It seems like you're going about the problem correctly except that obviously the splitting field won't be as described if the quadratic is reducible. Other than that, unless I'm forgetting something (which would hardly be surprising), what you're saying is completely true.
Er...aren't those fields are isomorphic? Take a map that's identity on the rationals, and root 2 mapped to root 3, right?
I mean, strictly speaking I don't think that the degree being the same is enough to guarantee an isomorphism (if I had to guess, I'd say the automorphism groups of the...
I don't think this is true unless you assume U and V have the same dimension. For T to be injective, it must be the case that dim V \geq dim U, but equality doesn't have to hold. However, there does have to be an S such that ST is the identity on U (namely you just take Sx to be the preimage...
Okay, so I'm sure you noticed that M = aX + bY. Then M^n = (aX + bY)^n, right? Now, if X and Y were just plain old real numbers (or variables if you like), what would you do to expand (aX + bY)^n
I believe a coset representative would be the a in aH. Of course, talking about "coset representatives" when H isn't a normal subgroup is a little odd. For one, if H isn't a subgroup, the "representative" might be unique, and that's bad for most things you want to do with cosets (like in your...
I was wondering about the classification of groups with a certain number of subgroups. I (sort of mostly I think maybe) get the ideas behind classification of groups of a certain (hopefully small) order, but I came across a question about classifying all groups with exactly 4 subgroups, and I...