Question On Intersections of Homogeneous Prime Ideals

Question

Let $ S $ be a graded ring, and let $ S_{+} $ denote the irrelevant ideal of $ S. $ And suppose that $ \mathfrak{a} $ is a homogeneous ideal of $ S $ contained in $ S_{+}. $

I am attempting to show that $ \bigcap_{\substack{\mathfrak{p} \supset \mathfrak{a} \\ \mathfrak{p} \in \text{Proj}(S)}}\mathfrak{p} = \bigcap_{\substack{\mathfrak{p} \supset \mathfrak{a} \\ \mathfrak{p} \in \text{Spec}(S)}}\mathfrak{p}. $

I see on the one hand that $ \bigcap_{\substack{\mathfrak{p} \supset \mathfrak{a} \\ \mathfrak{p} \in \text{Proj}(S)}}\mathfrak{p} \supset \bigcap_{\substack{\mathfrak{p} \supset \mathfrak{a} \\ \mathfrak{p} \in \text{Spec}(S)}}\mathfrak{p}. $ The reverse inclusion is proving more difficult to show.

Any hints would be appreciated.

EDIT: I have realised that the equality is not true in general.

Answer 1

You need to assume $\sqrt{\mathfrak a} \not \supseteq S_+$. For example, in $k[x]$ with $\mathfrak a = (x^2)$, this fails.

The intersection of all primes containing $\mathfrak a$ is equal the intersection of the minimal primes of $\mathfrak a$. If you show that the minimal primes of $\mathfrak a$ are homogeneous and don't contain $S_+$, then you will be done. To do this: suppose $\mathfrak a \subseteq \mathfrak p$ with $\mathfrak p$ prime; define the ideal $P\subseteq \mathfrak p$ to be generated by the homogeneous elements of $\mathfrak p$; since $\mathfrak a$ is homogeneous, $\mathfrak a \subseteq P$; show that $P$ is prime; thus all minimal primes of $\mathfrak a$ are homogeneous; show that none of them contain $S_+$ (notice that this must be true for some minimal - and thus also homogeneous - prime of $\mathfrak a$, since $\sqrt{\mathfrak a} \not \supseteq S_+$).

Answer 2

Let $S=k[x,y]$ with $x,y$ of degree $0,1$ respectively, and set $J=(xy)$. Then $J\subset S_+=(y)$, $x\notin \sqrt{J}$, $y\notin \sqrt J$. For any $p\in Z(J)\subset Proj S$, we have $y\notin p$, hence $x\in p$. We find that $x$ is in LHS but not RHS.