Sufficient condition for radicality of an homogeneous ideal: still true when grading over arbitrary monoid?

Question

Let $A$ be a graded commutative ring (if necessary, with unit). Suppose we have an homogeneous ideal $I\subset A$ that satisfies the property “for all homogeneous $f\in A$ such that $f^r\in I$ for some $r\geq 0$, it holds $f\in I$.” If $A$ is graded over the natural numbers, $I$ must be radical. Here I wrote a proof of this fact.

However, I wonder whether this is still true for graduation over more general monoids. If $A$ graded over $\mathbb{Z}$, then my proof can be adapted to show the result true. The induction should be done instead over the “absolute degree” of the element, that for $f=\sum_{i=-\infty}^\infty f_i\in A$, is defined to be $\operatorname{|deg|}f=\max\{|\deg f_i|:f_i\neq 0\}$. If $\operatorname{|deg|}f=d$, then from $f^r\in I$ one would deduce $f_d^r,f^r_{-d}\in I$, and so $f_d,f_{-d}\in I$. On the other hand, $g=f-(f_d+f_{-d})$ would be of absolute degree $<d$ and we could continue the proof as in the link.

Is there a general proof of this fact when $A$ is graded over an arbitrary monoid? Or are there counterexamples in the general case?

Answer 1

For a simple counterexample, consider $A=\mathbb{F}_2[x]$. This can be given a $\mathbb{Z}/2$-grading where $x$ has degree $1$, and the ideal $I=(x^2+1)$ is then homogeneous. The quotient $A/I$ has the property that no nonzero homogeneous element is nilpotent (the only nonzero homogeneous elements to check are $1$ and $x$), so $I$ satisfies your condition. However, $I$ is not radical, since $(x+1)^2=x^2+1\in I$ but $x+1\not\in I$.