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Chapter 1, section 2, question 4(b) in Algebraic Geometry says

An algebraic set $Y \subseteq \mathbb{P}^n$ is irreducible iff $I(Y)$ is a prime ideal.

I'm confused about the solution given in http://www.math.northwestern.edu/~jcutrone/Work/Hartshorne%20Algebraic%20Geometry%20Solutions.pdf

which says

Looking at the affine cone, this follows from Cor 1.4

I'm not quite sure what this means. In one direction, if $Y=Z(\mathfrak{a})$ for some homogenous, radical ideal $\mathfrak{a}$ then $I(Y)$ is the same as if we considered $Y$ to be in affine space. So if $I(Y)$ is prime, then considered in $\mathbb{A}^{n+1}$, $Z(\mathfrak{a})$ is irreducible, so $Z(\mathfrak{a})$ is certainly irreducible in projective space.

In the other direction, I don't see why $Z(\mathfrak{a})$ being irreducible in $\mathbb{P}^n$ means it is irreducible when considered in $\mathbb{A}^{n+1}$, but maybe I am thinking about this wrong. I'm happy that the proof of 1.4 applies to $\mathbb{P}^n$, but I'd like to understand how the result applies.

Kumquats
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1 Answers1

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For the other direction you don't need to consider the affine cone. The first paragraph of the proof of Cor. 1.4 applies word for word. That proof only uses properties that hold for both projective and affine algebraic sets, and these were proved in exercises #2.3 and #2.4 a)-c).

That solution set is very terse so it's unclear what they meant, but my guess is they considered the (irreducible $\Rightarrow$ prime) direction to be easier, and applied the affine cone trick to get (prime $\Rightarrow$ irreducible).

Zavosh
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  • The second paragraph applies word for word too, though? – Kumquats Sep 07 '13 at 12:43
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    Aside from being prime, $\mathfrak{p}$ has to be homogeneous and not equal to $S_+$ for it to make sense, so not exactly word for word, but close. – Zavosh Sep 22 '13 at 04:39