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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.

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    $\begingroup$ 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. $\endgroup$
    – KConrad
    Commented Jun 15, 2018 at 13:57

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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.

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  • $\begingroup$ Thank you for anwering the question. But I don't know what is the proposition 2.2.8 and Lemma 2.2.9. I guess that the proposition and the lemma you referred is not the proposition and lemma in the book by Gille and Szamuely. Can you please give me the exact reference? Thank you. $\endgroup$
    – gualterio
    Commented Jun 16, 2018 at 15:45
  • $\begingroup$ The reference I made are actually to the exact same book by Gille and Szamuely that you are reading. $\endgroup$
    – thierry stulemeijer
    Commented Jun 17, 2018 at 15:28
  • $\begingroup$ I've found the lemma and the proposition. They are in "the first edition" not in the second one. Anyway thank you again for your answer. $\endgroup$
    – gualterio
    Commented Jun 17, 2018 at 15:52
  • $\begingroup$ I'm really sorry but I can't help but ask one more question. Can you explain to me why the first cohomology of the final step is trivial? Actually I don't understand why the multiplicative group of invertible elements can be interpreted as the set of automorphism group of the "algebra". Instead I interpreted K* as the group of automorphisms of the k-vector space K equipped with a k-linear endomorphism(multiplication by a primitive element of K) and tried to apply the descent method, but I was not successful. Can you help me with the proof please? $\endgroup$
    – gualterio
    Commented Jun 20, 2018 at 7:51
  • $\begingroup$ 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 $1\to \tilde{K}\to \tilde{K}\times \tilde{K}\to \tilde{K}\to 1$. $\endgroup$
    – thierry stulemeijer
    Commented Jun 20, 2018 at 16:47

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