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Let $M_{2,2}(\Bbb Z)$ be the ring of $2 \times 2$ matrices with integer entries and $A=\left\{ \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \mid a,b \in \Bbb Z \right\}$ its subset. It is easily seen that $A$ is a commutative subring of $M_{2,2}(\Bbb Z)$. For a prime number $p$, I want to prove that the set $J_{p}=\left\{ \begin{pmatrix} mp & b \\ 0 & mp \end{pmatrix} \mid b,m \in \Bbb Z \right\}$ is a maximal ideal of $A$.

It seems that I could get a proof by proving that $A/J_p$ is a field. It suffices to find an identity $e$ for $A/J_p$ and prove every non-zero element in $A/J_p$ has inverse. Set $e=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. It's clear that $ex=xe=x$ for all $x \in A$. Furthermore, for non-zero $x=\begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \in A/J_p$ i.e. gcd$(a,p)=1$ there exists $y=\begin{pmatrix} a_1 & 0 \\ 0 & a_1 \end{pmatrix}\in A/J_p$ where $aa_1\equiv 1 \bmod p$ such that $ay=ya=1$ in $A/J_p$. So we are done.

Is the above proof correct? For another approach, can we find explicit form of the field $A/J_p=:F$? If it's possible to do that we may find an isomorphism between $A/J_p$ and $F$ induced from some surjective homomorphism $\phi: A \to F$ with $\ker \phi=J_p$.

Thanks in advance.

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    $\begingroup$ It's also worth noting that this ring is isomorphic to $\Bbb Z[x]/(x^2)$ via mapping $a+bx\to\left[\begin{smallmatrix}a&b\\0&a\end{smallmatrix}\right] $ $\endgroup$ – rschwieb Oct 9 '15 at 15:54
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The proof looks correct, but you are right that there is an alternative proof using the surjective homomorphism $\phi:A\to \mathbb{Z}_p$ (integers mod $p$), mapping the matrix $\begin{pmatrix} a & b \\ 0 & a\end{pmatrix}$ to $[a]$ (the equivalence class of $a$ mod $p$.)

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  • $\begingroup$ Thanks. I'm glad to know that we can solve it by both ways. $\endgroup$ – user Oct 9 '15 at 15:19

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