Given a complete discrete valuation field $L$, is there always a complete local field $K\subset L$? That is, I assume the residual field of $L$ is non-finite. Can I claim that it contains a complete field with finite residual field with respect to the same valuation?
Given a complete discrete valuation field $L$, is there always a complete local field $K\subset L$?
2
$\begingroup$
field-theory
extension-field
analytic-number-theory
p-adic-number-theory
valuation-theory
-
0I assume you want the residue field characteristic to be positive, though? Otherwise I see no chance to get to a finite residue field. – 2017-01-25
1 Answers
1
Following the lead of the comment by @Torsten Schoeneberg. I point out that $\Bbb R((t))$, the formal Laurent series over the real field, is certainly a complete discrete valuation field, but it contains no field that’s finite over a $\Bbb Q_p$. Why? Because fields of the latter type always have elements whose square is a negative integer, while $\Bbb R((t))$ contains no such element.