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Let $A$ be Noetherian ring that is an integral domain and let $\frak a$ be a proper ideal. I would like to know if it can happen that $\cap_{1}^{\infty}{\frak a}^n\ne 0$ and at the same time the sequence of ideals ${\frak a}^n$ does not stabilise? What would be a "natural example"?

(this question is motivated by trying to understand Krull intersection theorem).

PS. I was thinking of the following version of Krull intersection theorem:

Theorem. Let a $\frak a$ be be an ideal in a noetherian ring $A$. If $\frak a$ is contained in all maximal ideals of $A$, then $\cap_{1}^{\infty}{\frak a}^n=0$.

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  • $\begingroup$ Try with a product. $\endgroup$
    – Angelo
    Commented Oct 7, 2012 at 5:34
  • $\begingroup$ Product is often not integral? $\endgroup$
    – aglearner
    Commented Oct 7, 2012 at 5:37
  • $\begingroup$ To agleamer: sorry, I had not read your question carefully. As xbnv indicates, the answer to your question is negative. $\endgroup$
    – Angelo
    Commented Oct 7, 2012 at 5:57
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    $\begingroup$ The Krull intersection theorem gives the answer : if $x \in \cap \mathfrak{a}^n$ then $x \in x\mathfrak{a}$ so $(1+a)x=0$ with $a \in \mathfrak{a}$, thus $x=0$. $\endgroup$ Commented Oct 7, 2012 at 7:44
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    $\begingroup$ @aglearner : See for example Pete L. Clark's notes on commutative algebra, section 8.11.2 math.uga.edu/~pete/integral.pdf $\endgroup$ Commented Oct 7, 2012 at 15:28

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If $P$ is a prime ideal containing $a$ then $\cap a^n$ is contained in $\cap P^n$, whose image in $A_P$ vanishes, so $\cap a^n$ has vanishing image in $A_P$ for every point $P$ of Spec($A/a$). Such $P$ exist as long as $a$ isn't the unit ideal, so if $A$ is an integral domain then $\cap a^n = 0$ in $A$ since $A \rightarrow A_P$ is injective in such cases.

In contrast, for $A = k[x,y]/(xy) = k[x] \times_k k[y]$ for a field $k$, and $$a = (x-1)A = (x-1)k[x] \times_k k[y],$$ we have $\cap a^n = yk[y] \ne 0$ and $\{a^n\}$ doesn't stabilize. Thus, the "domain" condition cannot be relaxed to mere connectedness of the spectrum (even assuming reducedness).

[${\bf{Remark}}$: An earlier version of this answer had the following bogus "proof" that $\cap a^n = 0$ assuming that Spec($A$) is merely connected, and some of the comments refer to this bogus argument. So here it the incorrect proof of that incorrect generalization, with the mistake in the reasoning identified.

"Proof": Arguing as above, since $\cap a^n$ has vanishing image in the stalk $A_P$ at each point of Spec($A/a$), it vanishes over an open neighborhood $U$ of this closed set. On the other hand, the ideal sheaf $J = \cap (\widetilde{a})^n$ is the unit ideal on the open set $V$ of Spec($A$) complementary to the closed set Spec($A/a$). Since $\cap a^n$ is the ideal of global sections of $J$, if $J$ is quasi-coherent (equivalently, coherent) then $U$ and $V$ are disjoint open sets that cover the entire space, which would imply by connectedness that either $V$ is empty (in which case $\cap a^n = 0$ in $A$) or $U$ is empty (in which case $a$ is the unit ideal). "QED"

The error is that often the ideal sheaf $\cap (\widetilde{a})^n$ is not quasi-coherent (equivalently, isn't coherent), which is to say that it doesn't correspond to its ideal of global sections $\cap a^n$. This is an example of the fact that quasi-coherence can be destroyed by inverse limits, and more concretely inverse limits don't naturally commutes with tensor product or even just localization in general.

The failure of coherence for this ideal sheaf is seen rather concretely in the case $A = k[x,y]/(xy)$ with $a = (x-1)$ since the sheaf $\cap \widetilde{a}^n$ is the unit ideal sheaf away from the point $(1,0)$ whereas its ideal of global sections $\cap a^n = y k[y]$ is the ideal of the $x$-axis and so does not localize to the unit ideal at points $(c,0)$ with $c \ne 1$ (e.g., $c = 0$). This illustrates that the coherent ideal sheaf associated to $\cap a^n$ can be rather smaller than $\cap \widetilde{a}^n$.]

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  • $\begingroup$ xbnv, thank you for your answer. Sorry for a silly question, is this correct, that the spectrum of an integral domain is always connected? $\endgroup$
    – aglearner
    Commented Oct 7, 2012 at 5:52
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    $\begingroup$ @aglearner: Yes, it is even irreducible. If there are even two disjoint non-empty open sets then we can shrink them to be basic affine open, say for $A_f$ and $A_g$ with nonzero $f,g \in A$. The empty overlap corresponds to $A_{fg}$, so $A_{fg}=0$, absurd for a domain $A$. $\endgroup$
    – user27056
    Commented Oct 7, 2012 at 6:39
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    $\begingroup$ $Spec(A)$ is connected if and only if the only idempotents in $A$ are $0$ and $1$. $\endgroup$
    – user91132
    Commented Oct 7, 2012 at 10:00
  • $\begingroup$ Equivalently (for Noetherian rings $A$), $Spec(A)$ is connected if and only if $A$ is not the product of two proper subrings. $\endgroup$ Commented Oct 14, 2012 at 15:58
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    $\begingroup$ @Neil: in a product situation $A = A_1 \times A_2$ for nonzero commutative rings with identity $A_1$ and $A_2$, the $A_i$ are naturally quotients of $A$, not subrings (since the inclusions via $x \mapsto (x,0)$ and $y \mapsto (0,y)$ do not respect the multiplicative identity). Geometrically this corresponds to the fact that the disjoint union of the Spec($A_i$) does not naturally map to either of these Spec's (where would the points of the other Spec go?). $\endgroup$
    – user27056
    Commented Oct 16, 2012 at 12:01

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