I am studying "the local Langlands conjecture for GL(2)" of Bushnell and Henniart. I am having a hard time getting into the mechanics. And I have some problems with one of the first proposition.
Let $F$ be a local field, let $ \widehat{F}$ be the group of characters of $F$. Now, let $\psi \in \widehat{F}$ be a non trivial character of level $d$. So we have
- Let $a \in F$. The map $a\psi: x \mapsto \psi(ax)$ is a character of $F$ and if $a\neq 0$ then the character $a\psi$ has level $d-v_F(x)$.
- The map $a \mapsto a\psi$ is a group isomorphism $F \cong \widehat{F}$
It's clear to me that $a\psi$ is a character and that its level is $d-v_F(x)$, this is an obvious consequence of $$ x \in \ker(a\psi) \iff ax \in \ker{\psi}. $$ After that I have two problems with the isomorphism. The book says that $a \mapsto a\psi$ is an injective group homomorphism without proof. But it is difficult to me to accept this fact.
$a\psi=b\psi$ if and only if they agree on $x$ for all $x \in F$. So $$ \psi(ax)=\psi(bx) \qquad \forall x \in F $$ that is equivalent to $$ ax\equiv bx \bmod \ker\psi \qquad \forall x \in F $$ that is the same of $$ a\equiv b \bmod \ker\psi $$ So $$a\psi=b\psi \iff a\equiv b \bmod \ker\psi $$ that is clearly not the same of the book, so I am wrong. But why? And why this is an injective map?
The other problem is with the second part of the proof.
Let $w$ be a uniformizer of $F$ and let $\mathfrak{o}$ be the valuation ring of $F$ with maximal ideal $\mathfrak{p}$. If we take another non trivial character $\theta \in \widehat{F}$ with level $l$, clearly $uw^{d-l}\psi$ has level $l$ for all $u \in \mathfrak{o}^\times$. So $uw^{d-l}\psi$ and $\theta$ agree on $\mathfrak{p}^l$. But now, if we take $u,u' \in \mathfrak{o}^\times$, then $uw^{d-l}\psi$ and $u'w^{d-l}\psi$ agree on $\mathfrak{p}^{l-1}$ if and only if $u \equiv u' \bmod\mathfrak{p}$. It's not obvious how I can compute this.
Some hint to these two problems?
You said $\psi$ is nontrivial.
Rewrite $\psi(ax) = \psi(bx)$ as $1 = \psi(ax)\psi(bx)^{-1} = \psi(ax - bx) = \psi((a-b)x)$. This holds when $x$ runs over your local field, so unless $a = b$, we get $\psi(y) = 1$ for all $y$ in the field: $\psi$ is trivial. That contradicts the assumption that $\psi$ is nontrivial, so $a = b$.
Your mistake is that it is not true that the condition $ax \equiv bx \bmod \ker \psi$ for all $x$ in the field is equivalent to $a \equiv b \bmod \ker \psi$. That would be saying $(a-b)F \subset \ker \psi$ is equivalent to $a-b \in \ker \psi$. Do you see those are not generally equivalent?