Let R be an integral domain. Show that if the only ideals in R are {0} and R itself, R must be a field

Question

I know that if (x)={0} then the if 0=r0 such that r belongs to R therefor it's a field. Most likely I'm wrong but I need help with the second part if the ideal is R

Does $R$ contain a multiplicative identity? (I'm guessing so, because the question wouldn't make much sense otherwise.) — Alex Wertheim, Dec 07 '14 at 15:00
Ya, that's included in the def of integral domain, a commutative ring with identity..etc — Jessy White, Dec 07 '14 at 15:20

score 1 · Answer 1 · answered Dec 07 '14 at 14:59

1

Hint: if every nonzero ideal of $R$ is $R$, then every nonzero ideal contains $1$.

answered Dec 07 '14 at 14:59

Alex Wertheim

20,278

score 1 · Answer 2 · answered Dec 07 '14 at 15:15

1

If $R$ with unit.

Let $0\neq x\in R$ then $ 0\neq ( x)$ is an ideal of $R$ , hence $(x)=R$ so there exists $a$ in $R$ sush that $ax=1$.

answered Dec 07 '14 at 15:15

Hamou

6,745

Let R be an integral domain. Show that if the only ideals in R are {0} and R itself, R must be a field

2 Answers2