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I am trying to solve the following exercise:

Let $\mathcal{a}$ be a homogeneous ideal such that $\mathcal a \subset S = K[x_0,\dots,x_n].$ Show that the following affirmations are equivalent:

  1. $Z(a) = \emptyset$;
  2. $\sqrt{a} = S$ or $\sqrt{a} = S_+ = \langle x_0, \dots, x_n \rangle;$
  3. $a \supset S_d,$ for some $d>0,$ where $S_d$ is the group of the homogeneous polynomials of degree $d$.

I am following the resolution available here.

My concerns on the implication $\mathbf{(ii) \implies (iii)}$: None. I understood this proof.

My concerns on the implication $\mathbf{(iii) \implies (i)}$: I understand that the idea is to assume that $Z(a) \neq \emptyset$ and to reach an absurd. Keeping this in mind, assume $P \in Z(a).$ Then, $P \in Z(S_d)$ since $S_d \subset a \implies Z(a) \subset Z(S_d).$ This means that every homogeneous polynomial of degree $d$ vanishes at $P$. Until here, I understand everything. Now comes the part I don't understand:

"Since $P \neq 0$, this is absurd". I don't understand why $P$ must be different from zero and why this is absurd.

My concerns on the implication $\mathbf{(i) \implies (ii)}$: Basically everything. I don't understand how $Z(a) = \emptyset$ implies that, in $\mathbb A^{n+1}$, $Z(a) = \emptyset$ or $Z(a) = \{0\}$ and I also don't understand the follow up.

Any help is apreciatted in advance.

xyz
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    Z(a) is supposed to be a projective variety, right? So it’s points are the equivalence classes of nonzero points under multiplication and so nonzero. Is the last part not just due to Nullstellensatz? – Eric May 13 '23 at 13:55
  • Thanks for your help @Eric. I was now able to understand why $P$ must be different from zero. Why does this create an absurd though? I don't how I can apply the Nullstellensatz here! – xyz May 13 '23 at 14:03
  • https://math.stackexchange.com/questions/119434/projective-nullstellensatz – Eric May 13 '23 at 14:20

1 Answers1

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For $(iii)\implies (i)$, assume there is a $P=[p_0,\ldots,p_n]\in Z(a)$. By definition of the projective space, there must exist an $i$ such that $p_i\neq 0$. But note that $x_i^d\in S_d\subseteq a$. So $x_i^d$ must vanish at $P$, i.e. $p_i^d=0$, contradiction.

For $(i)\implies (ii)$, note that $$ Z_{\mathbb{P}^n}(a)=\{[p_0,\ldots,p_n]\in\mathbb{P}^n\ |\ \forall f\in a\text{ homogeneous}:\ f(p_0,\ldots,p_n)=0\} $$ and $$ Z_{\mathbb{A}^{n+1}}(a)=\{(p_0,\ldots,p_n)\in\mathbb{A}^{n+1}\ |\ \forall f\in a:\ f(p_0,\ldots,p_n)=0\}. $$ Now assume that $Z_{\mathbb{P}^n}(a)=\emptyset$ and assume by contradiction that there is $(p_0,\ldots,p_n)\in\mathbb{A}^{n+1}\setminus\{0\}$ such that $(p_0,\ldots,p_n)\in Z_{\mathbb{A}^{n+1}}(a)$. Then in particular every homogeneous element of $a$ vanishes at $(p_0,\ldots,p_n)$, so $[p_0,\ldots,p_n]\in Z_{\mathbb{P}^n}(a)$, which is a contradiction.

Therefore $Z_{\mathbb{A}^{n+1}}(a)\subseteq\{0\}$. If we have $Z_{\mathbb{A}^{n+1}}(a)=\emptyset$ then the Nullstellensatz gives $\sqrt{a}=S$ (which in fact implies $a=S$). If $Z_{\mathbb{A}^{n+1}}(a)=\{0\}$, the Nullstellensatz gives $\sqrt{a}=S_+$.

imtrying46
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