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Let $S$ be an $M$-graded $R$-algebra, where $M$ is some monoid, and $I\subset S$ an homogeneous ideal. The original, naïve, question, was: is it true that $\sqrt{I}$ is homogeneous? In this generality, the answer is no (see comments). However, if $M$ is a cancellative monoid with a total order, then the usual proof for $M=\mathbb{N}$ works, and indeed $\sqrt{I}$ is homogeneous. So:

  • Is there a natural class of monoids $M$ (larger, or different, from totally ordered cancellative monoids) such that in every $M$-graded algebra the radical of a homogeneous ideal is homogeneous?
  • The same question, for fixed $R$. In darij grinberg's example, it is relevant that the characteristic of the ring is 2. So, given a ring $R$, is there a natural class of monoids such that in every $M$-graded $R$-algebra the radical of a homogeneous ideal is homogeneous?

I am assuming everything in sight is commutative.

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Grade the $\mathbb F_2$-algebra $\mathbb F_2\left[x\right]/\left(x^2-1\right)$ by $\mathbb Z / 2\mathbb Z$ giving $\overline x$ the degree $\overline 1$. Now, $\overline{1+x}$ lies in the radical of the zero ideal... – darij grinberg Nov 22 '11 at 17:58
So this is a complete answer. – Martin Brandenburg Nov 22 '11 at 18:16
So the question was far too naïve, and it should have been: are there natural conditions on $M$ that guarantee that the radical of a homogeneous ideal is homogeneous? Is it good form to modify the question here, or should I start a new question? Thanks. – quim Nov 22 '11 at 18:51
Yes. Let $N$ be the monoid formed by the nonnegative integers and an element $i$ such that $i+n=n+1$ for all $n>0$. Give to $K[x,y]$ the grading in which elements of $K$ have degree 0, $x$ has degree 1 and $y$ has degree $i$. Then $(x^2+2xy+y^2)$ is homogeneous of degree 2 and its radical $(x+y)$ is not homogeneous. $N$ is not cancellative. – quim Nov 22 '11 at 22:05
Excercise 3.5 in Eisenbud's commutative algebra book may be related. It asks, among other things, to show that all associated primes are homogeneous as soon as the monoid is totally ordered abelian and shows a counterexample if not. This feature is independent of the characteristic. However, in the radical the non-homogeneous minimal primes can 'fit together' again. – Thomas Kahle Nov 23 '11 at 16:18

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