Chapter 1, Exercise 2.4: The projective fundamental correspondence (part 2)

(a) There is a 1-1 inclusion-reversing correspondence between algebraic sets in $\mathbb P^n$, and homogeneous radical ideals of $S$ not equal to $S_+$, given by $Y\mapsto I(Y)$ and $\mathfrak a\mapsto Z(\mathfrak a)$. Note: Since $S_+$ does not occur in this correspondence, it is sometimes called the irrelevant maximal ideal of $S$.

(b) An algebraic set $Y\subseteq\mathbb P^n$ is irreducible if and only if $I(Y)$ is a prime ideal.

(c) Show that $\mathbb P^n$ itself is irreducible.

(a) We know that both directions are inclusion-reversing by the preceding exercise. Thus we must show that both compositions of $I$ and $Z$ do nothing to their argument. For example, if $Y$ is algebraic (closed) then $Z(I(Y))=\bar Y = Y$. Similarly, if $\mathfrak a$ is a homogeneous radical ideal with $Z(\mathfrak a)\neq\emptyset$, then $I(Z(\mathfrak a))=\sqrt{\mathfrak a}=\mathfrak a$. Note that in the case $Z(\mathfrak a)=\emptyset$, since $\mathfrak a$ is radical we have either $\mathfrak a=S$ or $\mathfrak a=S_+$. The second case is omitted by hypothesis, and thus $S$ uniquely corresponds to $\emptyset$, completing the bijection.

(b) If $Y$ is irreducible and $f,g\in S^h,\,fg\in I(Y)$, then $(fg)\subseteq I(Y)$ so $Z(f)\cup Z(g)=Z(fg)\supseteq Y$. Decompose $Y$ as $Y=(Y\cap Z(f))\cup(Y\cap Z(g))$, both of which are closed, so $Y=Y\cap Z(f)$ or $Y=Y\cap Z(g)$. In the first case $Z(f)\supseteq Y$, so $f\in\sqrt{I(Y)}=I(Y)$. In the other case $g\in I(Y)$, so $I(Y)$ is prime.

Conversely, if $I(Y)$ is prime and $Y=Y_1\cup Y_2$, apply ideals to get $I(Y)=I(Y_1)\cap I(Y_2)$. Thus $I(Y)=I(Y_1)$ or $I(Y)=I(Y_2)$. In either case $Y=Y_i$ for one of the two $i$, so $Y$ is irreducible.

(c) We have $I(\mathbb P^n)=(0)$, which is prime, so $\mathbb P^n$ is irreducible.