Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In Conrad and Brinon's notes http://math.stanford.edu/~conrad/papers/notes.pdf, two uniformizers of $B_{dR}$ are produced: one is $\xi := [\tilde{p}]-p$ (bottom of p.58), where $\tilde{p} = (p, p^{1/p}, p^{1/p^2}, \ldots)$ is some choice of compatible $p^r$ roots of unity, and the other is $t := \log(\epsilon)$, where $\epsilon$ is some basis of the Tate module of $\mathbb{C}_K$.

Furthermore, we know that $B_{dR}^+ / (t) \simeq \mathbb{C}_K$. Therefore, there should be some $a_1 \in \mathbb{C}_K$ such that $\xi = a_1t + ...$. My question is: how can we describe more explicitly what this $a_1$ is?

share|improve this question
    
May be I miss something but both are uniformizers and both reduce to $0$ modulo $\mathrm{ker}(\theta)$. So the $a_1$ you write is $0$. All you can say is that they differ multiplicatively by some $u\in (B_{dR}^+)^\times$: observe that the action of Galois on both is different, and so is the action of $\varphi$ (although this $\varphi$ acts on them only as elements of $B_{cris}^+$) so this might be quite non-trivial a unit.... –  Filippo Alberto Edoardo Dec 21 '12 at 2:16
1  
$a_1$ is not the reduction of the uniformizers mod $ker(\theta)$, but the reduction of the ratio mod $ker(\theta)$, which as you note is nonzero. –  Will Sawin Dec 21 '12 at 4:19
add comment

1 Answer 1

What you're asking for is a description of $$a_1 = \theta(\frac{\[\tilde{p}]-p}{t}).$$ It is an element of $C_p$ and you can't really "write it down explicitly". What you can do is let $G_{Qp}$ act on it and see what happens. If $g \in G_{Qp}$ then $g(\tilde{p})=\tilde{p} \cdot \epsilon^{c(g)}$ where $c(\cdot)$ is the Kummer cocycle and $g(t)=\chi(g) t$ where $\chi(\cdot)$ is the cyclotomic character. This should imply that $$g(a_1) = \theta(\frac{\[\tilde{p}][\epsilon^{c(g)}]-p}{\chi(g) t}) = \frac{a_1}{\chi(g)} + p \frac{c(g)}{\chi(g)}.$$ Now that you see this formula, you should recognize it. Let $V$ be the semistable extension of $Q_p$ by $Q_p(1)$. It has Hodge-Tate weights $0$ and $1$ and $a_1$ is basically the period corresponding to weight $0$, ie the period that tells you that $(V \otimes_{Qp} C_p)^{G_{Qp}} \neq 0$.

share|improve this answer
    
Thank you for this discussion. In fact, you've reconstructed my original application, which is to describe the periods of the Hodge-Tate isomorphism for a Tate curve. I was just trying to discuss an example of how $B_{dR}$ lets you analyze this sort of question. I think I've now figured it out to my satisfaction. –  Tony Dec 23 '12 at 13:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.