# Romyar Sharifi's notes: Group and Galois Cohomology; ideal vs submodule?

Lecture Notes

Let $$G$$ be a group. Below, the author talks about strictly $$\Bbb{Z}$$-group rings.

Definition 1.1.3.
(i) The augmentation map is the homomorphism $$\varepsilon : \Bbb{Z}[G] \to \Bbb{Z}$$ given by $$\varepsilon(\sum\limits_{g \in G} a_g g) = \sum\limits_{g \in G} a_g.$$ (ii) The augmentation ideal $$I_G$$ is the kernel of the augmentation map $$\varepsilon$$.

It's intuitively clear to me that this is a $$\Bbb{Z}$$-module homomorphism, and not a ring homomorphism. Thus any kernel is therefore a $$\Bbb{Z}$$-submodule. So why here is it called an ideal?

Actually the augmentation map $$\varepsilon$$ is really a ring homomorphism, and thus its kernel is really an ideal. A nice way to see this is that $$\varepsilon$$ is the image of the trivial homomorphism $$G\to \{1\}$$ under the "group ring" functor. But you can also check it directly, of course.
To elaborate on Arnaud's fine (and more general "group ring functor") answer: $$\epsilon$$ is defined, on generators, by $$\epsilon : \gamma \mapsto 1,$$ for all $$\gamma \in G$$. In particular, if $$g$$ and $$h\in G$$, then $$\gamma = gh\in G$$, so that by definition $$\epsilon (gh) = 1$$, $$\epsilon (g) =1$$ and $$\epsilon (h) =1$$. Therefore $$\epsilon (gh) =1 = 1\cdot 1 = \epsilon(g)\cdot \epsilon (h),$$ and so on... giving that $$\epsilon$$ is a ring homomorphism.
By the way, if one starts with the fact that $$I_G$$ is generated by elements of the form $$(g -1)$$, then the identity in $$\mathbb Z[G]$$ $$h ( g - 1) = (hg -1) -(h-1)$$ gives another way to see that $$I_G$$ is a $$G$$-module. To verify the 'fact': clearly $$g-1 \in I_G$$, but on the other hand, if $$\sum a_g g \in I_G$$, we must have that $$\sum a_g = 0$$, so that $$\sum a_g g= \sum a_g g - \sum a_g = \sum a_g ( g -1).$$
• Arnaud's group ring functor answer is pretty much this: suppose $\lambda \colon G \to H$ is a group homomorphism. Then one also gets a $\mathbb Z$ homomorphism $\Lambda \colon {\mathbb Z}[G]\to {\mathbb Z}[H],$ defined, at the $\mathbb Z$-module level, by the formula $$\Lambda (\sum_g a_g g) = \sum a_g \lambda (g),$$ i.e., extend $\lambda$ linearly. However, since $\lambda (gh) = \lambda (g) \lambda (h)$, for all $g,h\in G$, one gets that $\Lambda$ also preserves the multiplicative structure of the rings, and $\Lambda$ is a ring-morphism (there is a formal verification to be made!) (cont) Aug 9, 2019 at 16:48
• (cont) - BTW, obviously there is nothing special about the base ring ${\mathbb Z}$. Aug 9, 2019 at 16:50