# Passing from a set of generators to a set of homogeneous generators

Consider a graded ring $$R=R_0\oplus R_1\oplus\dots$$. Suppose the irrelevant ideal $$R_1\oplus R_2\oplus \dots$$ is finitely generated. How come we can assume that it is generated by a finite number of homogeneous elements? I'm not satisfied with the answer "by taking homogeneous parts". I have this example in mind: $$(x^2+x)$$ is strictly contained in the ideal generated by its homogeneous parts $$(x,x^2)$$. Of course in this case the irrelevant ideal of the polynomial ring isn't finitely generated, but why can't the same situation be true for finitely generated irrelevant ideals?

Let $$I$$ be the irrelevant ideal. Notice that if $$x\in I$$, then every homogeneous part of $$x$$ is in $$I$$. Indeed, by definition of $$I$$, every nonzero homogeneous part of $$x$$ must have positive degree, and thus be in $$I$$. So, we can indeed just take the homogeneous parts of a set of generators for $$I$$.
So to be clear, the claim here is not that you can always just take the homogeneous parts of a set of generators for an ideal to get a homogeneous set of generators for the same ideal. Rather, we can do this specifically for the irrelevant ideal because it has the special property that if $$x\in I$$, every homogeneous part of $$x$$ is also in $$I$$.