Lemma 11.8.6. Let $k$ be a field. For a $k$-algebra $A$ the following are equivalent

1. $A$ is finite central simple $k$-algebra,

2. $A$ is a finite dimensional $k$-vector space, $k$ is the center of $A$, and $A$ has no nontrivial two-sided ideal,

3. there exists $d \geq 1$ such that $A \otimes _ k \bar k \cong \text{Mat}(d \times d, \bar k)$,

4. there exists $d \geq 1$ such that $A \otimes _ k k^{sep} \cong \text{Mat}(d \times d, k^{sep})$,

5. there exist $d \geq 1$ and a finite Galois extension $k'/k$ such that $A \otimes _ k k' \cong \text{Mat}(d \times d, k')$,

6. 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 11.2. Assume (1). By Proposition 11.8.5 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 11.4.5. 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 11.3.3. $\square$

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