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Prove that for any primes $p$, $q$, $p\neq q$, the ring $\mathbb{Z}_{pq}$ (the ring of integers modulo pq) is semisimple, and for $p=q$ the same ring is not semisimple.

I was told that the easiest way is to observe that it has global dimension 1, so it's hereditary, not semisimple. But I don't know how to prove this.

I'm sure it's not complicated, but it eludes my mind. Thanks in advance for any useful replies.

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    By $\mathbb{Z}_n$ for $n$ not prime do you mean the $n$-adic completion of $\mathbb{Z}$?2012-06-13
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    Oh, sorry. I forgot that this could cause some confusion. I mean the integers modulo n. I shall edit now.2012-06-13
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    You don't know how to prove $\mathbb{Z}_{pq}$ has global dimension 1, or you don't know why $R$ semisimple $\Leftrightarrow R$ has global dimension 0?2012-06-13
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    Something is wrong...if $p\neq q$ then $\mathbb{Z}/pq \cong \mathbb{Z}/p \oplus \mathbb{Z}/q$ as rings. This is the direct sum of two semisimple rings (indeed, two fields!) therefore semisimple (see Mariano's answer http://math.stackexchange.com/questions/36347/semisimple-rings?rq=1)2012-06-13
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    Dear Juan, I don't know how to prove that it has global dim=1. Dear @mt_, I have read Mariano's answer and I have to say that the same things were in ny mind too. BUT, our professor told us that although $\mathbb{Z}/p \times \mathbb{Z}/q$ is semisimple, $\mathbb{Z}/pq$ is not and the above isomorphism is only as abelian groups. I think your remarks are correct and thus, our prof was simply wrong (also having in view rschwieb's answer, which I agree with).2012-06-13
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    You can map $\mathbb{Z}/pq$ to $\mathbb{Z}/p \times \mathbb{Z}/q$ by $[n] \mapsto ([n],[n])$. This is a ring isomorphism: it obviously respects addition, also $[m]\cdot [n] = [mn]\mapsto ([mn],[mn])=([m],[m])\cdot([n],[n])$, and it is clearly injective therefore bijective.2012-06-13
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    Yes, my thoughts exactly. So the conclusion is that, for p,q primes $\mathbb{Z}/pq$ IS semisimple, either by showing that it is isomorphic to that direct sum, or by rschwieb's argument (essentially the same idea). Thank you.2012-06-13
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    @AdrianM right (but only for $p,q$ *distinct* primes - otherwise there's a non-split extension $\mathbb{Z}/p \to \mathbb{Z}/p^2 \to \mathbb{Z}/p$)2012-06-13
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    @mt_ Yes, I just finished arguing with a colleague of mine on this subject. Apparently, the original question was: "True or false: Z/pq is a semisimple ring (p,q primes)". And the answer was "false", the counterexample being exactly p=q (since this cade was allowed!). Alright, I am accepting rschwieb's answer and calling it a close. Thank you all!2012-06-13
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    @AdrianM Hehe, that's a little scurvy to write the question that way, but nevertheless you probably learned a lot, here.2012-06-13

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Here, I'll always be assuming $n>1$

The ring $\mathbb{Z}/n\mathbb{Z}$ is semisimple exactly when $n$ is squarefree. The easy way to see this is that the ideals generated by primes are maximal and have trivial pairwise intersections, and so the Chinese Remainder theorem says the ring is a finite product of fields.

If, on the other hand, $n$ is not squarefree, (say $p^2$ divides it), then $\frac{n}{p}\mathbb{Z}/n\mathbb{Z}$ is a nonzero nilpotent ideal, and so the ring is not semisimple.

The ring $\mathbb{Z}/n\mathbb{Z}$ is local iff $n$ is a power of a prime.

The ring $\mathbb{Z}/n\mathbb{Z}$ is always quasi-Frobenius.