Let $G$ be a finite group and $\phi:M\to N$ be a surjective homomorphism of $G$-lattices (i.e. finitely generated $\mathbb{Z}[G]$-modules, free as $\mathbb Z$-modules), with kernel $K$. For every $n\geq 1$, let $M_n:=\phi^{-1}(nN)$. Since $nN\cong N$, we have an exact sequence $0\to K\to M_n \to N\to 0$. I was wondering if it is possible to determine the class of this extension in terms of $n$ and the class of $0\to K\to M\to N\to 0$ inside the finite abelian group $Ext^1_G(N,K)$?

For example, is it true that $[M_n]=a+bn$, for some $a,b\in Ext^1_G(N,K)$?

  • By "free $\mathbb{Z}[G]$-modules" do you actually mean "free $\mathbb{Z}$-modules with a linear action of $G$"? – Qfwfq Sep 13 at 1:14
  • Edited, thanks. – Tetawo Sep 13 at 1:15
up vote 2 down vote accepted

It is what you think it is:

Lemma. The class of the short exact sequence $0 \to K \to M_n \to N \to 0$ in $\operatorname{Ext}^1_G(N,K)$ is $n$ times the class of the sequence $0 \to K \to M \to N \to 0$.

Proof. By construction, the class in $\operatorname{Ext}^1_R(C,A)$ of a short exact sequence $0 \to A \to B \to C \to 0$ of $R$-modules is the image of $\operatorname{id}_C$ under the boundary map $\operatorname{Hom}(C,C) \to \operatorname{Ext}^1_R(C,A)$ coming from the long exact sequence $$0 \to \operatorname{Hom}(-,A) \to \operatorname{Hom}(-,B) \to \operatorname{Hom}(-,C) \to \operatorname{Ext}^1_R(-,A) \to \ldots.$$ In our situation, we have a commutative diagram with exact rows $$\begin{array}{ccccccccc}0 & \to & K & \to & M_n & \to & N & \to & 0 \\ & & |\!| & & \downarrow & & \downarrow{\scriptsize n}\!\!\!\! & & \\ 0 & \to & K & \to & M & \to & N & \to & 0.\!\end{array}$$ The long exact sequences give a commutative diagram $$\begin{array}{ccccccccccc}0 & \to & \operatorname{Hom}(N,K) & \to & \operatorname{Hom}(N,M_n) & \to & \operatorname{Hom}(N,N) & \to & \operatorname{Ext}^1_G(N,K) & \to & \ldots \\ & & |\!| & & \downarrow & & \downarrow{\scriptsize n}\!\!\!\!\!\!\! & & |\!| & \\ 0 & \to & \operatorname{Hom}(N,K) & \to & \operatorname{Hom}(N,M) & \to & \operatorname{Hom}(N,N) & \to & \operatorname{Ext}^1_G(N,K) & \to & \ldots,\!\!\!\end{array}$$ hence the image of $\operatorname{id}_N$ in the first row maps to $n$ times the image of $\operatorname{id}_N$ in the second row. $\square$

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