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The follow statement can be found in one of the answers here.

Statement: A ring is a field if and only if its only prime ideal is $(0)$.

For this, let's assume that $R$ is a ring with $1$, and is commutative. Then does the statement still hold true?

I have been trying to find the proof of this statement, but I cannot find it anywhere.

If $R$ is a field, it is clear that $(0)$ is the only prime ideal. But converse is not clear to me.

Thank you!

user26857
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Phil
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    If the only prime ideal is $0$, then $0$ is a maximal ideal... – diracdeltafunk Mar 31 '21 at 23:42
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    It's quite simple – it results from the fact that the ideal generated by any nonzero element is equal to the ring if and only if any nonzero element is invertible. – Bernard Mar 31 '21 at 23:43
  • @rschwieb the OP stated iff every prime ideal is (0), I'm not sure how you feel about that being a trivial leap from the only ideals are (0) and (1) or not – no lemon no melon Apr 01 '21 at 21:32
  • @nolemonnomelon shrug the hypothesis immediately implies 0 is maximal, and then you're at the duplicate. I guess it's not perfect :( – rschwieb Apr 02 '21 at 21:23

2 Answers2

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Hint: $R$ itself is (by definition) not a prime ideal. If $P$ is a prime ideal and $a \in P$ is non-zero, is it possible that $ax = 1$ for some $x \in R$?

Rob Arthan
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I think this what @diracdeltafunk meant in the comments:

Every commutative ring $R$ with identity $1$ has a maximal ideal $M$,

If an ideal is maximal, then it is prime,

$M \in \{P|P\text{ is a prime ideal of }R\}=\{(0)\}$

because $(0)$ is the only prime ideal of $R$

thus $M=(0)$, and $R\cong R/(0)$ is a field.

no lemon no melon
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