# Elementary proof that a central simple algebra over a field having a maximal subfield is a cyclic algebra

I'm currently reading the book "Central Simple Algebras and Galois Cohomology" written by Philippe Gille and Tamas Szamuely.

In the book, I don't understand a computational proof of the theorem that a central simple algebra $A$ over a field $k$ of degree $m$ containing a cyclic extension $K/k$ of degree $m$ is a cyclic algebra, in the page 38~39.

In particular, I want to understand fully the Lemma 2.5.4. But I have a difficulty in understanding the line "Under the diagonal embedding ..."

Is the diagonal embedding the same as the algebra homomorphism induced(by taking tensor product) from the inclusion map of $K$ into $A$.

If not, is it necessarily an algebra homomorphism?

I hope someone can elaborate the argument of the book. Thank you.

• Lemma 2.5.4 is a special case the Skolem-Noether theorem. With that name, you should be able to find other proofs of the Skolem-Noether theorem in many places: the Wikipedia page on the Skolem-Noether theorem, books on noncommutative algebra, and so on. Jun 15 '18 at 13:57

The proof uses as an essential ingredient Proposition 2.2.8, which itself relies on Lemma 2.2.9 telling you that the k-algebras in $M_n(k)$ isomorphic to $k^n$ are conjugate to the subalgebra of diagonal matrices.
Now just follow the given maps of $\tilde{K}$-algebra $K\otimes_k \tilde{K}\to A\otimes_k \tilde{K}\cong M_n(\tilde{K})$. You indeed don't know a priori whether this composition is going to map $K\otimes_k \tilde{K}$ to the subalgebra of diagonal matrices but if it doesn't, just apply Lemma 2.2.9 to rectify the situation as you wish.
• That's fine, no need to be sorry. The fact that "the $K$-algebra automorphism group of $K$ (as a $K$-algebra) is $K^{\times}$" is straightforward: $\alpha \in \text{Aut}(K)$ is determined by $\alpha (1)\in K^{\times}$ (because $\alpha (x) = \alpha (x.1) = x.\alpha (1)$). There is also another way to prove that $H^1$ is trivial, using the fact that $H^1(\tilde{G},\tilde{K})$ is trivial by Hilbert 90, and then deducing it for $K\otimes_k \tilde{K}\cong \tilde{K}^m$ in general by using the long exact sequence obtained from the SES $1\to \tilde{K}\to \tilde{K}\times \tilde{K}\to \tilde{K}\to 1$. Jun 20 '18 at 16:47