# Morphism of algebras

Let $Q(i)$ be the extension of the rational numbers $Q$ obtained by adjoining a root i of the polynomial $X^2 + 1$. Consider the algebra B defined by the Hilbert symbol $(-2, -5)$ over $Q(i)$. So, by definition, $$B := Q(i)[[\alpha, \beta]]/(\alpha^2 = -2, \beta^2 = -5, \alpha\beta = -\beta\alpha)$$ here $[[\alpha, \beta]]$ means non-commutative polynomials in $\alpha, \beta$. The algebra $B$ is ramified at the primes of $Q(i)$ lying above $5$, and unramified at all other places.

My question: Does there exist an injection of $Q$-algebras $$B \to M_4(Q),$$ where $M_4(Q)$ is the ring of $4 \times 4$ matrices over $Q$.

-

No. Consider the matrix corresponding to $i$. It is semisimple with eigenvalues $i,i,-i,-i$ . So it's centralizer is $8$-dimensional - clearly, it is $M_2(\mathbb Q(i))$. Since this algebra is ramified, it does not inject into the split algebra.
I am not sure if I understand your answer, but are you aware of the following construction? Pick an isom of $Q$-vectorspaces $Q^4 \cong Q(i)^2$. Then $M_2(Q(i))$ acts by $Q(i)$-linear (hence also $Q$-linear) endomorphisms on $Q(i)^2$. Via $Q^4 \cong Q(i)^2$ we then get an injection $M_2(Q(i)) \rightarrow End_Q(Q(i)^2) = M_4(Q)$. – mnr Jul 4 '13 at 9:04
Yes. Moreover, that is the only quaterniom algebra of $\mathbb Q(i)$ with such an injection, because it is the centralizer of $i$, which is a representative of the unique conjugacy class of matrices satisfying $x^2+1=0$. – Will Sawin Jul 4 '13 at 14:35