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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?

user26857
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Makoto Kato
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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
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    The answer I gave is similar except with details. Besides, it also proves it for $\mathbb{Z}$ graded rings. I see Sanchez's answer is fine for $\mathbb{N}$ graded rings. – rschwieb Nov 16 '12 at 13:55
  • @rschwieb Let $S = \mathbb{Z}[x]$ be the polynomial ring over $\mathbb{Z}$. The OP of the other thread means by monomials elements of the form $x^a$. So, for example, $(2x)$ is a homogeneous ideal, but not a monomial ideal of $S$. – Makoto Kato Nov 16 '12 at 21:41
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    @rschwieb: Sanchez's answer (which I presume is the answer which is currently selected) also works for $\mathbb{Z}$-graded rings, since any given element has only finitely many homogenous components. – Ingo Blechschmidt Aug 16 '16 at 13:02
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    @IngoBlechschmidt Four years later, I have no idea why I thought it didn't work for $\Bbb Z$-grading. Maybe I thought there was something about powers of elements with negative grades that could interfere. I agree with you. \ (ツ)/¯ – rschwieb Aug 16 '16 at 13:10

2 Answers2

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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 $J$.

  • Could you please explain why $x_{k}^{n}\in I$? I think that $x^{n}=\sum_{i}u_{i}$, $u_{i}$ has degree i then compare the highest degree of this component to $x_{k}^{n}$, they have to be equal. Is that true? – Soulostar Sep 20 '17 at 05:42
  • For people who will chance upon this in the future: yes. Since $x^n_k$ is the degree-$n$ homogeneous part of $x^n \in I$, by the definition of a homogeneous ideal, $x^n_k \in I$. – Soham Chowdhury Sep 03 '18 at 07:23
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    @SohamChowdhury $x_k^n$ is homogeneous of degree-$nk$, no? – Abhiram Natarajan Apr 17 '23 at 12:12
  • This proof isn't really constructive, since what we get from $x-x_k\in J$ is that there exists some new $\ell$ such that $(x-x_k)^\ell\in I$. – ZKe Feb 06 '24 at 20:36
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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.

Keenan Kidwell
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