# Hartshorne, Chapter 1, Projective varieties, Question 4(b)

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.

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).
• 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 4:39