Prove that every prime ideal of $R$ is maximal.

Question

Let $R$ be a commutative ring with identity.For each $a\in R$ there exist $n(>1)\in \mathbb N$ such that $a^n=a$.

My try

Let $I$ be a prime ideal of $R$. Then $R/I$ is an integral domain. If I can show using the hypothesis that $R/I$ is a field then we are done.

Any hints to show this?

If we allow $n=1$, such $n$ always exists. So presumably there exists $n>1$ such that $a^n=a$? — Hagen von Eitzen, May 10 '15 at 13:26
HINT: $R/I$ is a domain satisfying the same hypothesis of $R$. — Crostul, May 10 '15 at 13:29

Najib Idrissi · Accepted Answer · 2015-05-10T13:37:24.010

8

Let $I \subset R$ be a prime ideal. Suppose $a \in R$, then $a^n - a = 0$ for some $n>1$, thus $a(a^{n-1} - 1) = 0 \in I$. Since $I$ is prime, then either $a \in I$, or $a^{n-1} - 1 \in I$. So if $\bar a \in R/I$ is nonzero, ie. $a \not\in I$, then $a^{n-1} - 1 \in I \implies \bar{a}^{n-1} = \bar{1}$ and $\bar{a} \in R/I$ is a unit (because $n-1 > 0$). It follows that $R/I$ is a field, hence $I$ is maximal.

edited May 10 '15 at 13:37

answered May 10 '15 at 13:29

Najib Idrissi

54,185

score 0 · Answer 2 · answered May 10 '15 at 16:52

Let: $a\in R$ \ $I$. Consider that $(a+I)^n=a^n+I=a+I$. But when $I$ is prime ideal, $\frac{R}I$ is an integral domain. So: $(a+I)((a+I)^{n-1}-(1+I))=0$, which implies that: $a+I=I$ or $(a+I)^{n-1}-(1+I)=I$. But $a\in R$ \ $I$, so: $(a+I)^{n-1}-(1+I)=I$ and from here: $(a+I)(a+I)^{n-2}=(a+I)^{n-2}(a+I)=1+I$

Prove that every prime ideal of $R$ is maximal.

My try

2 Answers2