Let $G$ be a cyclic finite group and $R$ a commutative unital ring. There is a morphism (I think it's called the augmentation morphism) $$\epsilon:R[G]\mapsto R\\ \sum\limits_{g\in G}c_gg\mapsto\sum\limits_{g\in G}c_g $$

The augmentation ideal $I(G)$ is defined as the kernel of $\epsilon$.

The problem is to that $I(G)$ is a principal ideal (it is generated by one element).

Since $G$ is cyclic and finite $\exists g\in G$ s.t. $G=\langle g\rangle=\{1_G,g,g^2,...,g^{|G|-1}$} and say $|G|=n$.

At first I tried to directly find the generating element by taking $I(G)\ni z=\sum\limits_{i=0}^{n-1}z_ig^i$ with the property that $h(z):=\max\{i:\ z_i\ne0\}$ is minimal. But then proving that every element of $I(G)$ is a multiple of $z$ is hard.

I also have a lead but don't know what to do with it:

If $x\in R$ and $\sum\limits_{g\in G}c_gg\in I(G)$ then $x=x-0=x-\sum\limits_{g\in G}c_g"="x-\sum\limits_{g\in G}c_g1_G$

  • $\begingroup$ The morphism is not called the inclusion (it would be weird to call it that, since it is not injective in general!) but augmentation. $\endgroup$ – Mariano Suárez-Álvarez Apr 5 '18 at 17:37
  • $\begingroup$ Notice that you used a rather complicated way to define an element $z$, but you can simply take $z=1-g$, which obviously satisfies your conditions. $\endgroup$ – Mariano Suárez-Álvarez Apr 5 '18 at 17:39
  • $\begingroup$ Is it the natural morphism? $\endgroup$ – John Cataldo Apr 5 '18 at 17:39
  • $\begingroup$ @MarianoSuárez-Álvarez like $z=1_R1_G+(-1_R)g$ where $g$ generates $G$? $\endgroup$ – John Cataldo Apr 5 '18 at 17:42
  • $\begingroup$ That is an awfully complicated way of writing $1-g$. $\endgroup$ – Mariano Suárez-Álvarez Apr 5 '18 at 17:43

As suggested by the comment of Mariano Suárez-Álvarez, the ideal is generated as an $R$-module by elements of the form $h - h'$. Since $G$ is cyclic we can write without loss of generality, for $m < n$, $$h - h' =g^m - g^n = \sum_{k=m}^{n-1} g^k - g^{k+1} = \sum_{k=m}^{n-1}g^k(1-g).$$ So the ideal is principal and generated by $1-g$.

  • $\begingroup$ Is it obvious though that every element in $I(G)$ is a sum of elements of the form $g^m-g^n$? Why is something$*(1_R-1_R)$ the only way to get $0$ in $R$? $\endgroup$ – John Cataldo Apr 5 '18 at 18:06
  • $\begingroup$ Slghtly easier is to note that if $\sum_{h\in G}a_h=0$, then $\sum_{h\in G}a_hh=\sum_{h\in G}a_h(h-1)$. $\endgroup$ – Mariano Suárez-Álvarez Apr 5 '18 at 18:07
  • $\begingroup$ @MarianoSuárez-Álvarez once we have that is it even necessary to prove that $g^m-g^n$ is a combination of $(1-g)$? $\endgroup$ – John Cataldo Apr 5 '18 at 18:11
  • $\begingroup$ Nice! That's definitely easier and is a complete proof. $\endgroup$ – Suhas Vijaykumar Apr 6 '18 at 18:27
  • $\begingroup$ I think that $I(G)$ is only generated by elements $h - g$ it's sufficient to show that the kernel of the homomorphism of $R$-modules $R^n \to R$ is generated by sums of idempotents $e_i - e_j$. $\endgroup$ – Suhas Vijaykumar Apr 6 '18 at 18:29

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