Chapter 1, Exercise 2.3: The projective fundamental correspondence (part 1)

(a) If $T_1\subseteq T_2$ are subsets of $S^h$, then $Z(T_1)\supseteq Z(T_2)$.

(b) If $Y_1\subseteq Y_2$ are subsets of $\mathbb P^n$, then $I(Y_1)\supseteq I(Y_2)$.

(c) For any two subsets $Y_1,Y_2$ of $\mathbb P^n$, $I(Y_1\cup Y_2)=I(Y_1)\cap I(Y_2)$.

(d) If $\mathfrak a\subseteq S$ is a homogeneous ideal with $Z(\mathfrak a)\neq\emptyset$, then $I(Z(\mathfrak a))=\sqrt{\mathfrak a}$.

(e) For any subset $Y\subseteq\mathbb P^n$, $Z(I(Y))=\bar Y$.

(a) If $P\in Z(T_2)$ then all polynomials in $T_2$ vanish at $P$, which includes polynomials in $T_1$, so $P\in Z(T_1)$.

(b) If $f\in I(Y_2)$ then $f$ vanishes on $Y_2$, hence on $Y_1$, so $f\in I(Y_1)$.

(c) If $f\in I(Y_1\cup Y_2)$ then $f$ vanishes on all points of $Y_1$ and $Y_2$, hence $f\in I(Y_1)$ and $f\in I(Y_2)$. Thus $f\in I(Y_1)\cap I(Y_2)$. Now, if $f\in I(Y_1)$ and $f\in I(Y_2)$, then $f$ vanishes on points of $Y_1$ and $Y_2$, hence on $Y_1\cup Y_2$, hence $f\in I(Y_1\cap Y_2)$.

(d) The inclusion $\sqrt{\mathfrak a}\subseteq I(Z(\mathfrak a))$ is trivial. We know that any non-constant $f\in I(Z(\mathfrak a))$ lies in $\sqrt{\mathfrak a}$ by Exercise 2.1. If $f$ is constant, then either $f\equiv 0$ or $f\not\equiv 0$. In the first case $f\in\sqrt{\mathfrak a}$ vacuously (since $\sqrt{\mathfrak a}$ is an ideal). In the second, we find $Z(\mathfrak a)=\emptyset$ which is disallowed by hypothesis and thus unimportant.

(e) The same argument works in this case as in the proof of Proposition 1.2 in the book, and thus we need not repeat it here.