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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.

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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 $J$.

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  • $\begingroup$ 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. $\endgroup$ Jan 16 at 22:28

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