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In the proof of Lemma $3.36$ in Algebraic Geometry and Arithmetic Curves, it is stated that, if $B=\oplus_{d\ge0}B_d$ is a graded algebra over a ring $A,$ and if $I$ is an ideal of $B,$ then $$V(I)\cap\operatorname{Proj}(B)=V_+(I^h),$$ where $I^h=\oplus_{d\ge0}I\cap B_d$ is the homogenized ideal of $I,$ thus the topology on $\operatorname{Proj} B$ is induced from that on $\operatorname{Spec}(B).$ The containment of the left side in the right one being clear, I don't see why this is an equality.

I have no idea how can one be sure that a prime ideal in $\operatorname{Proj}B$ contains $I$ iff it contains $I^h.$
Any hint or reference is well welcomed.

Edit: I thought that this came from the equality $$\sqrt I=\sqrt{I^h},$$ which is false, thanks to a comment by @user121097.
I changed the question according to the quoted comment, sorry for this.

P.S.: The title does not match the question exactly. Apology again.

Community
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awllower
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  • You are correct! I changed the question in accordance. Apology here. – awllower Feb 27 '14 at 15:35
  • awllower: I believe you are reading the first edition of the book. As pointed out by @user121097, the claim is not in the new edition. So you should be able to find the change in the errata. – Cantlog Feb 27 '14 at 20:27
  • @Cantlog Thanks for your pointer: I found it in the Errata indeed! Maybe you can post an answer stating this is false, and can be found in Errata, etc.? Thanks very much! – awllower Feb 28 '14 at 09:08
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    You can also find this at Bosch's "Algebraic Geometry and Commutative Algebra" p.408, but sadly no proof. It also disappeared in Liu's second edition. – user123454321 Jan 09 '15 at 11:21
  • @GYC Thanks for the attention, I shall have a look at the book afterwards. – awllower Jan 09 '15 at 13:46

3 Answers3

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$$V(I)\cap\operatorname{Proj}(B)=V_+(I^h)$$

This is false. Consider $I=(X+1)$ in $B=K[X,Y]$. Then $I^h=(0)$, so $V_+(I^h)=\operatorname{Proj}B$. On the other side, we have $V(I)\cap\operatorname{Proj}B=\emptyset$.

user26857
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So more importantly, if $f \in S$, write $f = f_{0} + f_{1} + \cdots + f_{n}$. Given $\mathfrak{p} \in \text{Proj}S$, we have $\mathfrak{p} \ni f$ if and only if $\mathfrak{p} \ni f_{0}, \cdots, f_{n}$. Thus $V(f) \cap \text{Proj}S = V_{+}(f_{0}) \cap \cdots \cap V_{+}(f_{n})$, and the topology is inherited!

user123454321
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Edit: As pointed out in the comments below, I was not using the correct definition of $I^h$. I do recall working with this construction some time ago: it is actually the largest homogeneous ideal contained in $I$. An interesting fact that may be useful is that, if $P$ is prime in $B$, then $P^h$ is a homogeneous prime ideal. If a solution to the actual question comes to me, I will of course let you know.

Old, incorrect answer:

This is clear once you realize that $I^h$ is the smallest homogeneous ideal containing $I$. Thus if $P$ is a homogeneous ideal, then $I \subseteq P \iff I^h \subseteq P$ (the reverse implication following from $I \subseteq I^h$). In my opinion, the equality follows easily from this.

To explicitly show that $V_+(I^h) \subseteq V(I)\cap\operatorname{Proj}(B)$ as you want, suppose that $P \in \operatorname{Proj}(B)$ with $I^h \subseteq P$. Then $I \subseteq I^h \subseteq P$, so $P \in V(I) \cap \operatorname{Proj}(B)$.

Manny Reyes
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  • Are you sure $I^h$ is the smallest homogeneous ideal containing $I$? From the definition $I^h=\oplus_{d\ge0}(I\cap B_d)$ this does not seem to be the case (and I have only ever heard of $I^h$ as being generated by the homogeneous elements contained in $I$) – zcn Feb 27 '14 at 19:42
  • @user115654, ah, I see that we are working with different definitions of $I^h$. I was thinking of the ideal generated by all homogeneous parts of all elements in $I$; call this $I'$ instead. Under this definition, it's clear that $I \subseteq I'$. I apologize for the confusion. – Manny Reyes Feb 27 '14 at 21:41
  • I thought that $I^h$ is the smallest homogeneous ideal containing $I$ as well, before I realised that $I^h\subset I.$ Still thanks for the response. – awllower Feb 28 '14 at 08:50