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I'm reading Central Simple Algebras and Galois Cohomology by Gille and Szamuely. I am stuck on this line in one of the proof. $k$ is a field. A quaternion algebra is called split if it is isomorphic to $M_{2}(k)$ as a $k$-algebra.

This is the statement

Proposition 1.2.3 Consider a quaternion algebra $A$ over $k$, and fix and element $a\in k^{*}/k^{*^{2}}$. The following statements are equivalent:

1) $A$ is isomorphic to the quarternion algebra $(a,b)$ for some $k^{*}$.

2) The $k(\sqrt{a})$-algebra $A\otimes_{k} k(\sqrt{a})$ is split.

There is another part but that doesn't concern me. This is where I am stuck on the proof

note that $(a,b)\otimes_{k}k(\sqrt{a})$is none but the quaternion algebra $(a,b)$ over $k(\sqrt{a})$.

I do not see this as obvious and can't seem to find the appropriate map. Any hints would be very much appreciated. Thanks.

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  • $\begingroup$ Did you mean "none but the quaternion algebra"? $\endgroup$
    – user122283
    Jun 4, 2014 at 1:40
  • $\begingroup$ @SanathDevalapurkar Yes, sorry for that. $\endgroup$ Jun 4, 2014 at 1:50

1 Answer 1

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I looked in the book it goes on to say "none but the quaternion algebra $(a,b)$ over the algebra $k(\sqrt{a})$". This is just a change of the ground field. In the tensor product one still has exactly the same defining relations for the quaternion algebra, but now one has also scalar multiplication from this larger base field.

Edit:Yes you included this in you question now.

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  • $\begingroup$ Great thanks. Simple misunderstanding of $(a,b)$. $\endgroup$ Jun 4, 2014 at 2:21

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