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Let $S = \sum_{n \in \mathbb{Z}} S_n$ be a commutative graded ring. Let $I$ be a homogeneous ideal of $S$. Let $J$ be the radical of $I$, i.e. $J = \{x \in S| x^n \in I$ for some $n > 0\}$. Is $J$ a homogeneous ideal?

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possible duplicate of the radical of a monomial ideal is also monomial The minor difference is that that post is concerned with the grading of polynomials. –  rschwieb Nov 15 '12 at 19:22
@rschwieb I don't see how to derive an answer for my question from the one you linked. –  Makoto Kato Nov 15 '12 at 20:32
To someone who voted to close, please explain the reason for the vote to close. –  Makoto Kato Nov 15 '12 at 20:40
What that person is calling a "monomial ideal" is just a homogeneous ideal in the graded ring of polynomials. –  rschwieb Nov 15 '12 at 20:41
@rschwieb But your answer is much longer than Sanchez'. –  Makoto Kato Nov 15 '12 at 20:46

2 Answers 2

up vote 10 down vote accepted

Yes. Let $x^n \in I$. Take the highest graded piece $x_k$ of $x$, then it's clear by looking at degrees that $x_k^n \in I$, i.e. $x_k \in J$. Now $x-x_k$ is again in $J$ so we can repeat the process to see that components of $x$ in each degree all lie in $J$.

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I will take for granted the following fact two facts:

(1) A homogeneous ideal $I$ in $S$ is prime if and only if for all homogeneous elements $a,b\in S$ with $ab\in I$, either $a\in I$ or $b\in I$.

(2) An intersection of homogeneous ideals is homogeneous (this is obvious from the characterization of homogeneity as the property of containing all the homogeneous components of all elements).

Now, for a prime ideal $\mathfrak{p}$, we let $\mathfrak{p}^h$ denote the ideal of $S$ generated by the homogeneous elements of $\mathfrak{p}$. It is visibly homogeneous, and it is prime by (1). The radical of an ideal $I$ is the intersection of the prime ideals containing $I$, and if $I$ is homogeneous, this coincides with the intersection of all homogeneous prime ideals containing $I$. It is clear that, regardless of the homogeneity of $I$, the first intersection is contained in the second. Conversely, suppose that $a\in S$ lies in every homogeneous prime ideal containing $I$. If $\mathfrak{p}$ is any prime containing $I$, then $\mathfrak{p}^h\subseteq\mathfrak{p}$ is a homogeneous prime which also contains $I$ (because it contains the homogeneous elements of $I$ and $I$ is generated by these elements by assumption). So $r\in\mathfrak{p}^h\subseteq\mathfrak{p}$. Thus $\sqrt{I}$ is homogeneous by (2).

Alternatively, if you're willing to invoke Zorn's lemma, which implies that every prime ideal is contained in a minimal prime ideal, then the radical of $I$ is the intersection of the prime ideals minimal over $I$, and since such a prime ideal $\mathfrak{p}$ contains $\mathfrak{p}^h$, also a prime containing $I$, every prime ideal minimal over a homogeneous ideal like $I$ must be homogeneous itself.

Actually, the characterization of the radical of $I$ as the intersection of prime ideals containing $I$ already uses Zorn's lemma, so one might as well use the shorter, second argument...unless you want to avoid Zorn altogether, in which case my answer is inadmissible.

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+1 This is nice. However, I prefer the Sanchez' answer. This is just my personal taste. –  Makoto Kato Nov 15 '12 at 21:07
Thanks. Just out of curiosity, by any chance is part of your preference due to the fact that my answer uses the axiom of choice? –  Keenan Kidwell Nov 15 '12 at 21:45
Sort of. Generally I prefer constructive proofs. –  Makoto Kato Nov 15 '12 at 21:49
Yeah. I like Sanchez' proof a lot too. –  Keenan Kidwell Nov 15 '12 at 23:02
I prefer this proof by far. Gives much more insight when one learns algebraic geometry. +1 –  Patrick Da Silva Sep 8 '14 at 20:08

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