From Silverman's AEC page 332:

I need to understand why the determination of the following local kernel $$ ker \Big( H^1(G_v, E[\phi]) \rightarrow WC(E/K_v)[\phi] \Big) $$

is straightforward. The book says that it is the same as answering the questions whether a curve has a point over a complete local field (which I understand).

This is further reduced by Hensel's Lemma to checking whether the curve has a point in some finite ring $R_v/ \mathcal{M}_v^e$ for some easily computable integer $e$.

Now, as far as I know and understand, Hensel's Lemma says that for a polynomial $f(x) \in R_v[x]$, if it has a root in $R_v/\mathcal{M}$, then it lifts to a unique root in $R_v$ and hence $K_v$. However, this version does not seem to be directly used here. I suspect that maybe some several variable version is being used? And where did that $e$ come from?

I would really appreciate it if someone could explain it to me. Thank you.


1 Answer 1


you really shouldn't crosspost. Anyway, you've slightly misstated Hensel's lemma, you left out the assumption that $f(x)$ has a simple root in $R_v/\mathcal{M}_v$. That's where the $e$ is coming from. In general, if $f(x)$ has a root of higher multiplicity in $R_v/\mathcal{M}_v$, then you need to work in $R_v/\mathcal{M}_v^e$. So if the curve is non-singular modulo $v$, then you can take $e=1$, and it really does reduce to the 1-variable Hensel lemma, since the curve has dimension 1. However, if the curve is singular and it has a singular point defined over $R_v/\mathcal{M}_v$, that's not enough to conclude that the point can be lifted, you need to work with a larger value of $e$.

  • $\begingroup$ Love it when the author replies himself. Thank you! $\endgroup$
    – Shreya
    Mar 24, 2020 at 10:59

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