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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?

wow
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2 Answers2

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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.

Dietrich Burde
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  • Unfortunately, this is the same explanation I gave the user on a previous question, so they must be interested in something more low power. We might actually have to explain what a maximal right ideal of the matrix ring looks like. – rschwieb Nov 11 '17 at 15:11
  • Is the property of Morita invariant also applicable for showing primness. That means can I say $R$ is prime if and only if $M_{n}(R)$ is prime? – wow Nov 11 '17 at 15:21
  • @wow Yes, being a prime ring is a Morita invariant property, and $R$ is prime iff $M_n(R)$ is. $M_n(\mathbb Z)$ and $\mathbb Z$ are indeed both prime rings. – rschwieb Nov 11 '17 at 15:35
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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.

egreg
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