Timeline for Good reduction of abelian varieties [S-T] -- Why is this ring henselian?

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Jun 28, 2012 at 17:30	comment	added	jmc		Ah. Because the polynomial is monic, the degree of the reduction does not go down. So there is a "kind of" 1-1 correspondence between the roots (if we take multiplicities into account). And thus roots reducing to simple roots are fixed by I. Got it.
Jun 26, 2012 at 17:43	comment	added	Will Sawin		Since the root under consideration is a simple root modulo $\bar{v}$ by assumption. Its residue doesn't change, and it he has to remain a root of the polynomial (since the polynomial is defined over $L$) so the root itself must be fixed.
Jun 26, 2012 at 13:45	comment	added	jmc		@Will: Actually I have onee question about your last sentence of the proof. I agree that the inertia group preserves residues of the roots. But how does it follow that therefore the roots are fixed?
Jun 26, 2012 at 6:44	history	edited	Will Sawin	CC BY-SA 3.0	deleted 2 characters in body
Jun 26, 2012 at 6:28	comment	added	Laurent Moret-Bailly		@Will: you mean "if $K$ is countable then $K_s$ is countable"!
Jun 25, 2012 at 19:31	comment	added	jmc		Thanks. Apperently the straightforward approach was the way I should have taken. I had not expected that. But then, I don't have very much experience with henselian rings (-;
Jun 25, 2012 at 19:30	vote	accept	jmc
Jun 25, 2012 at 18:55	history	answered	Will Sawin	CC BY-SA 3.0