a confusion on Mazur's discussion
问题内容
(I'm sorry for my English.)
Hello. I have been reading B.Mazur's article "An introduction to the deformation theory of Galois representations". I'm at the proof of proposition1 in §30, where he discusses I-ordinary deformation.
Let me write down the settings :
$A$ : a Noetherian local ring which is complete with respect to the maximal ideal-adic topology.
$G$ : a profinite group.
$I$ : a closed normal subgroup of G.
$\rho\colon G\to GL_2(A)$ : a (continuous) representation.
$V$ : $A^2$ with the $G$ - action by $\rho$ (i.e. a representation space for $\rho$)
$\overline{V}:= V \otimes_A k$ with the natural $G$ - action obtained from $\rho$
the proposition says that if
- the sub-$A$-module $V^I$ of $V$ which consists of $I$-invariant elements is a free-$A$-module of rank 1
- the natural mapping $V^I \to \overline{V}^I$ is surjective
- the action of $I$ on $\overline{V}/\overline{V}^I$ is not trivial
then $\rho$ is equivalent to a representation of upper-triangular representation whose (1,1)-entry is a unramified character, and (2,2)-entry is a character whose residual character is nontrivial on $I$.
Now, I have two questions : \ 1.From the hypotheses, can we deduce that $\operatorname{dim}_k \overline{V}^I = 1$?
(It is easy to show that $\operatorname{dim}_k \overline{V}^I\leq 1$ by the surjectivity of the natural mapping.)
2.If 1.is false,can we justify Mazur's discussion?
Thanks.
(Thank you for your comments. I refined my question.)
回答 (1)
The key point is to show that $(\overline V)^I\neq0$.
Let $\mathfrak m$ be the maximal ideal of $A$, let $k=A/\mathfrak m$, and write $L=V^I$. Note that $(\overline V)^I$ is not the same as $\overline{V^I}=V^I/\mathfrak mV^I$.
The natural map $L/\mathfrak mL\to(\overline V)^I$ is surjective, so $\dim_k(\overline V)^I\leq1$. We only need to rule out the zero-dimensional case.
Choose a generator $v$ of $L$, and write $v=(a,b)^T$ in some basis of $V$. Suppose $(\overline V)^I=0$. Then $a,b\in\mathfrak m$.
Let $S=\{(r,s)\in A^2:ra+sb=0\}$. For $\sigma\in I$, put $M_\sigma=\rho(\sigma)-1$. Since $v$ is $I$-invariant, every row of $M_\sigma$ lies in $S$.
Let $\overline S$ be the image of $S$ in $k^2$. I claim that $\overline S\neq k^2$. Otherwise $S+\mathfrak mA^2=A^2$, and Nakayama's lemma applied to $A^2/S$ gives $S=A^2$. Then $(1,0),(0,1)\in S$, so $a=b=0$, contradicting that $v$ generates the free rank-one module $V^I$.
Thus $\overline S$ is a proper subspace of $k^2$. Hence there exists $0\neq\overline w\in k^2$ annihilated by every vector in $\overline S$. Since every row of $\overline{M_\sigma}$ lies in $\overline S$, we get $\overline{M_\sigma}\overline w=0$ for all $\sigma\in I$. Therefore $0\neq\overline w\in(\overline V)^I$, a contradiction.
Hence $\dim_k(\overline V)^I=1$.
Therefore the surjection $V^I/\mathfrak mV^I\to(\overline V)^I$ is an isomorphism. A generator of $V^I$ has nonzero reduction, so one of its coordinates is a unit. Hence it extends to a basis of $V$, giving $V=V^I\oplus Ae_2$.
Because $I$ is normal in $G$, $V^I$ is $G$-stable. Indeed, for $g\in G$, $v\in V^I$, and $\sigma\in I$, we have $\rho(\sigma)\rho(g)v=\rho(g)\rho(g^{-1}\sigma g)v=\rho(g)v$.
Therefore, in a basis adapted to $V^I$, $\rho(g)$ has the form $\begin{pmatrix}\chi_1(g)&*\\0&\chi_2(g)\end{pmatrix}$. Since $I$ acts trivially on $V^I$, $\chi_1|_I=1$, so $\chi_1$ is unramified.
Finally, $(V/V^I)\otimes_Ak\simeq\overline V/(\overline V)^I$, so the residual character of $\chi_2$ is exactly the action of $I$ on this quotient. The last hypothesis gives that this character is nontrivial on $I$.