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I'm working on a problem in Sharp's Steps in Commutative Algebra, to be precise exercise 4.28 which states the following:

Let $K$ be a field and $R = K[X,Y]$ be the polynomial ring in the indeterminates $X$ and $Y$ over $K$. Let $I = (X^{3},XY)$.

(i) Show that the ideal $J_{n}:= (X^{3},XY,Y^{n})$ is primary for each $n \in \mathbb{N}$.

(ii) Show that $I = (X)\cap (X^{3},Y)$ is a minimal primary decomposition of $I$.

(iii) Construct infinitely many different primary decompositions of $I$.

I don't have trouble sorting (ii) out as I found this quite easy. I do however struggle to show that $(X^{3},XY,Y^{n})$ is primary. I have tried doing it directly and also to look at the quotient ring $R/J_{n}$ to show that every zero divisor is nilpotent. I also have not been able to explicitly find the radical of $J_{n}$ in general. Any help will be appreciated, thanks in advance.

user26857
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sqtrat
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3 Answers3

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The way I thought about this question goes like this:

i) Look at the quotient $R/J_n = K[X,Y,Z]/(X^3, XY,Y^n)$. It looks like $K$ adjoin elements $a,b$ with $a^3 = ab = b^n = 0$. Now $1, a,a^2,b,b^2,...,b^{n-1}$ form a $K$-basis for this algebra. Given an element in $R/J_n$ , either it has its constant coefficient (that of the basis element $1$) nonzero, in which case it's not a zerodivisor; or it has only terms in the powers of $a$ and $b$, which means it will be nilpotent. So all zerodivisors in the quotient are indeed nilpotent.

ii) You were okay with this bit: intersection of two primary ideals, neither contained in the other, different radicals. Cool cool.

iii) Just use part i). Given any $n$, $I = (X^3,XY) = (X)\cap(X^3,Y) = (X)\cap(X^3,Y)\cap (X^3,XY,Y)\cap...\cap(X^3,XY,Y^n) $ just intersecting with as many $J_n$ as you fancy, because they all contain $(X^3,Y)$ so can't affect what the intersection actually is. This gives infinitely-many primary decompositions, but of course they're not minimal. Hope I understood that part of the question right!

As for finding the radical of $J_n$, I'm looking for the smallest prime ideal $P$ containing $J_n$. So, the quotient $R/P$ must be an integral domain, in which all elements of $J_n$ from $R$ are zero. So, $X^3 = Y^n = 0$ implies $\bar{X}$ and $\bar{Y}$ are both zerodivisors, and so need to be zero. So, $P$ contains $X$ and $Y$. But of course, $R/(X,Y)$ is an integral domain - isomorphic to $K$ - so $P = (X,Y)$ must be the radical.

Hope this was helpful, in spite of the messiness.

Latimer Leviosa
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  • Please prove "$1, a,a^2,b,b^2,...,b^{n-1}$ form a $K$-basis for this algebra". – user26857 Apr 04 '15 at 12:01
  • They span the algebra over $K$, because the algebra is the $K$ span of products of powers of $a$ and $b$, and we're given $ab = 0$. Linear independence: no $K$ - linear combo of them is going to be zero, because the only polynomial relations the ideal $J_n$ gives us in $a$ and $b$ involve higher order terms. (Like I said, I suck at rigour, but I thought it was clearly enough a basis, unless I got something wrong.) – Latimer Leviosa Apr 04 '15 at 12:08
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Perhaps you are overthinking things, because finding this radical is not hard. The ideal contains $x^3$ and $y^n$, and therefore its radical, by definition, must contain $x$ and $y$. Can you see what the radical must be now?

Applying this lemma you will also see why the ideal is primary, and you are done.

Community
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rschwieb
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  • If "Finding the radical is exceptionally easy, in this case." then why don't leave the OP to find it? He was already told that the radical is $(X,Y)$, then why do you think he needs more assistance? – user26857 Apr 05 '15 at 08:05
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    Dear @user26857 then why don't leave the OP to find it? I think, in this case, that if the OP did not see what the radical is by now, then they are seriously stuck and need a straightforward answer. I recognize what you were trying to do by telling them the answer and explaining nothing, but IMO that strategy was poorly chosen in this case. Regards – rschwieb Apr 05 '15 at 11:35
  • I also softened the tone of that line now since it wasnt quite the message I was going for. – rschwieb Apr 05 '15 at 11:46
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you say "I also have not been able to explicitly find the radical of $J_n$". in fact if you solve this problem, (i) will be a result of Proposition 4.9 of the book. I prove this for $n\ge3$ (for $n\lt3 $ it is easy and similar). note that $(X,Y)^n\subset J_n\subset (X,Y)$. Taking radicals we have $(X,Y)\subset \sqrt {J_n}\subset (X,Y)$. so $\sqrt {J_n}= (X,Y)$.

(iii) $I=(X)\cap (X^{3},XY,Y^{n})$

user 1
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  • Thanks a lot for everybody's input, I have solved the entire question quite rigorously I must add, there is just one thing I'm wondering about, obviously the ideal $I$ can't be primary, otherwise that would be the minimal decomposition, and the radical of $I$ is the ideal generated by $X$. This leads me to the question that if $P \in Spec(R)$ and $\sqrt(I) = P$, is there a characterization in polynomial rings when $I$ will actually be primary? – sqtrat Apr 04 '15 at 12:43
  • you are welcome. you can accept one answer and ask this new question as a separate answer. (BTW. see for this new question Herzog-Hibi, 1.3) – user 1 Apr 04 '15 at 12:53