End}\,}}_F(I)$ given by left matrix multiplication is an isomorphism.

7 The Skolem–Noether theorem

We conclude this chapter with a fundamental result that characterizes the automorphisms of a simple algebra—and much more.

MainTheorem 7.7.1

(Skolem–Noether).  Let A, B be simple F-algebras and suppose that B is central. Suppose that $f,g :A \rightarrow B$ are homomorphisms. Then there exists $\beta \in B^\times$ such that $f(\alpha )=\beta ^{-1}g(\alpha )\beta$ for all $\alpha \in A$.

Proof. By Corollary 7.3.12, we have $B \simeq {{\,\mathrm{End}\,}}_D(V) \simeq {{\,\mathrm{M}\,}}_n(D^{op })$ where V is a simple B-module and $D={{\,\mathrm{End}\,}}_B(V)$ is a central F-algebra. Now the maps f, g give V the structure of an A-module in two ways. The A-module structure commutes with the D-module structure since $B\simeq {{\,\mathrm{End}\,}}_D(V)$. So V has two $A \otimes _F D$-module structures via f and g.

By Proposition 7.5.3, since D is central over F, we have that $A \otimes _F D$ is a simple F-algebra. By Corollary 7.3.9 and a dimension count, the two $A \otimes _F D$-module structures on V are isomorphic. Thus, there exists an isomorphism $\beta :V \rightarrow V$ of $A \otimes _F D$-modules; i.e. $\beta (f(\alpha )x)=g(\alpha )\beta (x)$ for all $\alpha \in A$ and $x \in V$, and $\beta (\delta x)=\delta \beta (x)$ for all $\delta \in D$ and $x \in V$. We have $\beta \in {{\,\mathrm{End}\,}}_D(V) \simeq B$ and so we can write $\beta f(\alpha ) \beta ^{-1}=g(\alpha )$ for all $\alpha \in A$, as claimed. $\square$

The following corollaries are immediate consequences (special cases) of the Skolem–Noether theorem.

Corollary 7.7.2

If $A_1,A_2$ are simple F-subalgebras of a central simple F-algebra B and is an isomorphism of F-algebras, then $\phi$ is induced by an inner automorphism of B.

Proof. Let $\iota _i :A_i \hookrightarrow B$ be the natural inclusions, and apply Main Theorem 7.7.1 to $f :=\iota _1$ and $g :=\iota _2 \circ \phi$: we conclude there exists $\beta \in B^\times$ such that $\iota _1(\alpha )=\alpha =\beta ^{-1}\iota _2(\phi (\alpha ))\beta$ or equivalently $\phi (\alpha )=\beta \alpha \beta ^{-1}$ for all $\alpha \in A_1$, as desired. $\square$

Corollary 7.7.3

If B is a central simple F-algebra and $\alpha _1,\alpha _2 \in B^\times$, then $\alpha _1,\alpha _2$ have the same irreducible minimal polynomial over F if and only if there exists $\beta \in B^\times$ such that $\alpha _2=\beta ^{-1}\alpha _1\beta$.

Proof. The implication ($\Leftarrow$) is immediate. Conversely ($\Rightarrow$), let $A_i :=F[\alpha _i] \simeq F[x]/(f_i(x))$ where $f_i(x) \in F[x]$ are minimal polynomials over F. Since these polynomials are irreducible, $A_i$ is a field hence simple, so Corollary 7.7.2 gives the result. $\square$

Corollary 7.7.4

The group of F-algebra automorphisms of a central simple algebra B is ${{\,\mathrm{Aut}\,}}(B) \simeq B^\times /F^\times$.

Proof. Taking $A=B$ in Main Theorem 7.7.1, we conclude that every automorphism of B as an F-algebra is inner, and an inner automorphism is trivial if and only if it is conjugation by an element of the center $F^\times$. $\square$

Example 7.7.5

By Corollary 7.7.4, we have a canonical isomorphism of groups

$$\begin{aligned} {{\,\mathrm{Aut}\,}}({{\,\mathrm{M}\,}}_n(F)) \cong {{\,\mathrm{GL}\,}}_n(F)/F^\times = {{\,\mathrm{PGL}\,}}_n(F). \end{aligned}$$

As a final application, we extend the splitting criterion of Main Theorem 5.4.4(i) $\Leftrightarrow$ (vi) to detect isomorphism classes of quaternion algebras.

Corollary 7.7.6

Let $K \supseteq F$ be a separable quadratic F-algebra, and let $b,b' \in F^\times$. Then

$$\begin{aligned} \displaystyle {\biggl (\frac{K,b}{F}\biggr )} \simeq \displaystyle {\biggl (\frac{K,b'}{F}\biggr )} \quad \Leftrightarrow \quad b/b' \in {{\,\mathrm{Nm}\,}}_{K|F}(K^\times ). \end{aligned}$$

Taking $b'=1$, we recover the previous splitting criteria.

Proof. For the implication $(\Leftarrow )$, if $b'/b = {{\,\mathrm{Nm}\,}}_{K|F}(\alpha )$ with $\alpha \in K^\times$, then an isomorphism is furnished as left K-vector spaces by sending $j \mapsto \alpha j$.

For the implication $(\Rightarrow )$, let be an isomorphism of F-algebras. If $K \simeq F \times F$ is not a field, then ${{\,\mathrm{Nm}\,}}_{K|F}(K^\times )=F^\times$ and the result holds. So suppose K is a field. Then $\phi (K) \subseteq B'$ isomorphic to K as an F-algebra, but need not be the designated one in B’; however, by the Skolem–Noether theorem, we may postcompose $\phi$ with an automorphism that sends $\phi (K)$ to the designated one, i.e., we may suppose that $\phi$ is a K-linear map (taking the algebras as left K-vector spaces). Let $\phi (j)=\alpha +\beta j'$ with $\alpha ,\beta \in K$. Then

$$\begin{aligned} \{0\}={{\,\mathrm{trd}\,}}(Kj)={{\,\mathrm{trd}\,}}(\phi (Kj))={{\,\mathrm{trd}\,}}(K\phi (j))={{\,\mathrm{trd}\,}}(K\alpha ) \end{aligned}$$

and thus $\alpha =0$ since K is separable. Consequently,

$$\begin{aligned} -b={{\,\mathrm{nrd}\,}}(j)={{\,\mathrm{nrd}\,}}(\phi (j))=-{{\,\mathrm{Nm}\,}}_{K|F}(\beta )b' \end{aligned}$$

and so $b/b'={{\,\mathrm{Nm}\,}}_{K|F}(\beta )$ as desired. $\square$

In the remainder of this section, we prove an important consequence of the Skolem–Noether theorem that compares centralizers of subalgebras to dimensions.

Definition 7.7.7

Let A be an F-subalgebra of B. Let

$$\begin{aligned} C_B(A) :=\{\beta \in B : \alpha \beta =\beta \alpha \text { for all }\alpha \in A\} \end{aligned}$$

be thecentralizer of A in B.

The centralizer $C_B(A)$ is an F-subalgebra of B.

Proposition 7.7.8

Let B be a central simple F-algebra and let $A \subseteq B$ a simple F-subalgebra. Then the following statements hold:

• (a) $C_B(A)$ is a simple F-algebra.

• (b) $\dim _F B = \dim _F A \cdot \dim _F C_B(A)$.

• (c) $C_B(C_B(A))=A$.

Part (c) of this proposition is called thedouble centralizer property.

Proof. First, part (a). We interpret the centralizer as arising from certain kinds of endomorphisms. We have that B is a left $A \otimes B^{op }$ module by the action $(\alpha \otimes \beta )\cdot \mu = \alpha \mu \beta$ for $\alpha \otimes \beta \in A \otimes B^{op }$ and $\mu \in B$. We claim that

$$\begin{aligned} C_B(A) = {{\,\mathrm{End}\,}}_{A \otimes B^{op }}(B). \end{aligned}$$

(7.7.9)

Any $\phi \in {{\,\mathrm{End}\,}}_{A \otimes B^{op }}(B)$ is left multiplication by an element of B: if $\gamma =\phi (1)$, then $\phi (\mu )=\phi (1)\mu =\gamma \mu$ by $1 \otimes B^{op }$-linearity. Now the equality

$$\begin{aligned} \gamma \alpha = \phi (\alpha )=\alpha \phi (1)=\alpha \gamma \end{aligned}$$

shows that multiplication by $\gamma$ is $A \otimes 1$-linear if and only if $\gamma \in C_B(A)$, proving (7.7.9).

By Proposition 7.5.3, the algebra $A \otimes B^{op }$ is simple. By the Wedderburn–Artin theorem, $A \otimes B^{op } \simeq {{\,\mathrm{M}\,}}_n(D)$ for some $n \ge 1$ and division F-algebra D. Since ${{\,\mathrm{M}\,}}_n(D)$ is simple, its unique simple left D-module is $V=D^n$, and ${{\,\mathrm{End}\,}}_{{{\,\mathrm{M}\,}}_n(D)}(V) \simeq D^{op }$. In particular, $B \simeq V^r$ for some $r \ge 1$ as an $A \otimes B^{op }$-module. So

$$\begin{aligned} C_B(A)={{\,\mathrm{End}\,}}_{A \otimes B^{op }}(B) \simeq {{\,\mathrm{End}\,}}_{{{\,\mathrm{M}\,}}_n(D)}(V^r) \simeq {{\,\mathrm{M}\,}}_r({{\,\mathrm{End}\,}}_{{{\,\mathrm{M}\,}}_n(D)}(V)) \simeq {{\,\mathrm{M}\,}}_r(D^{op }). \end{aligned}$$

Thus $C_B(A)$ is simple.

For part (b),

$$\begin{aligned} \dim _F C_B(A) = \dim _F {{\,\mathrm{M}\,}}_r(D^{op }) = r^2 \dim _F D \end{aligned}$$

and

$$\begin{aligned} \dim _F (A \otimes B^{op }) = \dim _F A \cdot \dim _F B = n^2 \dim _F D \end{aligned}$$

and finally

$$\begin{aligned} \dim _F B = \dim _F V^r = r \dim _F D^n = rn \dim _F D; \end{aligned}$$

putting these together gives $\dim _F A \cdot \dim _F C_B(A) = rn \dim _F D = \dim _F B$.

Finally, part (c) follows from (a) and (b):

$$\begin{aligned} \dim _F B = \dim _F C_B(A) \cdot \dim _F C_B(C_B(A)) = \dim _F A \cdot \dim _F C_B(A) \end{aligned}$$

so $\dim _F A = \dim _F C_B(C_B(A))$ and $A \subseteq C_B(C_B(A))$, therefore equality holds. $\square$

Example 7.7.10

We always have the two extremes $A=F$ and $A=B$, with $C_B(F)=B$ and $C_B(B)=F$, accordingly.

We note the following structurally crucial corollary of Proposition 7.7.8.

Corollary 7.7.11

Let B be a central simple F-algebra and let K be a maximal subfield. Then $[B:F]=[K:F]^2$.

Proof. If K is maximal, then $C_B(K)=K$ and $[B:F]=[K:F]^2$. $\square$

Corollary 7.7.11 generalizes the comparatively easier statement for quaternion algebras: the maximal subfields of a quaternion algebra are quadratic. Returning now to quaternion algebras, we conclude with a nice package of consequences of the above results concerning embeddings of quadratic fields into quaternion algebras.

7.7.12

Let B be a quaternion algebra over F and let $K \subseteq B$ be a quadratic separable F-subalgebra. Then the set of all embeddings of K in B is naturally identified with the set $K^\times \backslash B^\times$, as follows.

By the Skolem–Noether theorem (Corollary 7.1.5, and Exercise 7.11 for the case $K \simeq F \times F$), if $\phi :K \hookrightarrow B$ is another embedding, then there exists $\beta \in B^\times$ such that $\phi (\alpha )=\beta ^{-1} \alpha \beta$ for all $\alpha \in K$, and conversely. Such a conjugate embedding is the identity if and only if $\beta$ centralizes K. By Corollary 4.4.5, and Corollary 6.4.2 for characteristic 2, the centralizer of $K^\times$ in $B^\times$ is $K^\times$. Therefore, the set of embeddings of K in B is naturally identified with the set $K^\times \backslash B^\times$, with $K^\times$ acting on the left.

8 Reduced trace and norm, universality

In this last section, we consider notions of reduced trace and reduced norm in the context of semisimple algebras. Thr

7.8.1

Let B be a (finite-dimensional) central simple algebra over F, and let $F^{sep }$ denote a separable closure of F. By Exercise 7.24, we have an F-algebra homomorphism

$$\begin{aligned} \phi :B \otimes _F F^{sep } \simeq {{\,\mathrm{M}\,}}_n(F^{sep }) \end{aligned}$$

for some $n \ge 1$. By the Skolem–Noether theorem (Main Theorem 7.7.1), for any other isomorphism $\phi ':B \otimes _F F^{sep } \simeq {{\,\mathrm{M}\,}}_n(F^{sep })$, there exists $M \in {{\,\mathrm{GL}\,}}_n(F^{sep })$ such that $\phi '(\alpha ) = M \phi (\alpha ) M^{-1}$, so the characteristic polynomial of an element of $B \otimes _F F^{sep }$ is independent of the choice of $\iota$. In particular, from the canonical embedding $\iota :B \hookrightarrow B \otimes _F F^{sep }$ by $\alpha \mapsto \alpha \otimes 1$, we define thereduced characteristic polynomial of $\alpha \in B$ to be the characteristic polynomial of $(\phi \iota )(\alpha )$ as an element of $F^{sep }[T]$ and similarly thereduced trace andreduced norm of $\alpha$ to be the trace and determinant of $(\phi \iota )(\alpha )$ as elements of $F^{sep }$.

In fact, the reduced characteristic polynomial descends to F, as follows. The absolute Galois group ${{\,\mathrm{Gal}\,}}_F :={{\,\mathrm{Gal}\,}}(F{}^{sep }\,|\,F)$ acts on $B \otimes _F F^{sep } \simeq {{\,\mathrm{M}\,}}_n(F^{sep })$ by

$$\begin{aligned} \sigma (\alpha \otimes a)=\alpha \otimes \sigma (a) \end{aligned}$$

for $\sigma \in {{\,\mathrm{Gal}\,}}_F$, $\alpha \in B$, and $a \in F^{sep }$. Let $\sigma \in {{\,\mathrm{Gal}\,}}_F$. Since $\sigma (\alpha \otimes 1)=\alpha \otimes \sigma (1)=\alpha \otimes 1$, the reduced characteristic polynomials of $\iota (\alpha )$ and $\sigma (\iota (\alpha ))$ are the same. By comparison (see e.g. Reiner [Rei2003, Theorem 9.3]), if

$$\begin{aligned} f(\alpha ;T)=\det (T-\iota (\alpha ))=T^n+a_{n-1} T^{n-1} +\dots + a_0 \end{aligned}$$

is the reduced characteristic polynomial of $\iota (\alpha )$, then the reduced characteristic polynomial of $(\sigma (\iota ))(\alpha )$ is

$$\begin{aligned} \sigma (f)(\alpha ;T)=\det (T-\sigma (\iota (\alpha )))=T^n+\sigma (a_{n-1}) T^{n-1} + \dots + \sigma (a_0). \end{aligned}$$

And then since $f(\alpha ;T)=\sigma (f)(\alpha ;T)$ for all $\sigma \in {{\,\mathrm{Gal}\,}}_F$, by the fundamental theorem of Galois theory, $f(\alpha ;T) \in F[T]$. Therefore, the reduced norm and reduced trace also belong to F.

Alternatively, we may argue as follows. The characteristic polynomial of left multiplication by $\alpha$ on B is the same as left multiplication by $(\phi \iota )(\alpha )$ on ${{\,\mathrm{M}\,}}_n(F^{sep })$ (by extension of basis), and the latter is the nth power of the reduced characteristic polynomial by Exercise 3.13. Finally, if $f(T) \in F^{sep }[T]$ has $f(T)^n \in F[T]$ then in fact $f(T) \in F[T]$: see Exercise 7.25.

These definitions extend to a general semisimple algebra over F, but to do so it is convenient to give an alternate approach that avoids going to the separable closure and works in even more generality using universal elements; for more, see Garibaldi [Gar2004].

Let B be a (finite-dimensional) F-algebra with $n :=\dim _F B$, and choose a basis $e_1,\dots ,e_n$ for B. Let $F(x_1,\dots ,x_n)$ be a pure transcendental field extension of transcendence degree n, and let $\xi :=x_1e_1+\dots +x_ne_n \in B \otimes _F F(x_1,\dots ,x_n)$; we call $\xi$ theuniversal element of B in the given basis.

Definition 7.8.2

Theuniversal minimal polynomial of B (in the basis $e_1,\dots ,e_n$) is the minimal polynomial $m_B(\xi ;T)$ of $\xi$ over $F(x_1,\dots ,x_n)$.

For $\alpha =a_1e_1+\dots +a_ne_n \in B$ (with $a_i \in F$), the polynomial obtained from $m_B(\xi ;T)$ by the substitution $x_i \leftarrow a_i$ is called thespecialization of $m_B(\xi ;T)$ at $\alpha$.

The following example will hopefully illustrate the role of this notion.

Example 7.8.3

For ${{\,\mathrm{char}\,}}F \ne 2$ and $B=\displaystyle {\biggl (\frac{a,b}{F}\biggr )}$, in the basis 1, i, j, ij we have $\xi = t + x i + y j + z ij$ (substituting t, x, y, z for $x_1,\dots ,x_4$). We claim that the universal minimal polynomial is

$$\begin{aligned} m_B(\xi ;T) = T^2 - 2tT + (t^2-ax^2-by^2+abz^2). \end{aligned}$$

Indeed, we verify that $\xi$ satisfies $m_B(\xi ;\xi )=0$ by considering $\xi \in B \otimes _F F(t,x,y,z) = \displaystyle {\biggl (\frac{a,b}{F(t,x,y,z)}\biggr )}$ and computing that ${{\,\mathrm{trd}\,}}(\xi )=2t$ and ${{\,\mathrm{nrd}\,}}(\xi )=t^2-ax^2-by^2+abz^2$; and this polynomial is minimal because $\xi \not \in F(t,x,y,z)$ does not satisfy a polynomial of degree 1 over F(t, x, y, z).

7.8.4

If $B \simeq B_1 \times \dots \times B_r$, then in a basis for B obtained from the union of bases for the factors $B_i$ with universal elements $\xi _i$, we have

$$\begin{aligned} m_B(\xi ;T)=m_{B_1}(\xi _1;T)\cdots m_{B_r}(\xi _r;T). \end{aligned}$$

In the proofs that follow, we abbreviate by using multi-index notation, e.g. writing $F[x] :=F[x_1,\dots ,x_n]$.

Lemma 7.8.5

We have $m_B(\xi ;T) \in F[x_1,\dots ,x_n][T]$, i.e., the universal minimal polynomial has coefficients in $F[x_1,\dots ,x_n]$.

Proof. For part (a), we consider the map given by left multiplication by $\xi$ on $B \otimes _F F(x)$. In the basis $e_1,\dots ,e_n$, almost by construction we find that the matrix of this map has coefficients in F[x] (it is the matrix of linear forms obtained from left multiplication by $e_i$). We conclude that $\xi$ satisfies the characteristic polynomial of this matrix, which is a monic polynomial with coefficients in F[x]. Since $m_B(\xi ;T)$ divides this polynomial (over F(x)) by minimality, by Gauss’s lemma we conclude that $m_B(\xi ;T) \in F[x][T]$. $\square$

Proposition 7.8.6

For all $\alpha \in B$, the specialization of $m_B(\xi ;T)$ at $\alpha$ is independent of the choice of basis $e_1,\dots ,e_n$ and is satisfied by the element $\alpha$. Moreover, if $\phi \in {{\,\mathrm{Aut}\,}}(B)$ and $\alpha \in B$, then $\alpha$ and $\phi (\alpha )$ have the same specialized polynomials.

In view of Proposition 7.8.6, we write $m_B(\alpha ;T)$ for the specialization of $m_B(\xi ;T)$ at $\alpha \in B$; from it, we conclude that $m_B(\alpha ;\alpha )=0$.

Proof. Since $m_B(\xi ;\xi )=0$, by specialization we obtain $m_B(\alpha ;\alpha )=0$. For the independence of basis, let $e_1',\dots ,e_n'$ be another F-basis and $\xi '$ the corresponding universal element. Writing $e_i$ in the basis $e_i'$ allows us to write $\xi = \sum _{i=1}^n \ell _i(x)e_i'$ where $\ell _i(x) \in F[x]$ are linear forms; moreover, writing $\alpha =\sum _i a_i' e_i'$ we have $\ell _i(a)=a_i'$. The map $x_i \mapsto \ell _i(x)$ extends to an F-algebra automorphism $\phi$ of F[x] (repeat the construction with the inverse, and compose) with $\phi (c)(a)=c(a')$ for all $c(x) \in F[x]$. We let $\phi$ act on polynomials over F[x] by acting on the coefficients; by uniqueness of minimal polynomials, we have $\phi (m_B(\xi ;T))=m_B(\xi ';T)$. Therefore, looking at each coefficient, specializing $\phi (m_B(\xi ;T))$ at a is the same as specializing $m_B(\xi ';T)$ at $a'$, as claimed.

The second sentence follows by the same argument, as from $\phi \in {{\,\mathrm{Aut}\,}}(B)$ we have a new basis $e_i' :=\phi (e_i)$ so the specializations again agree. (This argument replaces the use of the Skolem–Noether theorem in the special case where B is a central simple algebra.) $\square$

Lemma 7.8.7

For any field extension $K \supseteq F$, we have $m_{B \otimes _F K}(\xi ;T)=m_K(\xi ;T)$.

Proof. First, because an F-basis for B is a K-basis for $B \otimes _F K$, the element $\xi$ (as the universal element of B), also serves as a universal element of $B \otimes _F K$. Since $K(x_1,\dots ,x_n) \subseteq F(x_1,\dots ,x_n)$, by minimality we have $m_{B_K}(\xi ;T) \mid m_B(\xi ;T)$. Conversely, let $F(x)[\xi ] \subseteq B \otimes _F F(x)$ be the subalgebra generated by $\xi$ over F(x); then $F(x)[\xi ] \simeq F(x)[T]/(m_B(\xi ;T))$. Tensoring with K gives $K(x)[\xi ] \simeq K(x)[T]/(m_B(\xi ;T))$ as the subalgebra of $(B \otimes _F K) \otimes _K K(x)$ generated by $\xi$. Thus $m_B(\xi ;T) \mid m_{B_K}(\xi ;T)$, so equality holds. $\square$

We conclude by relating this construction to more familiar polynomials.

Lemma 7.8.8

Let $\alpha \in B$. Then the following statements hold.

• (a) If $B=K \supseteq F$ is a separable field extension, then $m_K(\alpha ;T)$ is the characteristic polynomial of left multiplication by $\alpha$.

• (b) If B is a central simple F-algebra, then $m_B(\alpha ;T)$ is the reduced characteristic polynomial of $\alpha$.

Proof. For (a), we recall (as in the proof of Lemma 7.8.5 that $\xi$ satisfies the characteristic polynomial of left multiplication on K, a polynomial of degree $n=[K:F]$; on the other hand, choosing a primitive element $\alpha$, we see the specialization $m_K(\alpha ;T)$ is satisfied by $\alpha$ so has degree at least n, so equality holds and $m_K(\alpha ;T)$ is the characteristic polynomial. Therefore the universal minimal polynomial is the characteristic polynomial, and hence the same is true under every specialization.

For (b), it suffices to prove this when $F=F^{al }$ is algebraically closed, in which case $B \simeq {{\,\mathrm{M}\,}}_n(F)$; by Proposition 7.8.6, we may assume $B={{\,\mathrm{M}\,}}_n(F)$. By 7.8.1, we want to show that $m_B(\alpha ;T)$ for $\alpha \in {{\,\mathrm{M}\,}}_n(F)$ is the usual characteristic polynomial. But the universal element (in a basis of matrix units, or any basis) satisfies its characteristic polynomial of degree n, and a nilpotent matrix (with 1s just above the diagonal) has minimal polynomial $T^n$, so we conclude as in the previous paragraph. $\square$

In light of the above, we may make the following definition.

Definition 7.8.9

Let B be a semisimple F-algebra. For $\alpha \in B$, thereduced characteristic polynomial $f(\alpha ;T)=T^n- c_1 T^{n-1} + \dots + (-1)^n c_n \in F[T]$ is the specialization of the universal minimal polynomial $m_B(\xi ;T)$, and thereduced trace andreduced norm are the coefficients $c_1,c_n$, respectively.

Example 7.8.10

For a semisimple algebra $B \simeq {B_1 \times \dots \times B_r}$, with each $B_i$ central simple, we find that the reduced characteristic polynomial is just the product of the reduced characteristic polynomials on each simple direct factor $B_i$; this is well-defined (again) by the uniqueness statement in the Wedderburn–Artin theorem (Main Theorem 7.3.10).

Proposition 7.8.11

Let B be semisimple. Then the reduced trace ${{\,\mathrm{trd}\,}}:B \rightarrow F$ is F-linear and satisfies ${{\,\mathrm{trd}\,}}(\alpha \beta )={{\,\mathrm{trd}\,}}(\beta \alpha )$ for all $\alpha ,\beta \in B$; and the reduced norm ${{\,\mathrm{nrd}\,}}:B \rightarrow F$ is multiplicative, satisfying ${{\,\mathrm{nrd}\,}}(\alpha \beta )={{\,\mathrm{nrd}\,}}(\alpha ){{\,\mathrm{nrd}\,}}(\beta )$.

Proof. Consider (again) $V :=F(x)[\xi ] \subseteq B \otimes _F F(x)$ the subalgebra generated over F(x) by $\xi$; then $\xi$ acts on $V \simeq F(x)[T]/(m_B(\xi ;T))$ by left multiplication with characteristic polynomial $m_B(\xi ;T)$. It follows that the reduced trace and reduced norm are the usual trace and determinant in this representation, so the announced properties follow on specialization. $\square$

Remark 7.8.12. It is also possible to define the reduced characteristic polynomial on a semisimple algebra B by writing $B \simeq B_1 \times \dots \times B_r$ as a product of simple algebras; for details, see Reiner [Rei2003, §9].

9 Separable algebras

For a (finite-dimensional) F-algebra, the notions of simple and semisimple are sensitive to the base field F in the sense that these properties need not hold after extending the base field. Indeed, let $K \supseteq F$ be a finite extension of fields, so K is a simple F-algebra. Then $K \otimes _F F{}^{al }$ is simple only when $K=F$ and is semisimple if and only if $K \otimes _F F{}^{al }\simeq F{}^{al }\times \dots \times F{}^{al }$, i.e., K is separable over F.

It is important to have a notion which is stable under base change, as follows. For further reference, see Drozd–Kirichenko [DK94, §6], Curtis–Reiner [CR81, §7], Reiner [Rei2003, §7c], or Pierce [Pie82, Chapter 10].

Definition 7.9.1

Let B be a finite-dimensional F-algebra. We say that B is aseparable F-algebra if B is semisimple and Z(B) is a separable F-algebra.

In particular, a separable algebra over a field F with ${{\,\mathrm{char}\,}}F=0$ is just a semisimple algebra.

7.9.2

For a semisimple algebra $B \simeq {{\,\mathrm{M}\,}}_{n_1}(D_1) \times \dots \times {{\,\mathrm{M}\,}}_{n_r}(D_r)$, by Example 7.5.2 we have $Z(B) \simeq Z(D_1) \times \dots \times Z(D_r)$, and B is separable if and only if $Z(D_i)$ is separable for each $i=1,\dots ,r$.

Lemma 7.9.3

A finite-dimensional simple F-algebra is a separable algebra over its center K.

Proof. The center of B is a field $K=Z(B)$ and as a K-algebra, the center $Z(B)=K$ is certainly separable over K. (Or use Proposition 7.5.3 and Theorem 7.9.4(iii) below.) $\square$

The notion of separability in this context is quite robust.

Theorem 7.9.4

Let B be a finite-dimensional F-algebra. Then the following are equivalent:

• (i) B is separable;

• (ii) There exists a finite separable field extension K of F such that $B \otimes _F K \simeq M_{n_1}(K) \times \dots \times M_{n_r}(K)$ for integers $n_1,\dots ,n_r \ge 1$;

• (iii) For every extension $K \supseteq F$ of fields, the K-algebra $B \otimes _F K$ is semisimple;

• (iv) B is semisimple and the bilinear form

  $$\begin{aligned} B \times B&\rightarrow F \\ (\alpha ,\beta )&\mapsto {{\,\mathrm{trd}\,}}(\alpha \beta ) \end{aligned}$$

  is nondegenerate.

Moreover, if ${{\,\mathrm{char}\,}}F = 0$, then these are further equivalent to:

• (v) The bilinear form $(\alpha ,\beta ) \mapsto {{\,\mathrm{Tr}\,}}_{B|F}(\alpha \beta )$ is nondegenerate.

A separable F-algebra is sometimes calledabsolutely semisimple , in view of Theorem 7.9.4(iii).

Proof. First we prove (i) $\Rightarrow$ (ii). Let $B_i$ be a simple component of B; then $Z(B_i)$ is separable over F. Let $K_i \supseteq F$ be a separable field extension containing $Z(B_i)$ that splits $B_i$, so $B_i \otimes _{Z(B_i)} K_i \simeq {{\,\mathrm{M}\,}}_{n_i}(K_i)$. Let K be the compositum of the fields $K_i$. Then K is separable, and

$$\begin{aligned} B_i \otimes _F K \simeq {{\,\mathrm{M}\,}}_{n_i}(Z(B_i) \otimes _F K) \simeq {{\,\mathrm{M}\,}}_{n_i}(K) \times \dots \times {{\,\mathrm{M}\,}}_{n_i}(K) \end{aligned}$$

the number of copies equal to $[Z(B_i):F]$.

Next we prove (ii) $\Rightarrow$ (iii). Suppose $B \otimes _F K \simeq \prod _i {{\,\mathrm{M}\,}}_{n_i}(K)$ and let $L \supseteq F$ be an extension of fields. Let $M=KL$. On the one hand, $B \otimes _F M \simeq (B \otimes _F K) \otimes _K M \simeq \prod _i {{\,\mathrm{M}\,}}_{n_i}(M)$, so ${{\,\mathrm{rad}\,}}B \otimes _F M = \{0\}$; on the other hand, $B \otimes _F M \simeq (B \otimes _F L) \otimes _L M$ and ${{\,\mathrm{rad}\,}}(B \otimes _F L) \subseteq {{\,\mathrm{rad}\,}}(B \otimes _F L) \otimes _L M = \{0\}$, so $B \otimes _F L$ is semisimple.

For the implication (iii) $\Rightarrow$ (i), suppose B is not separable, and we show that there exists $K \supseteq F$ such that $B \otimes _F K$ is not semisimple. If B is not semisimple over F, we can just take $F=K$. Otherwise, Z(B) is not separable as an F-algebra, and there is a component of Z(B) which is an inseparable field extension K. Then $B \otimes _F K$ contains a nonzero nilpotent element in its center and this element generates a nonzero nilpotent ideal, so ${{\,\mathrm{rad}\,}}(B \otimes _F K) \ne \{0\}$ and $B \otimes _F K$ is not semisimple.

The implication (iii) $\Rightarrow$ (iv) holds for the following reason. We have $B \otimes _F F{}^{al }\simeq {{\,\mathrm{M}\,}}_{n_1}(F{}^{al }) \times \dots \times {{\,\mathrm{M}\,}}_{n_r}(F{}^{al })$, and the reduced trace pairing on each matrix ring factor is nondegenerate so the whole pairing is nondegenerate. By linear algebra we conclude that the bilinear form on B is nondegenerate.

The implication (iv) $\Rightarrow$ (i) holds with ${{\,\mathrm{char}\,}}F$ arbitrary: if $\epsilon \in {{\,\mathrm{rad}\,}}B$ then $\alpha \epsilon \in {{\,\mathrm{rad}\,}}B$ is nilpotent and ${{\,\mathrm{trd}\,}}(\alpha \epsilon )=0$ for all $\alpha \in B$, and by nondegeneracy $\epsilon =0$.

The final equivalence (iv) $\Leftrightarrow$ (v) follows when ${{\,\mathrm{char}\,}}F=0$ since the algebra trace pairing on each simple factor is a scalar multiple of the reduced trace pairing. $\square$

Exercises

Throughout the exercises, let F be a field.

$\triangleright$ 1.:

Prove that a quaternion algebra $B=\displaystyle {\biggl (\frac{a,b}{F}\biggr )}$ with ${{\,\mathrm{char}\,}}F \ne 2$ is simple by a direct calculation, as follows.

(a):

Let I be a nontrivial two-sided ideal, and let $\epsilon =t+xi+yj+zij \in I$. By considering $i\epsilon -\epsilon i$, show that $t+xi \in I$.

(b):

Arguing symmetrically and taking a linear combination, show that $t \in I$, and conclude that $t=0$, whence $x=y=z=0$.

Modify this argument to show that an algebra $B=\displaystyle {\biggl [\frac{a,b}{F}\biggr )}$ is simple when ${{\,\mathrm{char}\,}}F = 2$. [We proved these statements without separating into cases in 7.2.11.]

• 2. Let B be a quaternion algebra over F, and let $K \subseteq B$ be an F-subalgebra that is commutative. Show that $\dim _F K \le 2$.

• 3. Let B be a quaternion algebra. Exhibit an explicit isomorphism

  

  [Hint: see Exercise 2.11.]

$\triangleright$ 4.:

Let $B=\left\{ \begin{pmatrix} a &{} b \\ 0 &{} c \end{pmatrix} : a,b,c \in F\right\} \subseteq {{\,\mathrm{M}\,}}_2(F)$, and $V=F^2$ be the left B-module of column vectors. Show that V is indecomposable, but not simple, as a left B-module (cf. Example 7.2.5).

$\triangleright$ 5.:

This exercise proves basic but important facts about two-sided ideals in matrix algebras using matrix units.

(a):

Let D be a division F-algebra. Prove that ${{\,\mathrm{M}\,}}_n(D)$ is a simple F-algebra with center Z(D) for all $n \ge 1$. [Hint: Let $E_{ij}$ be the matrix with 1 in the $ij$th entry and zeros in all other entries. Show that $E_{ki}ME_{j\ell }=m_{ij}E_{k\ell }$ where $m_{ij}$ is the $ij$th entry of M.]

(b):

More generally, let R be a ring (associative with 1, but potentially noncommutative). Show that $Z({{\,\mathrm{M}\,}}_n(R))=Z(R)$ and that any two-sided ideal of ${{\,\mathrm{M}\,}}_n(R)$ is of the form ${{\,\mathrm{M}\,}}_n(I) \subseteq {{\,\mathrm{M}\,}}_n(R)$ where I is a two-sided ideal of R.

$\triangleright$ 6.:

Let F be a field, let B a simple algebra, and let I be a left B-module with $\dim _F I = \dim _F B$. Show that I is isomorphic to B as a left B-module, i.e., there exists $\alpha \in I$ such $I=B\alpha$.

• 7. In this exercise, we consider extensions of the Skolem–Noether theorem.

  • (a) Let B be a quaternion algebra over F and let $K_1,K_2 \subset B$ be F-subalgebras (not necessarily subfields). Suppose that is an isomorphism of F-algebras. Show that $\phi$ lifts to an inner automorphism of B. [Hint: repeat the proof of Corollary 7.1.5.]

  • (b) Show by example that Corollary 7.7.3 is false if the minimal polynomials are not supposed to be irreducible. In particular, provide an example of isomorphic algebras $K_1,K_2 \subseteq B$ that are not isomorphic by an inner automorphism of B.

• 8. Let B be a quaternion algebra over F, and let $K \subseteq B$ be a separable, quadratic F-subalgebra. Show that there exists $b \in F^\times$ such that $B \simeq ({K} \mid {b})$. [Hint: lift the standard involution on K via the Skolem–Noether theorem.]

$\triangleright$ 9.:

Let B be a finite-dimensional F-algebra. Show that if $\alpha \in {{\,\mathrm{rad}\,}}B$, then $1-\beta \alpha \in B^\times$ for all $\beta \in B$. [Hint: if $1-\beta \alpha$ is not left invertible then it belongs to a maximal left ideal; left invertible implies invertible.]

$\triangleright$ 10.:

Extend Corollary 7.1.5 to the case where $K=F \times F$ as follows: show that if $K_1,K_2 \subseteq B$ are F-subalgebras with $K_1 \simeq F \times F$, and is an isomorphism of F-algebras, then $\phi$ lifts to an inner automorphism of B.

• 11. Let $n \in \mathbb Z _{\ge 2}$ and let F be a field with ${{\,\mathrm{char}\,}}F \not \mid n$. Let $\zeta \in F$ be a primitive nth root of unity. Let $a,b \in F^\times$ and let $A=\displaystyle {\biggl (\frac{a,b}{F,\zeta }\biggr )}$ be the algebra over F generated by elements $i,j$ subject to

  $$ i^n=a, \quad j^n=b, \quad ji=\zeta ij. $$

  • (a) Show that $\dim _F A = n^2$.

  • (b) Show that A is a central simple algebra over F.

  • (c) Let $K=F[i] \simeq F[x]/(x^n-a)$. Show that if $b \in {{\,\mathrm{Nm}\,}}_{K|F}(K^\times )$ then $A \simeq {{\,\mathrm{M}\,}}_n(F)$. [Such algebras are calledcyclic algebras or sometimespower norm residue algebras.]

• 12. Generalize the statement of Proposition 7.5.3(a) as follows. Let A, B be F-algebras, and let $A' \subseteq A$ and $B' \subseteq B$ be F-subalgebras. Prove that

  $$\begin{aligned} C_{A \otimes B}(A' \otimes B') = C_A(A') \otimes C_B(B'). \end{aligned}$$

• 13. Let B be a finite-dimensional F-algebra. Show that the following are equivalent:

  • (i) B is separable;

  • (ii) B is semisimple and the center $K=Z(B)$ is separable;

  • (iii) $B \otimes _F B^{op }$ is semisimple.

• 14. Let $G \ne \{1\}$ be a finite group. Show that theaugmentation ideal, the two-sided ideal generated by $g-1$ for $g \in G$, is a nontrivial ideal, and hence F[G] is not simple as an F-algebra.

• 15. Let G be a finite group of order $n=\#G$. Show that F[G] is a separable F-algebra if and only if ${{\,\mathrm{char}\,}}F \not \mid n$ as follows. [This exercise is known as Maschke’s theorem.]

  • (a) Suppose first that ${{\,\mathrm{char}\,}}F = 0$ for a special but quick special case. Compute the trace pairing and conclude F[G] is separable.

  • (b) If ${{\,\mathrm{char}\,}}F \mid n$, show that $N=\sum _{g \in G} g$ is a nilpotent element in the center of F[G], so F[G] is not semisimple.

  • (c) Suppose that ${{\,\mathrm{char}\,}}F \not \mid n$. Let $B=F[G]$. Define the map of left B-modules by

    $$\begin{aligned} \phi :B&\rightarrow B \otimes _F B^{op } =:B^{e } \\ \phi (1)&= \frac{1}{n} \sum _{g \in g} g \otimes g^{-1} \end{aligned}$$

    so that $\phi (\alpha )=\alpha \phi (1)$ for all $\alpha \in B$. Give B the structure of a $B^{e }$-algebra by $(\alpha ,\alpha ^{o }) \cdot \beta \mapsto \alpha \beta \alpha ^{o }$. Show that $\phi$ is a homomorphism of $B^{e }$-modules, and that the structure map $\psi :B^{e } \rightarrow B$ has $\psi \circ \phi = {{\,\mathrm{id}\,}}_B$. Conclude that B is separable.

• 16. Let B be an F-algebra, and let $F{}^{al }$ be an algebraic closure of F. Show that B is simple if and only if $B \otimes _F F{}^{al }$ is simple.

• 17. Let D be a (finite-dimensional) division algebra over $F{}^{al }$. Show that $D=F{}^{al }$. Conclude that if B is a simple algebra over $F{}^{al }$, then $B \simeq {{\,\mathrm{M}\,}}_n(F{}^{al })$ for some $n \ge 1$ and hence is central.

$\triangleright$ 18.:

Let B be a (finite-dimensional) F algebra, and let $K \supseteq F$ be a finite separable extension of fields. Show that ${{\,\mathrm{rad}\,}}(B \otimes _F K)={{\,\mathrm{rad}\,}}(B) \otimes _F K$.

$\triangleright$ 19.:

Show that if B is a semisimple F-algebra, then so is ${{\,\mathrm{M}\,}}_n(B)$ for any $n \in \mathbb Z _{\ge 1}$.

$\triangleright$ 20.:

Let B be a (finite-dimensional) F-algebra with standard involution and suppose ${{\,\mathrm{char}\,}}F \ne 2$.

(a):

Show that ${{\,\mathrm{rad}\,}}B={{\,\mathrm{rad}\,}}{{\,\mathrm{nrd}\,}}$. Conclude B is semisimple if and only if ${{\,\mathrm{rad}\,}}{{\,\mathrm{nrd}\,}}= \{0\}$.

(b):

Suppose $B \ne F$ and B is central. Conclude that B is a quaternion algebra if and only if ${{\,\mathrm{rad}\,}}{{\,\mathrm{nrd}\,}}=\{0\}$.

• 21. Compute the Jacobson radical ${{\,\mathrm{rad}\,}}B$ of the F-algebra B with basis 1, i, j, ij satisfying

  $$\begin{aligned} i^2=a,\ j^2 = 0, \ \text { and }\ ij=-ji\end{aligned}$$

  for $a \in F$, and compute $B/{{\,\mathrm{rad}\,}}B$. In particular, conclude that such an algebra is not semisimple, so B is not a quaternion algebra. [Hint: restrict to the case ${{\,\mathrm{char}\,}}F \ne 2$ first.]

• 22. Give an example of (finite-dimensional) simple algebras A, B over a field F such that $A \otimes _F B$ is not simple. Then find A, B such that $A \otimes _F B$ is not semisimple.

$\triangleright$ 23.:

In Exercise 7.18, we saw that if D is a (finite-dimensional) division algebra over F then $D \otimes _F F{}^{al }\simeq {{\,\mathrm{M}\,}}_n(F{}^{al })$ for some $n \ge 1$. In this exercise, we show the same is true if we consider the separable closure. (We proved this already in Exercise 6.3 for D a quaternion algebra.)

Let F be aseparably closed field, so every nonconstant separable polynomial with coefficients in F has a root in F. Let D be a finite-dimensional central division algebra over F with ${{\,\mathrm{char}\,}}F = p$. For purposes of contradiction, assume that $D \ne F$.

(a):

Prove that $\dim _F D$ is divisible by p.

(b):

Show that the minimal polynomial of each nonzero $d \in D$ has the form $x^{p^e}-a$ for some $a \in F$ and $e \ge 0$.

(c):

Choose an $F{}^{al }$-algebra isomorphism . Show that ${{\,\mathrm{tr}\,}}\phi (x \otimes 1)=0$ for all $x \in D$.

(d):

Prove that D does not exist.

• 24. Let $K \supseteq F$ be a separable (possibly infinite) extension, and let $f(T) \in K[T]$. Suppose that $f(T)^n \in F[T]$ for some $n \in \mathbb Z _{\ge 1}$. Show that $f(T) \in F[T]$. [Hint: when $p={{\,\mathrm{char}\,}}F \mid n$, use the fact that $a^p \in F$ implies $a \in F$.]

• 25. Let B be a finite-dimensional F-algebra, let $\alpha \in B$, and let $f\!{}_{\textsf {\tiny {L}} }(\alpha ;T)$ and $f\!{}_{\textsf {\tiny {R}} }(\alpha ;T)$ be the characteristic polynomial of left and right multiplication of $\alpha$ on B, respectively.

  • (a) If B is semisimple, show that $f\!{}_{\textsf {\tiny {L}} }(\alpha ;T)=f\!{}_{\textsf {\tiny {R}} }(\alpha ;T)$.

  • (b) Give an example where $f\!{}_{\textsf {\tiny {L}} }(\alpha ;T) \ne f\!{}_{\textsf {\tiny {R}} }(\alpha ;T)$.

$\triangleright$ 26.:

Use the Skolem–Noether theorem to give another solution to Exercise 6.2: if $K \subset B$ is a separable quadratic F-algebra then $B \simeq ({K,b} \mid {F})$ for some $b \in F^\times$.

• 27. Give a direct proof of Corollary 7.7.4. [Hint: Use the fact that there is a unique simple left B-module.]

• 28. Let $B=({K,b} \mid {F})$ be a quaternion algebra. Show that the subgroup of ${{\,\mathrm{Aut}\,}}(B)$ that maps $K \subseteq B$ to itself is isomorphic to the group

  $$\begin{aligned} K^\times /F^\times \cup j (K^\times /F^\times ). \end{aligned}$$

  Show that the subgroup of ${{\,\mathrm{Aut}\,}}(B)$ that restricts to the identity on K (fixing K elementwise) is isomorphic to $K^\times /F^\times$.

$\triangleright$ 29.:

Use the Skolem–Noether theorem and the fact that a finite group cannot be written as the union of the conjugates of a proper subgroup to prove Wedderburn’s little theorem: a finite division ring is a field.

$\triangleright$ 30.:

Let B be a quaternion algebra over F. In this exercise, we show that the commutator subgroup

$$\begin{aligned}{}[B^\times ,B^\times ]=\langle \alpha \beta \alpha ^{-1}\beta ^{-1} : \alpha ,\beta \in B^\times \rangle \end{aligned}$$

is precisely

$$\begin{aligned}{}[B^\times ,B^\times ]=B^1=\{\gamma \in B^\times : {{\,\mathrm{nrd}\,}}(\gamma )=1\}={{\,\mathrm{SL}\,}}_1(B). \end{aligned}$$

(a):

Show that $[B^\times ,B^\times ] \le B^1$.

(b):

Show that $[{{\,\mathrm{GL}\,}}_2(F),{{\,\mathrm{GL}\,}}_2(F)]={{\,\mathrm{SL}\,}}_2(F)$ if $\#F>3$. [Hint: choose $z \in F$ such that $z^2-1 \in F^\times$, let $\gamma =\begin{pmatrix} z &{} 0 \\ 0 &{} z^{-1} \end{pmatrix}$, and show that for all $x \in F$ we have

$$\begin{aligned} \begin{pmatrix} 1 &{} x \\ 0 &{} 1 \end{pmatrix} = \left[ \gamma ,\begin{pmatrix} 1 &{} x(z^2-1)^{-1} \\ 0 &{} 1 \end{pmatrix}\right] \end{aligned}$$

and analogously for the transpose. See 28.3 for a review of elementary matrices.]

(c):

Suppose that B is a division algebra. Let $\gamma \in B^1$. Show that there exists $\alpha \in K=F(\gamma )$ such that $\alpha {\overline{\alpha }}^{-1}=\gamma$. [Hint: This is a special case of Hilbert’s theorem 90. Let $\alpha =\gamma +1$ if $\gamma \ne -1$, and $\alpha \in B^0 \smallsetminus \{0\}$ if $\gamma =-1$, with appropriate modifications if ${{\,\mathrm{char}\,}}F = 2$.] Conclude from the Skolem–Noether theorem that there exists $\beta \in B^\times$ such that $\beta \alpha \beta ^{-1}=\overline{\alpha }$, and thus $\gamma \in [B^\times ,B^\times ]$.

• 31. Show that every ring automorphism of $\mathbb H$ is inner. (Compare this with ring automorphisms of $\mathbb C$!)