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Let $R$ be a commutative ring. Let $P\subset R$ be a minimal prime ideal. Let $S=R-P$. Let $x\in P$ and $s \in S$. Is there any property saying that if $sx=0$, then $x=0$?

Should anyone helps me, I will be very grateful.

Thank you for all the comments. Maybe I was not asking what I wanted to ask. I am studying a proposition:

Let $R$ be a commutative ring. Let $P\subset R$ be a minimal prime ideal. Then every $x\in P$ is nilpotent.

Its proof is using localization and it is like this:

Let $x\in P$ be nonzero. There is only one prime ideal in $R_P$, namely $PR_P$, and $x/1$ is a nonzero element in it. Indeed, $x$ is nilpotent if and only if $x/1$ is nilpotent...

So here is the problem I don't understand. How do we know $x/1$ is nonzero in $R_P$ if $x$ is nonzero?

This proposition is on page 194 of Rotman's An Introduction to Homological Algebra. A proposition is proved beforehand:

If $S⊆R$ is a multiplicative set and $h: R → S^{-1}R$ is the localization map, then $\ker h = \{ r∈R : sr=0 \text{ for some } s∈S \}$.

So I was trying to know whether $P$ consists of all the zero divisors.

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  • $\begingroup$ This would be true for any prime ideal if R is an integral domain. $\endgroup$ – R_D Nov 19 '15 at 16:44
  • $\begingroup$ What if I drop this assumption that R is an integral domain. Is it true generally? $\endgroup$ – Deepleeqe Nov 19 '15 at 16:47
  • $\begingroup$ It won't be true in general. Consider $R=\mathbb Z_6$ and $P=\langle 2\rangle$. Let $s=3$ and $x=2$ $\endgroup$ – R_D Nov 19 '15 at 16:57
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There is something wrong with this proposition in Rotman. It is false whenever $R$ has more than one minimal prime, for example.

In general, the nilradical of a ring is the intersection of all prime ideals. It follows that a prime ideal can only be contained in the nilradical if there is a unique minimal prime ideal.

The text attempts to get around this by localization, but, as you have observed, the localization will not generally be injective, and so elements may be nilpotent in $R_\mathfrak{p}$ without being nilpotent in $R$.


What we should conclude instead is that every element $x\in\mathfrak{p}$ of a minimal prime ideal is a zero divisor. We can do that as follows:

First, observe that $x/1$ is nilpotent in $R_\mathfrak{p}$. Rotman's argument shows this, and it also follows from the fact that $\mathfrak{p}R_\mathfrak{p}$ is the unique prime, hence the nilradical, of $R_\mathfrak{p}$.

So $x^n / 1 = 0$ for some $n$, in other words $x^n$ is in the kernel of the localization map $R\to R_\mathfrak{p}$. As you mentioned, this means that there is some $s\in R\setminus\mathfrak{p}$ such that $sx^n = 0$. Since $s\neq 0$, we can conclude that $x$ is a zero divisor.

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  • $\begingroup$ Thanks a lot Slade!!! Your answer relieves my problem a lot. I really appreciate your help. I studied nothing about nilradical. So one more question. If we change "minimal prime ideal" to "unique minimal prime ideal", dose it fix the proposition? $\endgroup$ – Deepleeqe Nov 19 '15 at 18:36
  • $\begingroup$ @Deepleeqe Not quite, since there could be a unique minimal prime $\mathfrak{p}$ and some other prime ideal $\mathfrak{q}$ not containing $\mathfrak{p}$, as a prime need not contain a minimal prime. The correct condition is that $\mathfrak{p}$ should be a prime ideal contained in every other prime ideal. In fact, any element contained in every prime ideal is nilpotent, which is the usual characterization of the nilradical. $\endgroup$ – Slade Nov 19 '15 at 19:04
  • $\begingroup$ @Deepleeqe We could say it like this: This proposition is true exactly when $\mathfrak{p}$ is a minimum prime, in which case $\mathfrak{p}$ is exactly the set of nilpotent elements. Conversely, if the set of nilpotent elements is a prime ideal, then it is a minimum prime ideal. $\endgroup$ – Slade Nov 19 '15 at 21:50
  • $\begingroup$ @Slade "as a prime need not contain a minimal prime" Please check Reid, Undergraduate Commutative Algebra, Exercise 1.18. $\endgroup$ – user26857 Nov 19 '15 at 22:11
  • $\begingroup$ @user26857 You're absolutely right, somehow I was thinking that the intersection of prime ideals needn't be prime. So we can say that this is true iff $\mathfrak{p}$ is the unique minimal prime, in which case $\mathfrak{p}$ is also a minimum prime. $\endgroup$ – Slade Nov 19 '15 at 22:17
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It would even be true for any subset of $R$ if $R$ were an integral domain. If $R$ is not an integral domain it's not true. For a counterexample take $\mathbb{Z}/6$; the ideal generated by $3$ is prime (since the quotient by this ideal is an integral domain) and minimal (there are only 2 elements). Now $2 \notin (3)$ but $2 \cdot 3 = 0$

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