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Jun
17
answered Proving that a binomial coefficient involving a power of $2$ is even
Jun
17
comment Question from 14.6 “Galois Groups of Polynomials” from Dummit and Foote
He means that the field degree of the big field over the little field is $n!$. Since this number is the cardinality of your group, which is acting as a Galois group of the big over the little, it’s a direct application of FTGT.
Jun
17
answered Prove that $(G, *)$ is abelian group, for $ x * y = \tan^{-1}(tan(x) + tan(y))$
Jun
16
comment Galois action on torsion points of formal group
Tate’s explanation of this in the original paper is very clear, as I recall. I know that Serre’s treatment in the Brighton conference of 1967 is a mess by comparison (and that is a rare phenomenon, ’cause Serre is usually crystal clear).
Jun
15
comment $p$-Adic complex numbers, expansion into series
The more I look at this question, the more interesting it is. But your example works only over $\Bbb Q_5$. Have you expressed the same irrational quantity as an element of $\Omega_p$ for other values of $p$? No matter what, you should do lots of examples. By hand!
Jun
14
comment Find and show primitive element of field extension
Sorry, it’s now clear to me that the degree is $120$. Should be easy from here on in.
Jun
14
comment Find and show primitive element of field extension
The splitting field is certainly gotten by adjoining a single root $\lambda$ of $f$ and all the fifteenth roots of unity. As it happens, $[\Bbb Q(\zeta_{15})\colon\Bbb Q]=8$, so that the degree of your extension is at most $8\cdot15=120$. It’s not at all clear (to me), however, that this is the degree. My strategy would be to determine the degree, and then try $\lambda+\zeta_{15}$ for a primitive element. This trick works surprisingly often!
Jun
14
comment root of unity in a local field
The barebones fact is that if the $p^m$ roots of unity are there, then the ramification index (over $\Bbb Q_p$) must be divisible by $(p-1)p^{m-1}$. The converse is by no means true. I find knowledge of the Newton Polygon most useful here.
Jun
14
comment The class number and the inverse Galois problem
This will be an interesting question when you state it just that way. I’m not on intimate terms with many examples where $|G|>2$, but for the examples I know, it seems that your minimal number is $1$. But maybe you should look up tables of class numbers, to get some feel for the landscape.
Jun
12
comment Galois action on torsion points of formal group
I haven’t thought about this much for fifty years, but I think the reason is that you want your isomorphisms, when different $\pi$ are chosen, to fit together in such a way that any $\pi$ itself acts as Frobenius on the maximal unramified.
Jun
11
answered Why is $2^{32}-1=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)?$
Jun
11
comment Non-algebraic subfield intersection
When I was a grad student, this kind of question worried and confused me a lot.
Jun
11
answered Non-algebraic subfield intersection
Jun
8
comment Is $\mathbb{Q}(\sqrt{2})\subset\mathbb{Q}\left(\sqrt{2},i\sqrt[3]{2}\right)$ a Galois extension?
I advise answering @ÁlvaroLozano-Robledo’s question explicitly. I found the equation, and could see something very important just by looking at it.
Jun
8
answered Why do the equations $Ax + By + Cz= D$ represent planes in $\Bbb R^3$ and not lines?
Jun
7
comment Difference Aut(F:K) and G(F:K)
I certainly did not mean to disagree with anything you said, so I retreat only partially.
Jun
7
comment Difference Aut(F:K) and G(F:K)
Beyond that, @GeorgesElencwajg, OP did not specify that the extension was algebraic. We may (and do!) speak of the group of automorphisms of $k(t)$ over $k$, $t$ being a transcendental element, but this is nothing like a Galois group.
Jun
4
comment Does a field of transcendence degree n correspond to a variety?
The wikipedia article on Birational Geometry is a good quick overview of the story. Two varieties are birationally equivalent if their function-fields are isomorphic. But the varieties need not thereby be isomorphic. This is deep stuff.
Jun
4
comment Analytic function defined on unit disc with co-domain except negative real axis.
I disagree with these comments. Since $f$ maps into the slit disk (plane minus the nonpositive real axis), a domain on which the square root is well defined in such a way that $\sqrt1=1$, OP may indeed take the square root of $f$ to solve his problem.
Jun
4
comment Analytic function defined on unit disc with co-domain except negative real axis.
My comment was completely wrong, sorry. I’ll also comment below.