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 August 11th, 2012, 10:17 AM #1 Member   Joined: Nov 2010 Posts: 48 Thanks: 0 Prime Ideal Prove that for the ring $( \mathbb{Z} , +, .)$ , $I=m \mathbb{Z}$ is prime ideal if and only if $m$is prime.
 August 11th, 2012, 07:21 PM #2 Member     Joined: Aug 2011 From: Nouakchott, Mauritania Posts: 85 Thanks: 14 Math Focus: Algebra, Cryptography Re: Prime Ideal We suppose that $m$ is prime. Then $m\ne 1$ and thus $m{\mathbb Z}\ne {\mathbb Z}$. Now let $a$ and $b$ be two integers such that $ab\in m{\mathbb Z}$. Then we have $m|ab$. Since $m$ is prime, we have $m|a$ or $m|b$. That means : $a\in m{\mathbb Z}$ or $b\in m{\mathbb Z}$. Thus $m\mathbb Z$ is a prime ideal. Inversely, let $m$ be an integer such that the ideal $m\mathbb Z$ is prime. Let $p$ and $q$ be two integers such that $m=pq$. That means : $p|m$ and $q|m$. But : $pq=m\in m{\mathbb Z}$ and the ideal $m\mathbb Z$ is prime. So : $p\in m{\mathbb Z}$ or $q\in m{\mathbb Z}$ Hence : $m|p$ or $m|q$. If $m|p$, then $m=\pm p$ (because we already have $p|m$) and thus $q=\pm 1$. If $m|q$, then $m=\pm q$ (because we already have $q|m$) and thus $p=\pm 1$. Hence $m$ is prime.
 August 12th, 2012, 12:17 AM #3 Member   Joined: Nov 2010 Posts: 48 Thanks: 0 Re: Prime Ideal Thanks a lot.

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