# Nullstellensatz and nilpotence of a module

Let $$\nu : G \rightarrow H$$ be a surjective group homomomorphism with kernel $$N$$, $$H$$ abelian, and $$G$$ finitely generated.

The rational abelianization of $$N$$, $$H_1(N)$$ is a $$\mathbb{C}[H]$$-module, and we say that it is nilpotent if some power of the augmentation ideal $$I_H \subset \mathbb{C}[H]$$ is in the annihilator of the module.

In some paper I read that, as an easy consequence of the Hilbert Nullstellensatz, $$H_1(N)$$ is nilpotent if and only if $$\operatorname{Specm}(\mathbb{C} H) \cap \operatorname{supp}_{\mathbb{C} H}\left(H_{1} N\right) \subseteq\{1\},$$ where the support means the subset of prime ideals containing the annihilator, and "$$1$$" is the ideal $$I_H$$.

Can someone explain what version of the Nullstellensatz is applied here? I understand only the direct implication….

The result in question is Dimca, Hain, and Papadima - The abelianization of the Johnson kernel Lemma 3.3.

If we call the annihilator by $$J$$, then the statement reads

$$J$$ contains some power of the maximal ideal $$I_H$$ if and only if the intersection of the set of prime ideals containing $$J$$ with the set of maximal ideals is contained in $$\{I_H\}$$

which simplifies to

$$J$$ contains some power of the maximal ideal $$I_H$$ if and only if the set of maximal ideals containing $$J$$ is contained in $$\{I_H\}$$

The Nullstellensatz says:

The intersection of all maximal ideals containing $$J$$ is equal to the radical of $$J$$

which, if the set of maximal ideals containing $$J$$ is contained in $$\{I_H\}$$, implies the radical of $$J$$ is either the unit ideal or $$I_H$$. Since each generator of $$I_H$$ is contained in the radical of $$J$$, some power of each generator is contained in $$J$$, and since there are finitely many generators, some power of $$I_H$$ is contained in $$J$$.

• Thanks for this precise answer, it seems so clear now. What I thought was the Nullstellensatz is actually the "zariski lemma" which is a step in the proof of the Nullstellensatz, I guess. Jan 16 at 22:28