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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?

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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 $I$.

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