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In Atiyah-MacDonald Ch. 4 Problem 10, we have an arbitrary ring $A$ and various prime ideals $\mathfrak{p}$ and we are considering the kernels of the localization maps $A\rightarrow A_\mathfrak{p}$, which Atiyah and MacDonald call $S_\mathfrak{p}(0)$. Problem 10d asks us to show that

$$\bigcap_{\mathfrak{p}\in D(A)} S_\mathfrak{p}(0) = 0$$

where $D(A)\subset \operatorname{Spec}A$ is the set of prime ideals that are minimal over the annihilators of elements.

I see a clear path to this result in the case that $(0 )$ has a primary decomposition, for instance if $A$ is noetherian, but I haven't seen how to get it in the generality in which the problem is stated. Can you help?

My work so far: Let $a\in \bigcap_{\mathfrak{p}\in D(A)} S_\mathfrak{p}(0)$. We have to show $a=0$; it would be sufficient to show that $a$'s annihilator contains a non-zero-divisor.

Now $S_\mathfrak{p}(0)$ is the set of elements of $A$ that are annihilated by any element of $A$ outside $\mathfrak{p}$. Then for each $\mathfrak{p}$ in $D(A)$, there exists $s_\mathfrak{p}\in A\setminus\mathfrak{p}$ such that $as_\mathfrak{p}=0$. Consider the ideal $I$ generated by all the $s_\mathfrak{p}$'s. By construction, $I$ is not contained in any of the primes $\mathfrak{p}\in D(A)$, since $s_\mathfrak{p}\notin \mathfrak{p}$. But $I$ is contained in $a$'s annihilator. By a previous problem (problem 9), the union of the prime ideals in $D(A)$ is precisely the set of zero-divisors of $A$; thus if we could conclude that $I$ is not contained in $\bigcup_{\mathfrak{p}\in D(A)}\mathfrak{p}$, we would have shown that the annihilator of $a$ contains a non-zero-divisor and so would be done.

Now if $D(A)$ is finite, the prime avoidance lemma gives us exactly what we need: $I$ can't be contained in $\bigcup_{\mathfrak{p}\in D(A)}\mathfrak{p}$ without being contained in some $\mathfrak{p}$, which is ruled out by the construction. QED. And if $(0)$ has a primary decomposition, then $D(A)$ is finite because by a previous problem (also problem 9), in this case it is exactly the set of primes belonging to $(0)$.

But if $D(A)$ is infinite, there is no prime avoidance lemma. In general, an ideal can sit in an infinite union of primes without sitting inside any of the primes individually.

So I need a new approach. Is there a way to use the minimality of the $\mathfrak{p}$'s over annihilators of elements to deduce that $I$ is not contained in their union? Or is the construction of $I$ already barking up the wrong tree?

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  • $\begingroup$ "By construction, $I$ is not contained in any of the primes $\mathfrak{p}$", yes, and therefore $1 \in A$ and then obviously $a1=0$. $\endgroup$ Sep 16, 2013 at 8:48
  • $\begingroup$ @MartinBrandenburg - not all primes, just those minimal over annihilators. I've edited the post to clarify. YACP's hint does completely take care of it though. $\endgroup$ Sep 16, 2013 at 13:54

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Hint. If $x\in\bigcap_{\mathfrak{p}\in D(A)} S_\mathfrak{p}(0)$, $x\neq 0$ take $\mathfrak p$ a minimal prime over $(0:x)$.

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  • $\begingroup$ Right. Brilliant. ;) $\endgroup$ Sep 16, 2013 at 1:09

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