Lemma 11.4.1. 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'.
11.4 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.
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'. \square
Lemma 11.4.2. 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 11.3.3. By Lemma 11.4.1 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. \square
Lemma 11.4.3. 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. \square
Lemma 11.4.4. 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 11.4.3. \square
Lemma 11.4.5. 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 all of the summands e_{ii}Je_{jj} are equal to each other and are a two-sided ideal I of R. This proves (2). Part (3) is clear. \square
Lemma 11.4.6. 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 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 11.3.3 we can write A = \text{Mat}(n \times n, K) for some finite skew field extension K of k. By Lemma 11.4.5 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 11.4.5. 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. \square
Lemma 11.4.7. 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 11.3.3. Then the center of K' is k and we conclude that A \otimes _ k K' is simple by Lemma 11.4.4. Hence A \otimes _ k A' = \text{Mat}(n \times n, A \otimes _ k K') is simple by Lemma 11.4.5. \square
Lemma 11.4.8. The tensor product of finite central simple algebras over k is finite, central, and simple.
Lemma 11.4.9. 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'.
Lemma 11.4.10. 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 11.4.8 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. \square
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