1
$\begingroup$

I'm currently studying Iwasawa theory.

1) There are many $\mathbb{Z}_p$-modules on which some Galois groups act. So I often face some facts on the group representation over local fields or p-adic integer ring. But I can't find any references yet.

Of course, there are articles on p-adic representation. But I want references that are not too deep. I want references using just easy-to-follow arguments of algebra and representation theory.

Can you suggest any references?

2) Currently, I'm reading the paper "On the Iwasawa Invariants of Totally Real Number Fields" written by Ralph Greenberg. There I cannot understand a line which I have underlined with red line.

enter image description here

I'm afraid that there are many counter-examples against the line.(For example we can take cyclic group of order prime to the order of the group of units.) Can you please explain the line to me?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ math.u-psud.fr/~fontaine/galoisrep.pdf Is that the kind of thing you are looking for? $\endgroup$
    – Asvin
    Commented Aug 13, 2018 at 7:29
  • 1
    $\begingroup$ Also, Washington's Cyclotomic Fields book contains some Iwasawa theory and is very approachable. $\endgroup$
    – Asvin
    Commented Aug 13, 2018 at 7:29
  • $\begingroup$ I guess the notion of p-adic representation is what I've been looking for. Thank you very much. By the way, Do you know any other reference that is more elementary, self contained, which focus on the facts that are easy to check? $\endgroup$
    – gualterio
    Commented Aug 13, 2018 at 7:50

1 Answer 1

Reset to default
1
$\begingroup$

Your question has nothing to do with $p$-adic representations, it is a general fact about representations of finite groups. Given a group $\Delta$ and a field $F$ of characteristic $0$ (characteristic prime to $\vert\Delta\vert$ is enough) let $\{\chi_1,\dots,\chi_t\}$ be the set of its irreducible $F$-valued characters. Then $$ e_i=\frac{1}{\Delta}\sum_{g\in\Delta}\chi_i(g)g\in F[\Delta] $$ verify $\sum_{i}e_i=1$ because $$ \sum_{i=1}^t e_i=\frac{1}{\Delta}\Big(\sum_{e\neq g\in\Delta}\big(g\sum_{i=1}^t\chi_i(g)\big)+\sum_{i=1}^t\chi_i(e)\Big). $$ By the orthogonality relations of characters (see here), the sum over all characters of $\chi(g)$ verifies $$ \sum_{i=1}^t\chi_i(g)=\begin{cases} 0&\text{ if }g\neq e\\ \vert\Delta\vert&\text{ if }g=e \end{cases} $$ and thus $\sum e_i=1$.

ArithmeticGeometer's suggestion of reading Washington's book (expecially Chapter 2) is actually a very good one.

Improve this answer
$\endgroup$
5
  • $\begingroup$ First of all, thank you very much for the answer. But I guess the problem is not solved. Your general argument works only when the field is algebraically closed(with the assumption on the characteristic) or when the order of the group of roots of unity is a multiple of the order of the group $\delta$. For odd prime $l$, the group of roots of unity of $\mathbb{Q}_l$ has order $l-1$. If $\delta$ is a cyclic group of order prime to $l-1$, then the only character is the trivial one. Then the summation of idempotents is not $1$. $\endgroup$
    – gualterio
    Commented Aug 14, 2018 at 15:13
  • $\begingroup$ It needs only a small tweaking when the field is not algebraically closed. The group algebra (since char is zero) is a product of matrix algebras $A_i$ over division algebras over the field. Each of these matrix algebras has an identity $e_i$ whose projection to $A_j (j\neq i)$ is zero. Clearly the sum of the $e_i$ is identity on the product algebra. $\endgroup$
    – Venkataramana
    Commented Aug 14, 2018 at 16:13
  • $\begingroup$ Thank you for answering. I guess I need some time to fully understand your answer. I guess you are right in that for any group ring there are orthogonal idempotents forming a basis. But in the article, the idempotents are ones coming from 'representations'. If the field is not algebraically closed there may be representations less than the order of the group. $\endgroup$
    – gualterio
    Commented Aug 14, 2018 at 17:07
  • $\begingroup$ Well, that is how idempotents arise. If you have an irreducible group representation, the algebra spanned by the group elements in the representation is simple and is hence a matrix algebra over the "commutant' which in the irreducible case, is a division algebra. On the other hand, if you have a matrix algebra $M_r(D)$ as a factor of the group algebra, then $D^r$ is a representation of the group. $\endgroup$
    – Venkataramana
    Commented Aug 15, 2018 at 3:32
  • $\begingroup$ @Venkataramana Thank you again for answering again. But I cannot follow your argument for now. Can you please elaborate your argument? By the way, in the article, the group $\Delta$ is fixed. Do you mean that the idempotents from characters on the 'fixed' group, still form a basis? I'm sorry but I'm not used to the theory of group representations and simple algebras. $\endgroup$
    – gualterio
    Commented Aug 16, 2018 at 0:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .