3
$\begingroup$

Below is the problem with which I am struggling.

Let $G$ be a finite group.

Let $L/K$ be an extension of fields of characteristic $0.$

Let $M$ and $N$ be finitely generated $K[G]$-modules.

Claim. If $L\otimes_K M \cong L\otimes_K N$ as $L[G]$-modules, then $M\cong N$ as $K[G]$-modules.

First, I tried to prove it in the case where $L/K$ is finite by considering $\phi\circ\theta: M\to N$ where

$\bullet\; \theta:M\to L\otimes_K N$ is given by $\theta(m)=\phi(1\otimes_K m)$ where $\phi$ is the given isomorphism, and

$\bullet\; \phi:L\otimes_K N\to N$ is determined by $\phi(l\otimes_K n) = Tr_{L/K}(l)\cdot n.$

But this didn't work out...

I am aware that there are a few similar posts, such as:

  1. Can extending a finite ground field make modules isomorphic?

  2. https://mathoverflow.net/questions/28469/hilbert-90-for-algebras

My problem with (1) is that, the OP gives a "very easy argument" for my claim but I don't understand what "determinant" means when the ground ring is non-commutative.

As for (2), it proves a much stronger result and makes use of many things I don't know about yet.

I would be immensely grateful if someone could show me a proof in my particular setting!

  • 3
    Just a comment on the proof in (1). It is the determinant over the ground field (your $K$, not $K[G]$). Note that a morphism $K[G]$-modules is an isomorphism if and only if it is an isomorphism as a $K$-linear mapping. For the argument in (1) to work, we only need that $M$ and $N$ are finite dimensional over $K$. – 2017-02-19
  • 0
    @user350031: It would be nicer if you put the formulae in your title and in your claim in visual agreement (either drop or preserve $[G]$ in both of them). – 2017-02-19

0 Answers 0