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Suppose $A$ is a commutative ring with $1\neq0$ satisfying the property that $A_\mathbf{m}$ has no nonzero nilpotent elements for any maximal ideal $\mathbf{m}$, where $$A_\mathbf{m}=S^{-1}A\quad \text{ and }\quad S=A-\mathbf{m}.$$ Prove that $A$ has no nonzero nilpotent elements.

My attempt at the solution was to prove the contrapositive, i.e., suppose $A$ has a nonzero nilpotent element $x$. Then I would like to show that $A_\mathbf{m}$ has a nonzero nilpotent element for some maximal ideal. When I'm using the definition for the equivalence class of fractions, i.e., $$\frac{a}{b}=\frac{a'}{b'}\quad \text{if and only if}\quad (ab'-a'b)u=0$$ for some $u\in S$, I'm having trouble showing that $\frac{x}{s}\neq\frac{0}{1}$. This is because the definition reduces us to $x\cdot u=0$, and it very well seems that $u$ could be the element that makes $x$ a zero divisor. Is there some way to construct an ideal that contains no zero divisors of $x$? My original thought was to take $\mathbf{m}$ to be the maximal ideal containing all of the zero divisors of $x$, but the smallest ideal containing all of the zero divisors very well could be the whole ring (as in the case of $\mathbb{Z}/15\mathbb{Z}$).

Any help or ideas regarding the proof would be very helpful!

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It might be easier to do it by contradiction, i.e. suppose $A$ has a nonzero nilpotent $x$. This $x$ would be 0 in all $A_m$. Can you conclude that $x$ must then be 0 in $A$? –  user27126 Dec 5 '12 at 22:32
    
I'll think about that for a few minutes and see if I can come up with anything useful from it... Thanks for the suggestion! –  Clayton Dec 5 '12 at 22:35
    
Do you know that the nilradical of the localization is actually $S^{-1}\mathfrak{R}$? –  user38268 Dec 5 '12 at 23:23
    
@BenjaLim, I'm not familiar with nilradicals at all, so I can't say that I do know that information... –  Clayton Dec 5 '12 at 23:29
    
@Clayton It is a basic fact in commutative algebra that for any ring $A$, the set of all nilpotent elements (the nilradical) which is denoted $\mathfrak{R}$ is the intersection of all prime ideals in $A$. –  user38268 Dec 5 '12 at 23:35
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2 Answers

up vote 4 down vote accepted

Suppose $x$ is a non-zero nilpotent of $A$ like you do. You need a maximal ideal $m$ such that $x/1$ is non-zero in $A_m$. Define the ideal $I = \{s \in A : sx = 0\}$ (show this is an ideal). Now assume that $x/1$ is zero in $A_m$ for all maximal ideals $m$. What can you deduce about the ideal $I$ then? Hence what must $x$ be? Is this a contradiction?

What you actually need to prove, as @Sanchez points out, is that given a ring $A$, if $x \in A$ is nonzero, then $x/1$ must be nonzero in $A_m$ for some maximal ideal $m$.

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Also, you can frame this proof quite easily without contradiction. –  Rankeya Dec 5 '12 at 23:08
    
Okay, so it's easy to show this is an ideal, but I'm not sure what to deduce about $I$. I'm guessing the punchline is that $x$ was zero in the first place, which would imply that $A$ has no nonzero nilpotent elements, but I don't quite see it yet. –  Clayton Dec 5 '12 at 23:17
    
I feel like perhaps $I$ is trivial, and there $x=0$, which would be nice, but I just don't see it yet...? –  Clayton Dec 5 '12 at 23:27
    
Hint: $I$ is not contained in any maximal ideal if $x$ is zero in $A_m$ for all maximal ideals $m$. –  Rankeya Dec 5 '12 at 23:31
    
Okay, you need to know that any proper ideal is contained in a maximal ideal. So, I hope you are aware of this fact. –  Rankeya Dec 5 '12 at 23:34
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This is more a long comment to Rankeya's answer above than an answer. We recall more generally the following equivalences:

Proposition 3.8 (Atiyah - Macdonald): Let $M$ be an $A$ - module. Then the following are equivalent:

  1. $M = 0$

  2. $M_\mathfrak{p} = 0$ for all prime ideals $\mathfrak{p}$.

  3. $M_\mathfrak{m} = 0$ for all maximal ideals $\mathfrak{m}$.

Your problem is then solved by applying the proposition with $M = \mathfrak{R}$ the nilradical of $A$ and noting that the nilradical of $A_\mathfrak{m}$ is in fact $M_\mathfrak{m}$.

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