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Assume $R$ is commutative. Prove that if $P$ is a prime ideal of $R$ and $P$ contains no zero-divisors then $R$ is an integral domain.

Proof: let $ab \in P$ where $ab \not= 0$. that means $a \in P$ or $b \in P$. Which implies that $a+P=P$ or $b+P=P$ the only way that $a+P=P$. If $a =0$ similarly $b+P=P$ of $b=0$. Which is a contradiction thus $ab$ is an integral domain. I do not know if this is a right approach.

egreg
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user146269
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  • You are on the wrong track. $a+P=P$ does not imply $a=0$. However the proof is an easy one (just not this approach). – hardmath Feb 26 '15 at 18:07
  • I think you are a bit confused. What do you mean by $ab$ is an integral domain? And why are you asking $ab \neq 0$? – Crostul Feb 26 '15 at 18:11

3 Answers3

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Try to show that: if there exists a zero-divisor, then there is a zero-divisor in $P$.

This then shows that if there is no zero-divisor in $P$, then there is none at all and you are done.

To do this assume: $ab=0$ with non-zero $a,b$, that is assume a zero-divisor. Now, $0 \in P$ so $a \in P$ or $b \in P$ and you are essentially done.

quid
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  • i essentially did something like this but i did not see how $R$ is an integral domain because the definition of an integral domain is that $ab =0$ either $a=0$ or $b=0$ – user146269 Feb 26 '15 at 18:36
  • An equivalent way of expressing integral domain is that there do not exists non-zero $a,b$ such that $ab=0$. This is what I used. Indeed, this double negation is not very elegant but I though it might be more transparent. – quid Feb 26 '15 at 18:47
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Assuming $\mathfrak p$ is prime in $R$ and contains no nonzero zero divisors, take a minimal prime $\mathfrak q$ under $\mathfrak p$. Since $\mathfrak q$ consists entirely of zero divisors but contains no nonzero zero divisors, it must be the zero ideal. Since we've shown the zero ideal is prime, $R$ is an integral domain.

Of course, it remains to show minimal primes consist of zero divisors. You can find some discussion here.

A proof is sketched there: In $R_\mathfrak q$, we know that $\mathfrak q R_\mathfrak q$ is a maximal ideal, and that the primes of $R_\mathfrak q$ are those under $\mathfrak q$. Since $\mathfrak q $ is minimal, it follows the only prime ideal of this ring is $\mathfrak q R_\mathfrak q$, so the radical of this ring is $\mathfrak q R_\mathfrak q$. This means that for every $q\in \mathfrak q$ there is a smallest $n$ such that $q^n/1=0$, i.e. there is $s\notin \mathfrak q $ such that $sq^n=0$. Since $n$ was chosen smallest, $sq^{n-1}$ is not zero, and hence $q$ is a zero divisor.

Community
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Pedro
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    This is way more complicated than necessary. – Marc van Leeuwen Feb 26 '15 at 21:49
  • @MarcvanLeeuwen I agree. But it gives important information, to me. And the "trick" of localizing at minimal primes is something quite important. Knowing minimal primes consist of zero divisors is something that seems relevant to me, too. – Pedro Feb 26 '15 at 21:57
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Suppose $a,b\in R$ with $ab=0$, we must show that $a=0$ or $b=0$. Certainly the images $\overline a,\overline b$ of $a,b$ in $A/P$ satisfy $\overline a\overline b=0$, and since $R/P$ is an integral domain (by the definition of prime ideal) one of $\overline a,\overline b$ is zero, that is $a\in P$ or $b\in P$. Assuming by symmetry the former is the case, then either $b=0$ (in which case we are done), or else $b\neq0$ and the fact that $P$ does not contain zero divisors (of $R$) imply that $a=0$ (and we are done as well).

I think this is more or less the proof you wanted to give, but your sentences get confused from "the only way" on; I simply cannot decipher them.

Marc van Leeuwen
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