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

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

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 $c \in K$

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

is injective. Since both sides of the arrow have the same dimension we win. $\square$

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