## 11.5 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 11.5.1. 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 11.3.3) 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 11.4.6. Hence $A \cong B$ implies $K^{op} \cong (K')^{op}$ and we win.
$\square$

Given two finite central simple $k$-algebras $A$, $B$ the tensor product $A \otimes _ k B$ is another, see Lemma 11.4.8. 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 11.4.10 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 11.5.2. 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 11.4.9. 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 11.5.3. 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 10.35.19. Since $k$ is algebraically closed we conclude that $k[x] = k$. Since $x$ was arbitrary we conclude $k = K$.
$\square$

Lemma 11.5.4. 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 11.5.3.
$\square$

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