Chapter 1, Exercise 2.2: Homogeneous ideals with no zero set

For a homogeneous ideal $\mathfrak a\subseteq S$, show that the following conditions are equivalent:

• (i) $Z(\mathfrak a)=\emptyset$ (the empty set);
• (ii) $\sqrt{\mathfrak a}=$ either $S$ or the ideal $S_+=\bigoplus_{d>0}S_d$;
• (iii) $\mathfrak a\supseteq S_d$ for some $d>0$.

$(i)\implies(ii)$: Suppose $Z(\mathfrak a)=\emptyset$. From the previous proof we see that, in $\mathbb A^{n+1}$, either $Z(\mathfrak a)=\emptyset$ or $Z(\mathfrak a)=\{0\}$. In the first case, taking affine ideals yields $\mathfrak a=S$, so $\sqrt{\mathfrak a}=S$. In the second case, we find $\mathfrak a=I(\{0\})=S_+$, and $\sqrt{\mathfrak a}=S_+$.

$(ii)\implies(iii)$: If $\sqrt{\mathfrak a}=S$ then $1\in\sqrt{\mathfrak a}$, so $1\in\mathfrak a$ and $\mathfrak a=S$ so $\mathfrak a\supseteq S_d$ for any $d$. If $\sqrt{\mathfrak a}=S_+$ then we can find some $r>0$ such that $x_i^r\in\mathfrak a$ for all $i$. By the pigeonhole principle, each polynomial in $S_{r(n+1)}$ has some $x_i^r$ in every term, so $\mathfrak a\supseteq S_{r(n+1)}$.

$(iii)\implies(i)$: Suppose $\mathfrak a\supseteq S_d$ and $P\in Z(\mathfrak a)$. Then $P\in Z(S_d)$ and thus every homogeneous polynomial of degree $d>0$ vanishes at $P$. Since $P\neq(0,\dots,0)$ this is absurd.