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Let $A$ be a $\mathbb{Z}^{\geq0}$ graded ring, and $f$ an element of positive degree. It is well known that the homogenous prime ideals of $A_f$ are in bijection with prime ideals of $(A_f)_0$, that is the prime ideals of the sub ring of degree zero elements in $A_f$. The way I know how to show this bijection is outlined in Vakil's notes, and makes sense to me.

However, in this question, one of the answers states that the way Vakil goes about it is perhaps not the best way to think about it. Indeed, the answer states the better way to do this is to take $\mathfrak{p}_0\subset (A_f)_0$ and map it to $\sqrt{\mathfrak{p}_0 A_f}$. Now I interpret this notation to mean that we take $\mathfrak{p}_0$ as a subset of $A_f$, define a new ideal $I$ generated by $\mathfrak{p}_0$ over $A_f$, and take it's radical.

Now, I think it should be clear that if $\iota:(A_f)_0\rightarrow A_f$ is the inclusion map, then $\iota^{-1}(\sqrt{\mathfrak{p}_0A_f})=\mathfrak{p}_0$, as clearly $\mathfrak{p}_0\subset \iota^{-1}(\sqrt{\mathfrak{p}_0A_f})$, and if $a/f^k\in \iota^{-1}(\sqrt{\mathfrak{p}_0A_f})$, then there is some $r$ such that $a^r/f^{kr}\in \mathfrak{p}_0A_f$. Since $a^r/f^{kr}$ is degree $0$, we must have that $a^r/f^{kr}\in \mathfrak (p_0 A_f)_0=\mathfrak{p}_0$, hence $a/f^k\in \mathfrak{p}_0$ as $\mathfrak{p}_0$ is prime. It follows that $\iota^{-1}(\sqrt{\mathfrak p_0}A_f)=\mathfrak{p}_0$.

Moreover, since $\mathfrak{p}_0A_f$ is generated by degree zero elements, it must be a homogenous ideal, so $\sqrt{\mathfrak{p}_0A_f}$ is also homogenous. However, what I struggle with is showing that this is actually a prime ideal, and I can't seem to make any headway on showing this.

So, if $\mathfrak{p}_0\subset (A_f)_0$ is prime, how do we show that $\sqrt{\mathfrak{p}_0A_f}$ is prime?

Chris
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1 Answers1

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To save some typing, I will write $I={\mathfrak p}_0 A_f$ and $J=\sqrt{{\mathfrak p}_0 A_f}$. As you said, both are homogeneous ideals of $A_f$. We want to show that $J$ is prime if ${\mathfrak p}_0\subset (A_f)_0$ is prime.

Since $J$ is homogeneous, we only need to show if $a\in (A_f)_k$ and $b\in (A_f)_l$ are homogenous, we have $ab\in J$ implies that $a\in J$ or $b\in J$. (See for example Showing that a homogenous ideal is prime.)

By the definition of radical, $\exists r\geq 1$ such that $(ab)^r\in I$. Let $j=\deg f\geq 1$. Then $$ (ab)^{jr}/f^{(k+l)r}\in I\cap (A_f)_0={\mathfrak p}_0, $$ since its degree is 0, and $(ab)^r$ divides it. Also, clearly $({\mathfrak p}_0A_f)\cap (A_f)_0={\mathfrak p}_0$ by degree consideration and using ${\mathfrak p}_0\subset (A_f)_0$ is an ideal.

Now $$ (ab)^{jr}/f^{(k+l)r} = a^{jr}/f^{kr} \cdot b^{jr}/f^{lr} \in {\mathfrak p}_0, $$ and both $a^{jr}/f^{kr}$ and $b^{jr}/f^{lr}$ belong to $(A_f)_0$, so by ${\mathfrak p}_0$ is a prime ideal in $(A_f)_0$, we see one of them belongs to ${\mathfrak p}_0$.

Say, $a^{jr}/f^{kr}\in {\mathfrak p}_0$, then $a^{jr}\in {\mathfrak p}_0A_f=I$, and so $a\in \sqrt{{\mathfrak p}_0 A_f}=J$. We are done.

I do want to give some additional remarks.

  1. In the original document of Vakil, he defined this ideal as $$ P = \oplus Q_i, $$ and $a\in Q_i$ iff $a^{j}/f^i\in {\mathfrak p}_0$. (Recall $j=\deg f$.)

It can be worked out that $$ P = J = \sqrt{{\mathfrak p}_0 A_f}. $$ So Vakil just gave a more concrete description of this more abstract version.

  1. It seems if $j=\deg f=1$, then we don't need the radical. $I={\mathfrak p}_0 A_f$ would be prime already. For then we don't need to raise things to the degree of $f$, and so $a/f^i\in (A_f)_0$ if $a\in (A_f)_i$. (In general we needed $a^j/f^i\in (A_f)_0$.)
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