# Showing $M_2 \cong \mathbb{Q} [x] /\langle x^2\rangle$

As the title says, I am trying to show the following isomorphic as rings $M_2 \cong \mathbb{Q} [x] / \langle x^2\rangle$. Here, $M_2$ denotes the set $$\left\{\begin{bmatrix} a_0 && 0 \\ a_1 && a_0\end{bmatrix} : a_0, a_1\in\mathbb{Q}\right\} \, .$$

Intuitively, this is confusing. As I understand it, $\mathbb{Q} [x] / \langle x^2\rangle$ has four elements while $M_2$ has infinitely many. Off the bat, how can this be true? (I am asked to use the Chinese Remainder Theorem in this proof if that is a helpful hint.)

• $\mathbb{Q}[x] / \langle x \rangle$ as a $\mathbb{Q}$-vector space is spanned by $\{1, x\}$. So it has infinitely many elements. – Joppy Apr 26 '17 at 1:25
• Ah, ok! I think you mean $x^2$, but I take your point. – user322548 Apr 26 '17 at 1:29
• That quotient does not have four elements. – Mariano Suárez-Álvarez Apr 26 '17 at 1:32

Hint: The ring $\mathbb{Q}[x] / \left< x^2 \right>$ can be identified with the ring
$$\{ a_0 + a_1 x \, | \, a_0, a_1 \in \mathbb{Q} \}$$
$$(a_0 + a_1x)(a_0' + a_1'x) = a_0 a_0' + (a_0 a_1' + a_1' a_0)x$$
(so $x^2 = 0$). Can you find a natural map between this ring and $M_2$ and show that it is a ring isomorphism?