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](https://arxiv.org/abs/1101.1392) Lemma 3.3.



Nullstellensatz and nilpotence of a module
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 $J$.


