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.