Question on primitive ring

Question

I wanted to show that the ring $R=M_{2}(\mathbb{Z})$ is a not primitive.

Here is what I did.

Suppose for contradiction it is primitive. Then there is a faithful and irreducible $R-$ module $M$.

Let $$0\neq x\in M$$

Then from the irreduciblity of $M$, $$Rx=M$$

Define a module homomorphism $$\phi:R\rightarrow M, r\mapsto rx$$

Then $$ker(\phi)=ann_{R}(M)={0}$$ (here is one of my doubt) by the faithfulness of the module. Moreover $\phi$ is subjective. Therefore $\phi$ is an isomorphism. But we know that $R$ is not simple module as an $R$ module.Let its the non trivial sub module be given by $V$. Therefore contradiction( as $\phi(V)$ is a non trivial sub module of $M$, which was claimed as irreducible).

Is my argument correct. Or is there another approach?

Answer 1

Suppose that $M_2(\mathbb Z)$ is primitive. Being right primitive is a Morita invariant property, so $\mathbb Z$ is also primitive. But a commutative primitive ring is a field, so $\mathbb Z$ is not primitive, a contradiction.

Answer 2

No, it doesn't work, because you could apply the same argument to any ring.

The kernel of $\phi$ is the annihilator of $x$, which is not generally the same as the annihilator of $Rx$ when $R$ is not commutative.

For commutative rings the argument works to prove that primitive is the same as simple, that is, a commutative ring is primitive if and only if it is a field.