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$\mathbb{Z}_n$ is integers modulo $n$. Local ring is a commutative ring if it has a unique maximal ideal. Please help me prove the claim.

math
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  • I don't understand why would anyone want to close this question. Please give a reason! – math Dec 05 '14 at 22:57
  • what do you call $;\Bbb Z_n;$ to? The integers modulo $;n;$? The $;n$-adics...? The close vote is too harsh imo, but it probably comes from the fact you show no self effort in your our question's solution. – Timbuc Dec 05 '14 at 22:58
  • integers modulo n, sorry for not explaining this earlier. – math Dec 05 '14 at 22:59
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    @math the reason is because you don't show effort, primarily. (not my vote) – apnorton Dec 05 '14 at 23:03
  • What relationship do the ideals of $\mathbb{Z}_n$ and of $\mathbb{Z}$ have? – hardmath Dec 06 '14 at 00:27

3 Answers3

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You probably know (and if you don't prove it: it is a nice exercise) that a unitary commutative ring $\;R\;$ is a local one iff the set of all non units in the ring is an ideal, and in this case exactly this very ideal is the unique maximal one there is.

Well, show this is the case in $\;\Bbb Z_n:=\Bbb Z/n\Bbb Z\;$ iff $\;n\;$ is a prime power.

Timbuc
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If two distinct primes $p$ and $q$ divide $n$, then $$ n\mathbb{Z}\subseteq p\mathbb{Z} $$ and $$ n\mathbb{Z}\subseteq q\mathbb{Z} $$ so $p\mathbb{Z}/n\mathbb{Z}$ and $q\mathbb{Z}/n\mathbb{Z}$ are distinct maximal ideals of $\mathbb{Z}/n\mathbb{Z}$.

Why are they distinct? (Hint: consider the quotient ring.)

The converse is easy: the only maximal ideal in $\mathbb{Z}/p^k\mathbb{Z}$ ($p$ a prime, $k>0$) is $p\mathbb{Z}/p^k\mathbb{Z}$.

egreg
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HINT: Use the Chinese Remainder Theorem: it tells you that $$ \frac{\Bbb Z}{n\Bbb Z}\simeq \frac{\Bbb Z}{p_1^{e_1}\Bbb Z}\times\cdots\frac{\Bbb Z}{p_k^{e_k}\Bbb Z} $$ if $n=\prod_{i=1}^kp_i^{e_i}$ is the primary decomposition of $n$. Now, what are the maximal ideals in a product of rings?

Andrea Mori
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