Show that two rings of matrices are not isomorphic

Question

Let $p$ be a prime number and $$ A_p = \{\left( \begin{matrix} a & bp \\ b & a \\ \end{matrix} \right)|\ a, b \in \mathbb{Z} \} $$

Show that $A_2$ and $A_3$ are not isomorphic. The problem had 2 other questions, but this is the one where I got stuck.

I tried by writing $\left( \begin{matrix} 4 & 0 \\ 0 & 4 \\ \end{matrix} \right)$ as$\left( \begin{matrix} 2 & 0 \\ 0 & 2 \\ \end{matrix} \right) + \left( \begin{matrix} 2 & 0 \\ 0 & 2 \\ \end{matrix} \right)$ and as $\left( \begin{matrix} 2 & 0 \\ 0 & 2 \\ \end{matrix} \right) \left( \begin{matrix} 2 & 0 \\ 0 & 2 \\ \end{matrix} \right)$ as with $\mathbb{2Z}$ and $\mathbb{3Z}$, but I don't know how to continue.

Answer 1

Building on the observation of Lord Shark the Unknown, we see that in $A_2$ there is an element $$\left( \begin{matrix} 0 & 2 \\ 1 & 0 \\ \end{matrix} \right)$$ whose square is $2I$. In $A_3$ the square of an arbitrary element takes the form $$\left( \begin{matrix} a^2+3b^2 & 6ab \\ 2ab & a^2+3b^2 \\ \end{matrix} \right)$$ For such an element to equal $2I$ we must have $a=0$ or $b=0$ in which case there is no solution to $a^2+3b^2=2$, so $A_3$ has no element whose square is $2I$. $A_2$ and $A_3$ cannot be isomorphic.

Answer 2

Here's another way to see it. If $\phi$ is an isomorphism, then $a$ and $\phi(a)$ satisfy the same polynomials with integer coefficients, so they have the same eigenvalues. But the eigenvalues of $A_p$ are $a\pm b\sqrt{p}.$