Prime ideals and integral domains

abstract-algebraidealsring-theory

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?

Best Answer

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)
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