0

EDIT: In this proof I suppose that $R$ is not a prime ideal of $R$.

Let $R$ be a commutative ring with unity. For every $x\in R$ there exists a natural number $n_i>1$ such that $x^{n_i}=x$. Prove that every prime ideal of $R$ is maximal.

My Attempt:

Suppose by contradiction that there exists a prime ideal $P\subseteq R$ such that $P$ is not a maximal ideal. this means that there exists an ideal $M\subseteq R$ such that $P\subseteq M$.

We'll notice that for every $x\in P: x=x^{n_i}\in P$, which means that for every $1\le j\le n_i: x^j \in P$, this follows from the definition of a prime ideal and the fact that $R$ is commutative.

Now, by the assumption $P$ is not maximal therefore there exists $t\in M: t\notin P$. and again, we conclude that for every $1 \le j\le (n_i): t^j \in M \land t^j \notin P$.

But this doesn't really bring me into any contradiction.

I'm not sure what am I missing.

user26857
  • 53,190
Noa Even
  • 2,875

1 Answers1

5

Assume for a moment that $R$ is also an integral domain. We'll show this implies $R$ is a field. Indeed, let $0\ne x\in R$. By assumption there is some $n>1$ such that $x^n=x$. Then $x(x^{n-1}-1)=0$. Since $R$ is an integral domain it follows that $x^{n-1}=1$, and so $x$ is invertible. So $R$ is a field.

Now it's easy to solve your problem. Let $P\subseteq R$ be a prime ideal. Then $R/P$ is an integral domain, and obviously it has the same property as $R$. As we have shown this implies $R/P$ is a field, and so $P$ is a maximal ideal.

Mark
  • 43,582