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Prove that if an integral domain contains exactly one prime ideal, then it's a field.

My attempt : Since the ring is an integral domain, $(0) $ is a prime ideal. So no other prime ideal exists. Now, if $I $ is a proper ideal, then either $I $ is maximal or we can get a proper ideal containing it. If I can somehow get a proper maximal ideal, then it is prime. But the problem is that the ring $R $ need not be finite. So, is my approach correct? Otherwise, how should I proceed?

cqfd
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    Can you see what needs to be proved? Only that any non-zero element is invertible, so take $;0\neq r\in R;$ and look at $;\langle r\rangle;$ ... – DonAntonio Nov 14 '18 at 11:45
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    Your approach is correct, I suggest that your try to find all maximal ideals of this ring, in this case there are a only maximal ideal. Because every maximal is a prime ideal, so every element different to zero is a unit then R is a field. – carlos arturo Hurtado Nov 14 '18 at 11:47
  • @DonAntonio I tried to prove $(r) =R$ by contradiction, but couldn't succeed. Anyway, I found out a proof saying an integral domain has a maximal ideal. So I think I can proceed as state above. Thanks. – cqfd Nov 14 '18 at 12:56
  • @ThomasShelby Every proper ideal in every ring with identity is contained in a maximal ideal. It doesn't have to do with being an integral domain, really... – rschwieb Nov 14 '18 at 15:11

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If $R$ has exactly one prime ideal then its nilradical $N$ must be that ideal because it is the intersection of all the prime ideals. Then $R/N$ is a field, see here:

Each element of a ring is either a unit or a nilpotent element iff the ring has a unique prime ideal

Here we have $N=0$ by assumption, see your argument, so that $R$ is a field.

Dietrich Burde
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The path of least resistance is:

  1. $\{0\}$ is the only prime ideal (exactly as you have reasoned)
  2. All maximal ideals of $R$ are prime (easy exercise proven all over the site, e.g. here) and hence the only possible maximal ideal is $\{0\}$.
  3. Therefore $\{0\}$ is the only maximal ideal of the ring, and the only proper ideal of the ring.
  4. A ring with exactly two ideals (the zero ideal and the whole ring) is a field (proven all over the site, e.g. here)
rschwieb
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Hint $\ a\neq 0\Rightarrow (a)=1 $ (else some max($\Rightarrow$prime) $M \supseteq (a),$ contra $0 =$ unique prime ideal)

Bill Dubuque
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