Introduction
A reference is the lectures by Serre in the Seminaire Cartan, see [Serre-Cartan]. Serre in turn refers to [Deuring] and [ANT]. We changed some of the proofs, in particular we used a fun argument of Rieffel to prove Wedderburn’s theorem. Very likely this change is not an improvement and we strongly encourage the reader to read the original exposition by Serre.
Noncommutative algebras
Let \(k\) be a field. In this chapter an algebra \(A\) over \(k\) is a possibly noncommutative ring \(A\) together with a ring map \(k \to A\) such that \(k\) maps into the center of \(A\) and such that \(1\) maps to an identity element of \(A\). An \(A\)-module is a right \(A\)-module such that the identity of \(A\) acts as the identity.
Definition
Let \(A\) be a \(k\)-algebra. We say \(A\) is finite if \(\dim_k(A) < \infty\). In this case we write \([A : k] = \dim_k(A)\).
Definition
A skew field is a possibly noncommutative ring with an identity element \(1\), with \(1 \not = 0\), in which every nonzero element has a multiplicative inverse.
A skew field is a \(k\)-algebra for some \(k\) (e.g., for the prime field contained in it). We will use below that any module over a skew field is free because a maximal linearly independent set of vectors forms a basis and exists by Zorn’s lemma.
Definition
Let \(A\) be a \(k\)-algebra. We say an \(A\)-module \(M\) is simple if it is nonzero and the only \(A\)-submodules are \(0\) and \(M\). We say \(A\) is simple if it is nonzero and the only two-sided ideals of \(A\) are \(0\) and \(A\).
Definition
A \(k\)-algebra \(A\) is central if the center of \(A\) is the image of \(k \to A\).
Definition
Given a \(k\)-algebra \(A\) we denote \(A^{op}\) the \(k\)-algebra we get by reversing the order of multiplication in \(A\). This is called the opposite algebra.
Wedderburn’s theorem
The following cute argument can be found in a paper of Rieffel, see [Rieffel]. The proof could not be simpler (quote from Carl Faith’s review).
Lemma
Let \(A\) be a possibly noncommutative ring with \(1\) which contains no nontrivial two-sided ideal. Let \(M\) be a nonzero right ideal in \(A\), and view \(M\) as a right \(A\)-module. Then \(A\) coincides with the bicommutant of \(M\).
Proof
Let \(A' = \text{End}_A(M)\), so \(M\) is a left \(A'\)-module. Set \(A'' = \text{End}_{A'}(M)\) (the bicommutant of \(M\)). We view \(M\) as a right \(A''\)-module1. Let \(R : A \to A''\) be the natural homomorphism such that \(mR(a) = ma\). Then \(R\) is injective, since \(R(1) = \text{id}_M\) and \(A\) contains no nontrivial two-sided ideal. We claim that \(R(M)\) is a right ideal in \(A''\). Namely, \(R(m)a'' = R(ma'')\) for \(a'' \in A''\) and \(m\) in \(M\), because left multiplication of \(M\) by any element \(n\) of \(M\) represents an element of \(A'\), and so \((nm)a'' = n(ma'')\) for all \(n\) in \(M\). Finally, the product ideal \(AM\) is a two-sided ideal, and so \(A = AM\). Thus \(R(A) = R(A)R(M)\), so that \(R(A)\) is a right ideal in \(A''\). But \(R(A)\) contains the identity element of \(A''\), and so \(R(A) = A''\).
Lemma
Let \(A\) be a \(k\)-algebra. If \(A\) is finite, then
\(A\) has a simple module,
any nonzero module contains a simple submodule,
a simple module over \(A\) has finite dimension over \(k\), and
if \(M\) is a simple \(A\)-module, then \(\text{End}_A(M)\) is a skew field.
Proof
Of course (1) follows from (2) since \(A\) is a nonzero \(A\)-module. For (2), any submodule of minimal (finite) dimension as a \(k\)-vector space will be simple. There exists a finite dimensional one because a cyclic submodule is one. If \(M\) is simple, then \(mA \subset M\) is a sub-module, hence we see (3). Any nonzero element of \(\text{End}_A(M)\) is an isomorphism, hence (4) holds.
Theorem
Let \(A\) be a simple finite \(k\)-algebra. Then \(A\) is a matrix algebra over a finite \(k\)-algebra \(K\) which is a skew field.
Proof
We may choose a simple submodule \(M \subset A\) and then the \(k\)-algebra \(K = \text{End}_A(M)\) is a skew field, see Lemma 0746. By Lemma 0745 we see that \(A = \text{End}_K(M)\). Since \(K\) is a skew field and \(M\) is finitely generated (since \(\dim_k(M) < \infty\)) we see that \(M\) is finite free as a left \(K\)-module. It follows immediately that \(A \cong \text{Mat}(n \times n, K^{op})\).
Lemmas on algebras
Let \(A\) be a \(k\)-algebra. Let \(B \subset A\) be a subalgebra. The centralizer of \(B\) in \(A\) is the subalgebra \[C = \{y \in A \mid xy = yx \text{ for all }x \in B\}.\] It is a \(k\)-algebra.
Lemma
Let \(A\), \(A'\) be \(k\)-algebras. Let \(B \subset A\), \(B' \subset A'\) be subalgebras with centralizers \(C\), \(C'\). Then the centralizer of \(B \otimes_k B'\) in \(A \otimes_k A'\) is \(C \otimes_k C'\).
Proof
Denote \(C'' \subset A \otimes_k A'\) the centralizer of \(B \otimes_k B'\). It is clear that \(C \otimes_k C' \subset C''\). Conversely, every element of \(C''\) commutes with \(B \otimes 1\) hence is contained in \(C \otimes_k A'\). Similarly \(C'' \subset A \otimes_k C'\). Thus \(C'' \subset C \otimes_k A' \cap A \otimes_k C' = C \otimes_k C'\).
Lemma
Let \(A\) be a finite simple \(k\)-algebra. Then the center \(k'\) of \(A\) is a finite field extension of \(k\).
Proof
Write \(A = \text{Mat}(n \times n, K)\) for some skew field \(K\) finite over \(k\), see Theorem 0747. By Lemma 0749 the center of \(A\) is \(k \otimes_k k'\) where \(k' \subset K\) is the center of \(K\). Since the center of a skew field is a field, we win.
Lemma
Let \(V\) be a \(k\) vector space. Let \(K\) be a central \(k\)-algebra which is a skew field. Let \(W \subset V \otimes_k K\) be a two-sided \(K\)-sub vector space. Then \(W\) is generated as a left \(K\)-vector space by \(W \cap (V \otimes 1)\).
Proof
Let \(V' \subset V\) be the \(k\)-sub vector space generated by \(v \in V\) such that \(v \otimes 1 \in W\). Then \(V' \otimes_k K \subset W\) and we have \[W/(V' \otimes_k K) \subset (V/V') \otimes_k K.\] If \(\overline{v} \in V/V'\) is a nonzero vector such that \(\overline{v} \otimes 1\) is contained in \(W/(V' \otimes_k K)\), then we see that \(v \otimes 1 \in W\) where \(v \in V\) lifts \(\overline{v}\). This contradicts our construction of \(V'\). Hence we may replace \(V\) by \(V/V'\) and \(W\) by \(W/(V' \otimes_k K)\) and it suffices to prove that \(W \cap (V \otimes 1)\) is nonzero if \(W\) is nonzero.
To see this let \(w \in W\) be a nonzero element which can be written as \(w = \sum_{i = 1, \ldots, n} v_i \otimes k_i\) with \(n\) minimal. We may right multiply with \(k_1^{-1}\) and assume that \(k_1 = 1\). If \(n = 1\), then we win because \(v_1 \otimes 1 \in W\). If \(n > 1\), then we see that for any \(c \in K\) \[c w - w c = \sum\nolimits_{i = 2, \ldots, n} v_i \otimes (c k_i - k_i c) \in W\] and hence \(c k_i - k_i c = 0\) by minimality of \(n\). This implies that \(k_i\) is in the center of \(K\) which is \(k\) by assumption. Hence \(w = (v_1 + \sum k_i v_i) \otimes 1\) contradicting the minimality of \(n\).
Lemma
Let \(A\) be a \(k\)-algebra. Let \(K\) be a central \(k\)-algebra which is a skew field. Then any two-sided ideal \(I \subset A \otimes_k K\) is of the form \(J \otimes_k K\) for some two-sided ideal \(J \subset A\). In particular, if \(A\) is simple, then so is \(A \otimes_k K\).
Proof
Set \(J = \{a \in A \mid a \otimes 1 \in I\}\). This is a two-sided ideal of \(A\). And \(I = J \otimes_k K\) by Lemma 074B.
Lemma
Let \(R\) be a possibly noncommutative ring. Let \(n \geq 1\) be an integer. Let \(R_n = \text{Mat}(n \times n, R)\).
The functors \(M \mapsto M^{\oplus n}\) and \(N \mapsto Ne_{11}\) define quasi-inverse equivalences of categories \(\text{Mod}_R \leftrightarrow \text{Mod}_{R_n}\).
A two-sided ideal of \(R_n\) is of the form \(IR_n\) for some two-sided ideal \(I\) of \(R\).
The center of \(R_n\) is equal to the center of \(R\).
Proof
Part (1) proves itself. If \(J \subset R_n\) is a two-sided ideal, then \(J = \bigoplus e_{ii}Je_{jj}\), and the sets of \((i,j)\)-entries occurring in \(e_{ii}Je_{jj}\) are all equal to a two-sided ideal \(I\) of \(R\). This proves (2). Part (3) is clear.
Lemma
Let \(A\) be a finite simple \(k\)-algebra.
There exists exactly one simple \(A\)-module \(M\) up to isomorphism.
Any finite \(A\)-module is a direct sum of copies of a simple module.
Two finite \(A\)-modules are isomorphic if and only if they have the same dimension over \(k\).
If \(A = \text{Mat}(n \times n, K)\) with \(K\) a finite skew field extension of \(k\), then \(M = K^{\oplus n}\) is a simple \(A\)-module and \(\text{End}_A(M) = K^{op}\).
If \(M\) is a simple \(A\)-module, then \(L = \text{End}_A(M)\) is a skew field finite over \(k\) acting on the left on \(M\), we have \(A = \text{End}_L(M)\), and the centers of \(A\) and \(L\) agree. Also \([A : k] [L : k] = \dim_k(M)^2\).
For a nonzero finite \(A\)-module \(N\) the algebra \(B = \text{End}_A(N)\) is a matrix algebra over the skew field \(L\) of (5). Moreover \(\text{End}_B(N) = A\).
Proof
By Theorem 0747 we can write \(A = \text{Mat}(n \times n, K)\) for some finite skew field extension \(K\) of \(k\). By Lemma 074D the category of modules over \(A\) is equivalent to the category of modules over \(K\). Thus (1), (2), and (3) hold because every module over \(K\) is free. Part (4) holds because the equivalence transforms the \(K\)-module \(K\) to \(M = K^{\oplus n}\). Using \(M = K^{\oplus n}\) in (5) we see that \(L = K^{op}\). The statement about the center of \(L = K^{op}\) follows from Lemma 074D. The statement about \(\text{End}_L(M)\) follows from the explicit form of \(M\). The formula of dimensions is clear. Part (6) follows as \(N\) is isomorphic to a direct sum of copies of a simple module.
Lemma
Let \(A\), \(A'\) be two simple \(k\)-algebras one of which is finite and central over \(k\). Then \(A \otimes_k A'\) is simple.
Proof
Suppose that \(A'\) is finite and central over \(k\). Write \(A' = \text{Mat}(n \times n, K')\), see Theorem 0747. Then the center of \(K'\) is \(k\) and we conclude that \(A \otimes_k K'\) is simple by Lemma 074C. Hence \(A \otimes_k A' = \text{Mat}(n \times n, A \otimes_k K')\) is simple by Lemma 074D.
Lemma
The tensor product of finite central simple algebras over \(k\) is finite, central, and simple.
Proof
Lemma
Let \(A\) be a finite central simple algebra over \(k\). Let \(k'/k\) be a field extension. Then \(A' = A \otimes_k k'\) is a finite central simple algebra over \(k'\).
Proof
Lemma
Let \(A\) be a finite central simple algebra over \(k\). Then \(A \otimes_k A^{op} \cong \text{Mat}(n \times n, k)\) where \(n = [A : k]\).
Proof
By Lemma 074G the algebra \(A \otimes_k A^{op}\) is simple. Hence the map \[A \otimes_k A^{op} \longrightarrow \text{End}_k(A),\quad a \otimes a' \longmapsto (x \mapsto axa')\] is injective. Since both sides of the arrow have the same dimension we win.
The Brauer group of a field
Let \(k\) be a field. Consider two finite central simple algebras \(A\) and \(B\) over \(k\). We say \(A\) and \(B\) are similar if there exist \(n, m > 0\) such that \(\text{Mat}(n \times n, A) \cong \text{Mat}(m \times m, B)\) as \(k\)-algebras.
Lemma
Similarity.
Similarity defines an equivalence relation on the set of isomorphism classes of finite central simple algebras over \(k\).
Every similarity class contains a unique (up to isomorphism) finite central skew field extension of \(k\).
If \(A = \text{Mat}(n \times n, K)\) and \(B = \text{Mat}(m \times m, K')\) for some finite central skew fields \(K\), \(K'\) over \(k\) then \(A\) and \(B\) are similar if and only if \(K \cong K'\) as \(k\)-algebras.
Proof
Note that by Wedderburn’s theorem (Theorem 0747) we can always write a finite central simple algebra as a matrix algebra over a finite central skew field. Hence it suffices to prove the third assertion. To see this it suffices to show that if \(A = \text{Mat}(n \times n, K) \cong \text{Mat}(m \times m, K') = B\) then \(K \cong K'\). To see this note that for a simple module \(M\) of \(A\) we have \(\text{End}_A(M) = K^{op}\), see Lemma 074E. Hence \(A \cong B\) implies \(K^{op} \cong (K')^{op}\) and we win.
Given two finite central simple \(k\)-algebras \(A\), \(B\) the tensor product \(A \otimes_k B\) is another, see Lemma 074G. Moreover if \(A\) is similar to \(A'\), then \(A \otimes_k B\) is similar to \(A' \otimes_k B\) because tensor products and taking matrix algebras commute. Hence tensor product defines an operation on equivalence classes of finite central simple algebras which is clearly associative and commutative. Finally, Lemma 074I shows that \(A \otimes_k A^{op}\) is isomorphic to a matrix algebra, i.e., that \(A \otimes_k A^{op}\) is in the similarity class of \(k\). Thus we obtain an abelian group.
Definition
Let \(k\) be a field. The Brauer group of \(k\) is the abelian group of similarity classes of finite central simple \(k\)-algebras defined above. Notation \(\text{Br}(k)\).
For any map of fields \(k \to k'\) we obtain a group homomorphism \[\text{Br}(k) \longrightarrow \text{Br}(k'),\quad A \longmapsto A \otimes_k k'\] see Lemma 074H. In other words, \(\text{Br}(-)\) is a functor from the category of fields to the category of abelian groups. Observe that the Brauer group of a field is zero if and only if every finite central skew field extension \(k \subset K\) is trivial.
Lemma
The Brauer group of an algebraically closed field is zero.
Proof
Let \(k \subset K\) be a finite central skew field extension. For any element \(x \in K\) the subring \(k[x] \subset K\) is a commutative finite integral \(k\)-sub algebra, hence a field, see Algebra, Lemma 00GS. Since \(k\) is algebraically closed we conclude that \(k[x] = k\). Since \(x\) was arbitrary we conclude \(k = K\).
Lemma
Let \(A\) be a finite central simple algebra over a field \(k\). Then \([A : k]\) is a square.
Proof
This is true because \(A \otimes_k \overline{k}\) is a matrix algebra over \(\overline{k}\) by Lemma 074M.
Skolem-Noether
Theorem
Let \(A\) be a finite central simple \(k\)-algebra. Let \(B\) be a simple \(k\)-algebra. Let \(f, g : B \to A\) be two \(k\)-algebra homomorphisms. Then there exists an invertible element \(x \in A\) such that \(f(b) = xg(b)x^{-1}\) for all \(b \in B\).
Proof
Choose a simple \(A\)-module \(M\). Set \(L = \text{End}_A(M)\). Then \(L\) is a skew field with center \(k\) which acts on the left on \(M\), see Lemmas 0746 and 074E. Then \(M\) has two \(B \otimes_k L^{op}\)-module structures defined by \(m \cdot_1 (b \otimes l) = lmf(b)\) and \(m \cdot_2 (b \otimes l) = lmg(b)\). The \(k\)-algebra \(B \otimes_k L^{op}\) is simple by Lemma 074F. Since \(B\) is simple, the existence of a \(k\)-algebra homomorphism \(B \to A\) implies that \(B\) is finite. Thus \(B \otimes_k L^{op}\) is finite simple and we conclude the two \(B \otimes_k L^{op}\)-module structures on \(M\) are isomorphic by Lemma 074E. Hence we find \(\varphi : M \to M\) intertwining these operations. In particular \(\varphi\) is in the commutant of \(L\) which implies that \(\varphi\) is multiplication by some \(x \in A\), see Lemma 074E. Working out the definitions we see that \(x\) is a solution to our problem.
Lemma
Let \(A\) be a finite central simple \(k\)-algebra. Any \(k\)-algebra automorphism of \(A\) is inner. In particular, any \(k\)-algebra automorphism of \(\text{Mat}(n \times n, k)\) is inner.
Proof
The Skolem-Noether theorem (Theorem 074Q) applies.
The centralizer theorem
Theorem
Let \(A\) be a finite central simple algebra over \(k\), and let \(B\) be a simple subalgebra of \(A\). Then
the centralizer \(C\) of \(B\) in \(A\) is simple,
\([A : k] = [B : k][C : k]\), and
the centralizer of \(C\) in \(A\) is \(B\).
Proof
Throughout this proof we use the results of Lemma 074E freely. Choose a simple \(A\)-module \(M\). Set \(L = \text{End}_A(M)\). Then \(L\) is a skew field with center \(k\) which acts on the left on \(M\) and \(A = \text{End}_L(M)\). Then \(M\) is a right \(B \otimes_k L^{op}\)-module and \(C = \text{End}_{B \otimes_k L^{op}}(M)\). Since the algebra \(B \otimes_k L^{op}\) is simple by Lemma 074F we see that \(C\) is simple (by Lemma 074E again).
Write \(B \otimes_k L^{op} = \text{Mat}(m \times m, K)\) for some skew field \(K\) finite over \(k\). Then \(C = \text{Mat}(n \times n, K^{op})\) if \(M\) is isomorphic to a direct sum of \(n\) copies of the simple \(B \otimes_k L^{op}\)-module \(K^{\oplus m}\) (the lemma again). Thus we have \(\dim_k(M) = nm [K : k]\), \([B : k] [L : k] = m^2 [K : k]\), \([C : k] = n^2 [K : k]\), and \([A : k] [L : k] = \dim_k(M)^2\) (by the lemma again). We conclude that (2) holds.
Part (3) follows because (2) applied to \(C \subset A\) shows that \([B : k] = [C' : k]\) where \(C'\) is the centralizer of \(C\) in \(A\) (and the obvious fact that \(B \subset C')\).
Lemma
Let \(A\) be a finite central simple algebra over \(k\), and let \(B\) be a simple subalgebra of \(A\). If \(B\) is a central \(k\)-algebra, then the multiplication map \(B \otimes_k C \to A\) is an isomorphism, where \(C\) is the (central simple) centralizer of \(B\) in \(A\).
Proof
We have \(\dim_k(A) = \dim_k(B \otimes_k C)\) by Theorem 074T. By Lemma 074F the tensor product is simple. Hence the natural map \(B \otimes_k C \to A\) is injective hence an isomorphism.
Lemma
Let \(A\) be a finite central simple algebra over \(k\). If \(K \subset A\) is a subfield, then the following are equivalent
\([A : k] = [K : k]^2\),
\(K\) is its own centralizer, and
\(K\) is a maximal commutative subring.
Proof
Theorem 074T shows that (1) and (2) are equivalent. It is clear that (3) and (2) are equivalent.
Lemma
Let \(A\) be a finite central skew field over \(k\). Then every maximal subfield \(K \subset A\) satisfies \([A : k] = [K : k]^2\).
Proof
Special case of Lemma 074V.
Splitting fields
Definition
Let \(A\) be a finite central simple \(k\)-algebra. We say a field extension \(k'/k\) splits \(A\), or \(k'\) is a splitting field for \(A\) if \(A \otimes_k k'\) is a matrix algebra over \(k'\).
Another way to say this is that the class of \(A\) maps to zero under the map \(\text{Br}(k) \to \text{Br}(k')\).
Theorem
Let \(A\) be a finite central simple \(k\)-algebra. Let \(k'/k\) be a finite field extension. The following are equivalent
\(k'\) splits \(A\), and
there exists a finite central simple algebra \(B\) similar to \(A\) such that \(k' \subset B\) and \([B : k] = [k' : k]^2\).
Proof
Assume (2). It suffices to show that \(B \otimes_k k'\) is a matrix algebra. We know that \(B \otimes_k B^{op} \cong \text{End}_k(B)\). Since \(k'\) is the centralizer of \(k'\) in \(B^{op}\) by Lemma 074V we see that \(B \otimes_k k'\) is the centralizer of \(k \otimes k'\) in \(B \otimes_k B^{op} = \text{End}_k(B)\). Of course this centralizer is just \(\text{End}_{k'}(B)\) where we view \(B\) as a \(k'\) vector space via the embedding \(k' \to B\). Thus the result.
Assume (1). This means that we have an isomorphism \(A \otimes_k k' \cong \text{End}_{k'}(V)\) for some \(k'\)-vector space \(V\). Let \(B\) be the commutant of \(A\) in \(\text{End}_k(V)\). Note that \(k'\) sits in \(B\). By Lemma 074U the classes of \(A\) and \(B\) add up to zero in \(\text{Br}(k)\). From the dimension formula in Theorem 074T we see that \[[B : k] [A : k] = \dim_k(V)^2 = [k' : k]^2 \dim_{k'}(V)^2 = [k' : k]^2 [A : k].\] Hence \([B : k] = [k' : k]^2\). Thus we have proved the result for the opposite to the Brauer class of \(A\). However, \(k'\) splits the Brauer class of \(A\) if and only if it splits the Brauer class of the opposite algebra, so we win anyway.
Lemma
A maximal subfield of a finite central skew field \(K\) over \(k\) is a splitting field for \(K\).
Proof
Lemma
Consider a finite central skew field \(K\) over \(k\). Let \(d^2 = [K : k]\). For any finite splitting field \(k'\) for \(K\) the degree \([k' : k]\) is divisible by \(d\).
Proof
By Theorem 074Z there exists a finite central simple algebra \(B\) in the Brauer class of \(K\) such that \([B : k] = [k' : k]^2\). By Lemma 074K we see that \(B = \text{Mat}(n \times n, K)\) for some \(n\). Then \([k' : k]^2 = n^2d^2\) whence the result.
Proposition
Consider a finite central skew field \(K\) over \(k\). There exists a maximal subfield \(k \subset k' \subset K\) which is separable over \(k\). In particular, every Brauer class has a finite separable splitting field.
Proof
Since every Brauer class is represented by a finite central skew field over \(k\), we see that the second statement follows from the first by Lemma 0750.
To prove the first statement, suppose that we are given a separable subfield \(k' \subset K\). Then the centralizer \(K'\) of \(k'\) in \(K\) has center \(k'\), and the problem reduces to finding a maximal subfield of \(K'\) separable over \(k'\). Thus it suffices to prove, if \(k \not = K\), that we can find an element \(x \in K\), \(x \not \in k\) which is separable over \(k\). This statement is clear in characteristic zero. Hence we may assume that \(k\) has characteristic \(p > 0\). If the ground field \(k\) is finite, then the result is clear as well (because extensions of finite fields are always separable). Thus we may assume that \(k\) is an infinite field of positive characteristic.
To get a contradiction assume no element of \(K\) outside \(k\) is separable over \(k\). By the discussion in Fields, Section 037H this means the minimal polynomial of any \(x \in K\) is of the form \(T^q - a\) where \(q\) is a power of \(p\) and \(a \in k\). Since it is clear that every element of \(K\) has a minimal polynomial of degree \(\leq \dim_k(K)\) we conclude that there exists a fixed \(p\)-power \(q\) such that \(x^q \in k\) for all \(x \in K\).
Consider the map \[(-)^q : K \longrightarrow K\] and write it out in terms of a \(k\)-basis \(\{a_1, \ldots, a_n\}\) of \(K\) with \(a_1 = 1\). So \[(\sum x_i a_i)^q = \sum f_i(x_1, \ldots, x_n)a_i.\] Since multiplication on \(K\) is \(k\)-bilinear we see that each \(f_i\) is a polynomial in \(x_1, \ldots, x_n\) (details omitted). The choice of \(q\) above and the fact that \(k\) is infinite shows that \(f_i\) is identically zero for \(i \geq 2\). Hence we see that it remains zero on extending \(k\) to its algebraic closure \(\overline{k}\). But the algebra \(K \otimes_k \overline{k}\) is a matrix algebra (for example by Lemmas 074H and 074M), which implies there are some elements whose \(q\)th power is not central (e.g., \(e_{11}\)). This is the desired contradiction.
The results above allow us to characterize finite central simple algebras as follows.
Lemma
Let \(k\) be a field. For a \(k\)-algebra \(A\) the following are equivalent
\(A\) is a finite central simple \(k\)-algebra,
\(A\) is a finite dimensional \(k\)-vector space, \(k\) is the center of \(A\), and \(A\) has no nontrivial two-sided ideal,
there exists \(d \geq 1\) such that \(A \otimes_k \bar k \cong \text{Mat}(d \times d, \bar k)\),
there exists \(d \geq 1\) such that \(A \otimes_k k^{sep} \cong \text{Mat}(d \times d, k^{sep})\),
there exist \(d \geq 1\) and a finite Galois extension \(k'/k\) such that \(A \otimes_k k' \cong \text{Mat}(d \times d, k')\),
there exist \(n \geq 1\) and a finite central skew field \(K\) over \(k\) such that \(A \cong \text{Mat}(n \times n, K)\).
The integer \(d\) is called the degree of \(A\).
Proof
The equivalence of (1) and (2) is a consequence of the definitions, see Section 073Y. Assume (1). By Proposition 0752 there exists a separable splitting field \(k \subset k'\) for \(A\). Of course, then a Galois closure of \(k'/k\) is a splitting field also. Thus we see that (1) implies (5). It is clear that (5) \(\Rightarrow\) (4) \(\Rightarrow\) (3). Assume (3). Then \(A \otimes_k \overline{k}\) is a finite central simple \(\overline{k}\)-algebra for example by Lemma 074D. This trivially implies that \(A\) is a finite central simple \(k\)-algebra. Finally, the equivalence of (1) and (6) is Wedderburn’s theorem, see Theorem 0747.
This means that given \(a'' \in A''\) and \(m \in M\) we have a product \(m a'' \in M\). In particular, the multiplication in \(A''\) is the opposite of what you’d get if you wrote elements of \(A''\) as endomorphisms acting on the left.↩︎