Definition 11.8.1. 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'.
11.8 Splitting fields
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 11.8.2. 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 11.7.3 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 11.7.2 the classes of A and B add up to zero in \text{Br}(k). From the dimension formula in Theorem 11.7.1 we see that
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. \square
Lemma 11.8.3. A maximal subfield of a finite central skew field K over k is a splitting field for K.
Lemma 11.8.4. 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 11.8.2 there exists a finite central simple algebra B in the Brauer class of K such that [B : k] = [k' : k]^2. By Lemma 11.5.1 we see that B = \text{Mat}(n \times n, K) for some n. Then [k' : k]^2 = n^2d^2 whence the result. \square
Proposition 11.8.5. 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 spitting 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 11.8.3.
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 is separable over k. By the discussion in Fields, Section 9.28 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
and write it out in terms of a k-basis \{ a_1, \ldots , a_ n\} of K with a_1 = 1. So
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 11.4.9 and 11.5.3), which implies there are some elements whose qth power is not central (e.g., e_{11}). This is the desired contradiction. \square
The results above allow us to characterize finite central simple algebras as follows.
Lemma 11.8.6. Let k be a field. For a k-algebra A the following are equivalent
A is 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 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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