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

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 $$ 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?

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 $$x+1\not\in I$$.