The main question I want to eventually wind up answer is part b of Corollary 2.6 in Chapter 1 of Silverman's "The Arithmetic of Elliptic Curves". The proposition is

Let $V$ be a projective variety. Then $V \cap \mathbb{A}^n$ is an affine variety and either $V \cap \mathbb{A}^n = \emptyset$ or $V = \overline{V \cap \mathbb{A}^n}$ where the bar denotes a projective closure.

Silverman claims this follows directly from Corollary 2.4 in Hartshorne, but I don't see it. I think I have a proof of this statement, using information from the proof in the aforementioned corollary in Hartshorne, and using Silverman's claim that for a projective algebraic set $V$, $V \cap \mathbb{A}^n$ is an affine algebraic set

$\mathcal{I}_a(V \cap \mathbb{A}^n) = \alpha_i(\mathcal{I}_p(V))$

where $\alpha$ denotes the dehomogenization map with respect to the variable $X_i$, and $\mathcal{I}_a$ and $\mathcal{I}_p$ are the affine ideal and projective ideal of the sets of points $V \cap \mathbb{A}^n$ and $V$ respectively (though I don't see why exactly this is true either in the case of a projective algebraic set). However, when I try to prove this claim, I wind up having to use part 2 of corollary 2.6, or else I get stuck. Here is essentially what I have:

The inclusion $\alpha_i(\mathcal{I}_p(V)) \subseteq \mathcal{I}_a(V \cap \mathbb{A}^n) = \mathcal{I}_a(\varphi_i(V \cap U_i))$ is easy ($\varphi_i$ is the homeomorphism that identifies $\mathbb{A}^n$ with each open subset $U_i$ of $\mathbb{P}^n$). Now if I take some $f \in \mathcal{I}_a(\varphi_i(V \cap U_i))$, then one can show that the homogenization of $f$ with respect to $X_i$, call it $F_i$, vanishes on every point $Q \in V \cap U_i$. However, how can one prove that this implies that $F_i$ also vanishes at every $P \in V$? I know that the only place where we potentially would have a problem would be $V \cap H_i$, where $H_i = Z_p(X_I)$, which is a Zariski closed subset. However I can't see any reason why I might not wind up with a polynomial $f$ whose homogenization might not disappear on that closed set. I've only seen one proof of this fact and the proof doesn't even make mention of this and says that the preceding statement, that $F_i(Q) = 0$ for all $Q \in V \cap U_i$ suffices to show that $F \in \mathcal{I}_p(V)$ and I'm just not seeing it. I think I must be missing something incredibly basic or something because I can't see this being that difficult.

Now, I believe I can show this using topological arguments maybe easier than algebraic arguments, but for what I'm doing I'd really like to be able to show this purely algebraically. Also I would like to stick to classical methods as well as I'm not particularly well versed in the language of schemes and sheaves. Thanks in advance to anyone that can help clear this up.


1 Answer 1


I hope the following is what you are looking for in a classical and purely algebraic argument. If you don't like it, I'll delete it. It just really didn't fit in a comment.

Let $A$ be the homogeneous coordinate ring of $V$. It is a positively graded $\Bbbk$-algebra which is a quotient of $\Bbbk[X_0,\ldots,X_n]$ by the homogeneous ideal $I=\mathcal I_p(V)$, where $\Bbbk$ is your base field. Denote by $x_i$ the image of $X_i$ in $A$. Then, the affine coordinate coordinate ring of $V\cap D_+(X_i)$ is the ring $(A_{x_i})_0$, and $U_i:=D_+(X_i)\cong \mathbb A^n$ is the $i$-th standard embedding of affine space. Let us assume wlog that $i=0$ satisfies $V\cap U_0\ne \emptyset$.

Set $Y_i:=\frac{X_i}{X_0}$ and let $J:=(I_{X_0})_0=(\Bbbk[X_0,\ldots,X_n]_{X_0}\otimes I)_0$. Since localization is flat, $I_{X_0}$ is a homogeneous ideal of $\Bbbk[X_0,\ldots,X_n]_{X_0}$ and since the degree $0$ component of a graded ring is a subring, $J$ is an ideal of $(\Bbbk[X_0,\ldots,X_n]_{X_0})_0=\Bbbk[Y_1,\ldots,Y_n]$. It is the dehomogenization of $I$. Finally, we claim that


Indeed, define a map $\phi\colon\Bbbk[Y_1,\ldots,Y_n]\to (A_{x_0})_0$ by mapping $Y_i$ to $\frac{x_i}{x_0}$. This morphism is surjective. For a polynomial $f\in\Bbbk[Y_1,\ldots,Y_n]$, assume that $f(\frac{x_1}{x_0},\ldots,\frac{x_n}{x_0})=\phi(f)=0$. Let $F=X_0^{\deg(f)}\cdot f\in\Bbbk[X_0,\ldots,X_n]$ be the homogenization of $f$, then we must also have $F(x_0,\ldots,x_n)=x_0^{\deg(f)}\cdot\phi(f)=0$. This, however, is the case if and only if $F\in I$, which is the case if and only if $f=X_0^{-\deg(f)}\cdot F\in J$.

Hence, we have $\ker(\phi)=J$ and the claim follows from the universal property of the quotient. Now this proves that $J$ is the vanishing ideal of $V\cap U_0$.

  • $\begingroup$ Holy smokes, thank you! This is more or less exactly what I was looking for, though apparently my question is a bit trickier than I thought lol. $\endgroup$
    – user169368
    Commented Aug 13, 2014 at 17:48
  • 1
    $\begingroup$ So I do have one more question here. Silverman claims that $\mathcal{I}_a(\varphi(V \cap U)) = \alpha(\mathcal{I}_p(V))$ for any projective algebraic set $V$, not just varieties. Then can we use this to show that $\overline{(V \cap \mathbb{A}^n)} = \overline{\varphi^{-1}(\varphi(V \cap U))} = \overline{V \cap U} = V$ for any projective algebraic set $V$? This seems like a very strong statement for any projective algebraic set, though I imagine it should be right because in your argument, I don't see you relying on $V$ being a variety anywhere. $\endgroup$
    – user169368
    Commented Aug 14, 2014 at 1:03
  • $\begingroup$ If a projective algebraic set is just the vanishing of any homogeneous ideal (i.e. the vanishing of a certain number of homogeneous polynomials) then yes, this is in no way dependent on the fact that $V$ is irreducible. $\endgroup$ Commented Aug 14, 2014 at 6:50
  • $\begingroup$ That said, this does not feel too strong at all: A projective algebraic set is just a finite union of varieties, so you could easily deduct the statement topologically for projective algebraic sets if you knew it was true for varieties. $\endgroup$ Commented Aug 14, 2014 at 7:09
  • $\begingroup$ Good point. Cheers, thanks again for your help! $\endgroup$
    – user169368
    Commented Aug 14, 2014 at 8:40

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