Show that $\mathbb{Z}_n$ is local ring iff $n$ is a power of a prime number

Question

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

Answer 1

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 $p\mathbb{Z}/p^k\mathbb{Z}$.

Answer 2

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.

Answer 3

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?