# Prove that a nonzero integer p is prime if and only if the ideal (p) is maximal in Z.

I know that if (p) is maximal then Z/(p) is a field which implies its an integral domain which then implies that (p) is a prime ideal and p is prime. I'm having trouble going the forward direction assuming that p is prime and then showing it's a prime ideal.

• Use Euclid's lemma. – user121880 Apr 28 '14 at 0:18
• How would that help me? – user141447 Apr 28 '14 at 0:23
• If $a,b\in\Bbb Z$ and $ab\in (p)$, then $p\mid ab$, so $p\mid a$ or $p\mid b$, hence $a\in (p)$ or $b\in (p)$. – user121880 Apr 28 '14 at 0:31

In general, if $(a)$ is an ideal of $\mathbb Z$ then the ideals contained between $(a)$ and $\mathbb Z$ correspond to the divisors of $a$. For example, if $a=10$ then $\mathbb Z=(1)\supset(5)\supset(10)$ and $\mathbb Z=(1)\supset(2)\supset(10)$. This is a consequence of the more general fact that $(a)\supseteq(b)\iff a\mid b$. Now what are the divisors of a prime integer?

• I haven't notice that $(1)=\mathbb Z...$ :) – user441848 Jul 22 '17 at 3:46

One proof would be to use the fact that $\mathbb{Z}/n\mathbb{Z}$ is an integral domain iff it is a field. It is a basic fact from number theory that the units in $\mathbb{Z}/n\mathbb{Z}$ are exactly the numbers coprime to $n$ and the non units are always zero divisors $\mod n$.

Hint  Notice that for principal ideals: $\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$,  i.e. $(a)\supseteq (b)\iff a\mid b,\,$ thus

$\qquad\quad\begin{eqnarray} (p)\,\text{ is maximal} &\iff&\!\!\ (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (d)\\ &\iff&\ p\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ d\\ &\iff&\ p\ \ \text{ is irreducible}\\ \end{eqnarray}$

• Is $(p)$ has no proper container $(d)$ equivalent to say $(p)$ has no proper elements $(d)$? – user441848 Jul 14 '17 at 2:29
• @AnneliseToft $\$ What do you mean by "$(p)$ has no proper elements $(d)$"? $\ \$ – Bill Dubuque Jul 14 '17 at 2:37
• I dont understand the english term 'container', is there another way to express the term? – user441848 Jul 14 '17 at 2:40
• @AnneliseToft Read that as "divisor". Bill is saying that containment equals division for principal ideals. – Pedro Tamaroff Jul 14 '17 at 2:41
• @AnneliseToft $\ (a)\,$ contains $(b)$ means that $(a)\supseteq (b).\,$ The idea is that that a nontrivial ("proper") container $(d)$ of $(a)$ corresponds to a nontrivial divisor $d$ of $a$ using "contains = divides" in a PID, since $\ (1)\supsetneq (a) \supsetneq (d)\iff 1\mid a\mid d\,$ properly (i.e $\,a\nmid 1,\, d\nmid a).\ \$ – Bill Dubuque Jul 14 '17 at 3:02