# Showing the $V(\mathfrak{a})$ give us a topology on Proj$S$

I'm a bit confused about the proof of Lemma 2.4 on page 76 of Hartshorne's Algebraic Geometry:

Lemma 2.4

(a) If $\mathfrak{a}$ and $\mathfrak{b}$ are homogeneous ideals in $S$, then $V(\mathfrak{a}\mathfrak{b})=V(\mathfrak{a})\cup V(\mathfrak{b})$.

(b) If $\{ \mathfrak{a}_i\}$ is any family of homogeneous ideals of $S$, then $V\left(\sum\mathfrak{a}_i\right)=\cap V(\mathfrak{a}_i)$.

Proof

The proofs are the same as for (2.1a,b), taking into account the fact that a homogeneous ideal $\mathfrak{p}$ is prime if and only if for any two homogeneous elements $a,b \in S$, $ab \in \mathfrak{p}$ implies $a \in \mathfrak{p}$ or $b \in \mathfrak{p}$.

Now I don't see why we need to use this fact about homogeneous primes at all. Can't we just define $V'(\mathfrak{a})=\{\text{primes } \mathfrak{p} \text { of }S\;|\; \mathfrak{p}\supseteq \mathfrak{a}\}$, and then $V(\mathfrak{a})=\text{Proj }S \cap V'(\mathfrak{a})$, so by Lemma 2.1

$$\begin{array}{rll}V(\mathfrak{a}\mathfrak{b})&=&\text{Proj }S \cap V'(\mathfrak{a}\mathfrak{b}) \\ \text{(by 2.1)}&=&\text{Proj }S \cap \left( V'(\mathfrak{a}) \cup V'(\mathfrak{b}) \right) \\ &=&\left(\text{Proj }S \cap V'(\mathfrak{a})\right) \cup\left(\text{Proj }S \cap V'(\mathfrak{b}) \right) \\ &=& V(\mathfrak{a}) \cap V(\mathfrak{b})\end{array}$$

and similarly for (b)?

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Note we are talking about homogenous ideals here, not just ideals with a set of homogenous generators, so some care must be taken. –  Hagen von Eitzen Apr 26 '13 at 6:34
Hmm. I don't understand. On pg 92 of Matsumura's Commutative Ring Theory he writes "A submodule $N \subset M$ is called a homogeneous submodule if it can be generated by homogeneous elements." –  porkramen Apr 26 '13 at 7:15

For example, see (EGA, II, 2.3), where Grothendieck defines $V_+(E)$ as the intersection of $V(E)$ with $\mathrm{Proj}(S)$ and gets the "lemma" in question immediately. –  Adeel Khan Apr 26 '13 at 10:42