A commutative ring whose all proper ideals are prime is a field.

Question

Let $R$ be a commutative ring with $1$. Suppose that all ideals $I \neq R$ are prime. Prove that $R$ is a field.

Help me some hints.

Thanks a lot.

https://yutsumura.com/if-every-proper-ideal-of-a-commutative-ring-is-a-prime-ideal-then-it-is-a-field/ — qwr, Nov 19 '18 at 04:21
Hint $,(0)$ prime $\Rightarrow R$ domain. $R$ has a nonunit $,r!\neq! 0 \Rightarrow$ nonunit $r^2!\neq 0,$ is a reducible prime $\Rightarrow!\Leftarrow$ — Bill Dubuque, Sep 18 '23 at 14:37
The definition of primality comes handy. ${0}$ is a prime ideal immediately implies that $R$ does not have zero divisors. For any non-zero $r\in R$, by existence of $1$, we have: $r^2\in\langle r^2\rangle\implies r\in \langle r^2\rangle\tag*{}$ (by primality) which means that there exists $r'\in R$ such that $r'r^2=r$. Hence, every non-zero $r$ is a unit. — Nothing special, Dec 31 '23 at 13:31

score 16 · Accepted Answer · answered Jan 03 '14 at 09:24

16

First show that $R$ is a domain, by a suitable choice of ideal.

Then let $x \ne 0$, and deduce consequences from the fact that $xx \in x^2 R$.

answered Jan 03 '14 at 09:24

user118829

Can you explain more precise? – Truong Jan 03 '14 at 12:15
1

Did you imply we can choose a noninvertible $x \in R$, take ideal $(x^2)$? – Truong Jan 04 '14 at 01:16
1

Since $(x^2)$ is prime, we have $x \in (x^2)$ – Truong Jan 04 '14 at 01:17
1

That means $x=rx^2,r \in R$ – Truong Jan 04 '14 at 01:18
1

Since $(0)$ is prime, we have $1=rx$, a contradiction – Truong Jan 04 '14 at 01:19
How the converse? – Truong Jan 04 '14 at 01:22
Let $x \ne 0$. You want to prove that $x$ is invertible. The first possibility is that $(x^2)$ could be $R$, so you need to look at that case. If not, then $(x^2)$ is prime, and what you wrote also proves that $x$ is invertible. Your original question didn't ask for a converse, can you give the exact statement you're referring to? – user119003 Jan 04 '14 at 05:20
Ehm, of course. – Truong Jan 05 '14 at 00:38

score 6 · Answer 2 · answered Jan 03 '14 at 09:23

6

Hint: Consider some $r\in R$ and examine the principal ideal generated by $r^2$.

answered Jan 03 '14 at 09:23

Philip Hoskins

Can you explain more precise? – Truong Jan 03 '14 at 09:26
1

Well, first you'll need to verify that the cancellation law holds in this ring (how?). Assuming you've done that, suppose for the moment that the principal ideal satisfies $(r^2)=R$. In particular, this says $r$ is a multiple of $r^2$. This implies $r$ is invertible (why?). If $(r^2)\neq R$, then what may be deduced? – Philip Hoskins Jan 03 '14 at 09:36

2 Answers2