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Let $\cal{Q}$ be the field of rationals and $L=\cal{Q}(\sqrt 2, \sqrt 3)$ be a Galois extension of degree 4 ($[L:\cal{Q}]=4$).

Using theory of Drinfeld twists on the galois group of $L$, I have produced a noncommutative algebra $(\cal{Q}(\sqrt 2, \sqrt 3),\star)$, where $\star$ is the associative product defined on vector space $\cal{Q}(\sqrt 2, \sqrt 3)$ determined by; $$\sqrt 2 \star \sqrt 2=2, \hspace{0.5cm}\sqrt 3 \star \sqrt 3=3, \hspace{0.5cm}\sqrt 2 \star \sqrt 3=-\sqrt 6,\hspace{0.5cm}\sqrt 3 \star \sqrt 2=\sqrt 6 $$

I want to determine if this algebra is semisimple, and if so what the structure is.

What would be the best approach to this?

From the comments and some reading around I understand the following;

This algebra is a quaternion algebra, $(2,3)$ using notation from Gille and Szamuely's "Central simple algebras and galois cohomology". It follows then that this is either a division algebra or is the matrix algebra $M_2(\cal{Q})$.

Again, by a result in the mentioned book, it is a matrix algebra iff $3=x^2-2y^2$ has solutions in $Q$. I don't think this has any rational solutions so I assume that it isn't a matrix algebra, and therefore is a division algebra.

Inspector gadget
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  • When you write $\sqrt{a}$ do you mean $a$ is a nonnegative rational? – rschwieb Dec 06 '22 at 15:35
  • Hi, I've edited it now for a=2, b=3. I was trying to go for a more general problem to develop my understanding but looking at it again I don't think it adds any value.

    My go to would be trying to use the Artin Wedderburn theorem and calculating the nilradical of the space but it's been a while since I've used it and I can't find any explicit computations of nilradicals online.

     – Inspector gadget Dec 06 '22 at 17:22
  • @Inspectorgadget more standard would be to write $i^2=2$, $j^2= 3$, and $ij=-ji$. – peter a g Dec 06 '22 at 17:40
  • Originally you wrote $\sqrt{a}\sqrt{b}=-\sqrt{a}b$ which is pretty different from $\sqrt{a}\sqrt{b}=-\sqrt{ab}$. I'm glad you made your edit or we would not have known. – rschwieb Dec 06 '22 at 18:36
  • I've heard of hamiltonians/quaternions but wasn't familiar with quaternion algebras. Reading up on them, I see the link now. The algebra in the example would be quaternion algebra $(2,3)_Q$ (as per wikipedia's notation of quaternion algebras) - is this correct? It is then a "central simple algebra" and is either isomorphic to a division algebra or the 2x2 matrix algebra over Q. So now I need to check if it is a division algebra or not to complete the classification. Is – Inspector gadget Dec 06 '22 at 20:46

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