What does ramification have to do with separability? Does ramification have anything to do with inseparability? It feels like an extension of Q in which p ramifies should somehow correspond to an extension of F_p(t) Does totally ramified lt;--> pur
What does the ramified in ramified type theory mean? What does ramify in ramified type theory mean? I've taken the liberty of looking up ramify in a dictionary, and there it mentions that it means branching edit @David Roberts: I've hardly seen in mentioned in the set theory books I've looked; but not being a specialist in set theory, I haven't looked that far @Gerald Edgar: Thanks So, I've
What prevents a cover to be Galois? - MathOverflow Subnormal subgroups need not be normal However, if, say, $\pi_1 (X)< \pi_1 (Y)$ is a characteristic subgroup then it will be normal in $\pi_1 (Z)$ This is the simplest condition I know to ensure that composition of regular covering maps is again regular The same applies to ramified covers once you remove branch points and their preimages
What is a tamely-ramified Weil-Deligne representation? Similarly, taking ramified principal series which may now be wildly ramified, their explicit construction shows that the conductor in the automorphic sense is the product (or sum if you take the exponent) of the conductors of $\chi_1$ and $\chi_2$ And similarly, the conductor on the "Galois" side is the same
ramifications in compositum number fields - MathOverflow This is a purely local phenomenon at the prime $2$ In other words, although the quadratic extensions $ {\mathbf Q}_2 (\sqrt {-1})$ and $ {\mathbf Q}_2 (\sqrt {3})$ are both (totally) ramified over $ {\mathbf Q}_2$, their compositum $ {\mathbf Q}_2 (\sqrt {-1},\sqrt3)$ is not totally ramified, for it contains the unramified quadratic extension $ {\mathbf Q}_2 (\sqrt {5})$ For more examples
When is the composition of two totally ramified extension totally ramified? Take the base field Q_2 and consider the quadratic extensions generated by square roots of -1 and 2, with uniformizers 1+i and sqrt (2) Their norms 2 and -2 differ by -1, which is a norm from the second, but nor the first extension; yet the compositum is totally ramified Is there something I don't see?
ramification - Ramified quaternion algebras - MathOverflow I'm trying to better understand the connection between the concepts of ramification of a field extension, and ramification of a quaternion algebra I'm also trying to build a better understanding