Jacobson Radical of integer matrices

Question

Let $R = M_2(\mathbb Z)$

Find $J(R)$

The reason I'm asking is that I know by Artin Wedderburn that $M_n(D)$ is semisimple for any division ring D and hence $J(D)=0$. But here $\mathbb Z$ is of course not a division ring.

Answer 1

For definiteness, let's consider left $R$-modules. First choose a prime $p\in\mathbb{Z}$, and consider the set $A_p=\left(\mathbb{Z}\big/p\mathbb{Z}\right)^2$ as a space of $2\times1$ column vectors. A matrix $(m_{ij})_{ij}\in R$ acts on an element $(\overline{a},\overline{c})\in A_p$ by sending it to $(\overline{m_{11}a}+\overline{m_{12}c}, \overline{m_{21}a}+\overline{m_{22}c})$.

Exercise: Show that $A_p$ is simple. $\square$

Thus $J(R)\subseteq\operatorname{ann}A_p$. What form does this latter ideal take? Well, if $(m_{ij})_{ij}\in\operatorname{ann}A_p$, then in particular it sends $(\overline{1},\overline{0})$ and $(\overline{0},\overline{1})$ to $(\overline{0},\overline{0})$. Thus we must have $\overline{m_{ij}}=\overline{0}$ for each $i$ and $j$, and hence $m_{ij}\in p\mathbb{Z}$ for each $i$ and $j$. Conversely, if an element of $R$ is of that form (ie has all entries in $p\mathbb{Z}$), then it annihilates $A_p$. (Why?) So the annihilator of $A_p$ is precisely the left ideal $M_2(p\mathbb{Z})$. Now, we know that $$J(\mathbb{Z})=\bigcap_{p\text{ prime}}p\mathbb{Z}=\{0\}.$$ What can we thus conclude about $J(R)$, which is a subset of $\bigcap_{p\text{ prime}}\operatorname{ann} A_p$?