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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?

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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 $$\sum a_g g= \sum a_g g - \sum a_g = \sum a_g ( g -1). $$

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  • $\begingroup$ Thank you for the elaboration. I can understand it now. For some reason it didn't make sense before. Now it does. I will get to the second part of your answer after I go over the next part of the notes. $\endgroup$ – Shine On You Crazy Diamond Aug 9 at 16:20
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    $\begingroup$ 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) $\endgroup$ – peter a g Aug 9 at 16:48
  • $\begingroup$ (cont) - BTW, obviously there is nothing special about the base ring ${\mathbb Z}$. $\endgroup$ – peter a g Aug 9 at 16:50
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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.

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  • $\begingroup$ This is a little high level for me, but surely I will get to understand it as I progress through the notes. $\endgroup$ – Shine On You Crazy Diamond Aug 9 at 16:21

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