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(Vakil 5.5.R.) Let $A=\mathbb C[x,y]$. $Q_1=(y-x^2)$, $Q_2=(x-1,y-1)$ and $Q_3=(x-2,y-2)$. Let $$I=Q_1^{3}\cap Q_2^{15} \cap Q_3.$$ I want to show that the associated primes of $A/I$ are precisely $Q_1, Q_2, Q_3$.

It is easy to verify that the radical of the powers of a prime ideal is the prime itself. But I don't know how to show that this prime ideal is an associated prime of $A/I$ (an annihilator of an element of $A/I$, how to find it?).

On the other hand, I know every associated prime, as prime, must contain some $Q_i$. But why are there no other associated primes?

In respond to comments below:

$Q_1=ann((y-x^2)^2+I)$ doesn't seem to be true, because $$(y-x^2)((y-x^2)^2+I)\neq I, since~ (y-x^2)^3\not \in I$$

No One
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  • For your first question, what happens if you take $(y-x^2)^2$ or $(x-1)^{14}$ in this case? Can you see how they're annihilated by an element of $Q_1$ or $Q_2$, respectively? – KReiser Nov 19 '18 at 19:36
  • @KReiser I think it makes no difference whether it is $(y-x^2)^2$ or $(y-x^2)^3$, $(x-1)^{14}$ or $(x-1)^{15}$, is that right? – No One Nov 20 '18 at 03:15
  • It shows that $Q_i$ are associated primes by showing that each of the mentioned nonzero element in $A/I$ is annihilated by the corresponding prime. – KReiser Nov 20 '18 at 03:23
  • When have a primary decomposition of and ideal, then their radicals are the associated primes. (This is a standard fact in Commutative Algebra.) The only thing you have to show is that $Q_1^3$, $Q_2^{15}$, and $Q_3$ are primary. – user26857 Nov 20 '18 at 10:00
  • @user26857 I think the radicals are CALLED "associated primes of primary ideals. ". On the other hand, associated primes are defined to be annihilators of $A$ modules ($A/I$ in this case). My question is about why $Q_i$'s are annihilators of elements of $A/I$. – No One Nov 20 '18 at 18:16
  • @KReiser Oh... I probably misunderstood you. You meant polynomials instead of ideals (parenthesis could be ambiguous). – No One Nov 20 '18 at 18:22
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    @NoOne Let me reformulate: when have a primary decomposition of an ideal $I$, then their radicals are the associated primes of $I$ in the sense that these are the associated primes of the $A$-module $A/I$. (This is proved in every Commutative Algebra textbook.) – user26857 Nov 20 '18 at 18:39
  • @user26857 Sorry, I can't find it in A-M's commutative algebra book (at least in Chapter 4: primary decomposition). Could you show me where I can find it, or just add a proof below? – No One Nov 20 '18 at 19:37
  • @NoOne I was struggling with this too until I found this question. If you're still interested and you have access to Eisenbud's commutative algebra book, check out Theorem 3.10. Basically you just have to show that $Q_1^3$, $Q_2^{15}$ and $Q_3$ are "primary", which doesn't seem hard since $Q_2$ and $Q_3$ are maximal, and $Q_1$ is principle. Since I don't think Vakil has talked about primary decomposition at this point (i could be wrong), he probably has something more ad hoc in mind, but user26857's comment is the right generality. And I don't see how to do an ad hoc argument. – Tim kinsella May 08 '19 at 19:49
  • Oh he says as much in "Aside 5.5.13" just below the comment: he says primary ideals aren't used in the text. So I'm curious how he wants the reader to do this exercise. – Tim kinsella May 08 '19 at 19:55

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