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Jul
23
comment Neukirch's motivation for $p$-adic numbers
What disturbs me most about trying to give meaning to a particular ”coefficient” in a $p$-adic expansion is that these coefficients depend so strongly on your choice of representatives in $\Bbb Z$ or $\Bbb Z_p$ for the elements of $\Bbb F_p$, the residue field. For $\Bbb Z_5$, for instance, maybe you want to use $\{-2,-1,0,1,2\}$ for your possible “coefficients” of your expansion. Makes perfectly good sense. Unless you want to go all the way into abstraction and represent a $p$-adic integer by a Witt vector, it seems to me that the analogy with power series is extremely weak.
Jul
22
answered Problem on Galois theory and irreducible polynomial
Jul
22
comment Problem on Galois theory and irreducible polynomial
OP has explained that he meant the product of $\sqrt[p]2$ and $\sqrt[q]2$ to generate the big field, so I’m not sure your analysis is completely on the mark.
Jul
22
comment Problem on Galois theory and irreducible polynomial
I’ve edited to put in a central dot. Thanks for the clarification.
Jul
22
revised Problem on Galois theory and irreducible polynomial
changed a low dot (period) to a \cdot, as explained by OP
Jul
22
comment Problem on Galois theory and irreducible polynomial
In lightly editing your question, I see that you have a dot between the two radicals. Did you mean a comma (virgule) or did you mean the product of the two radicals?
Jul
22
revised Problem on Galois theory and irreducible polynomial
Changed plain Q to blackboard bold, and replaced the word “annulus” with the word “ring”
Jul
22
comment Problem involving cubic field extensions
Your proof in the algebraic case looks right to me, but I’m not at all sure that the statement is true in the general case. Do you have reason to think that it is?
Jul
22
comment How to evolve an expression with two denominators
I like this. Plus one point for you.
Jul
22
answered How to evolve an expression with two denominators
Jul
22
comment Priority of the 3 axioms of groups
I don’t think it makes sense to quote $g_3$ before $g_2$.
Jul
20
revised Existence of $\sqrt{-1}$ in $5$-adics, show resulting sum is convergent.
corrected an embarrassing typo
Jul
20
answered Existence of $\sqrt{-1}$ in $5$-adics, show resulting sum is convergent.
Jul
20
comment What to call a polynomial that with no roots in $\mathbb{Q}$ but does in $p$-adics
There actually is a (fairly) standard way of saying that $\Bbb Q_p$ contains an isomorphic copy of a field $K$: you say that there is an embedding of $K$ into $\Bbb Q_p$.
Jul
20
comment Why the extension dimension of $x^3-2$ equal to $6$?
@Travis’s answer puts his finger on it: adjoining a (primitive) cube root of $1$ is not a cubic extension but a quadratic one.
Jul
20
comment a maximum of 128 independent rules
This puts it all in good context.
Jul
19
comment Reference/advice for infinite Kummer extension $\mathbb{Q}_p(\sqrt[p^{\infty}]{p}, \zeta_{p^{\infty}})$
The proof I use involves Higher Ramification Theory, but I do wonder about the existence of an elementary proof. Please don’t hesitate to contact me by e-mail on the issue.
Jul
19
answered Reference/advice for infinite Kummer extension $\mathbb{Q}_p(\sqrt[p^{\infty}]{p}, \zeta_{p^{\infty}})$
Jul
18
comment Reference/advice for infinite Kummer extension $\mathbb{Q}_p(\sqrt[p^{\infty}]{p}, \zeta_{p^{\infty}})$
I ought to be able to help. Let me think on things.
Jul
18
comment Galois Group of the splitting field of the polynomial of $x^{11} - 7$ over $\mathbb{Q}$
Normality: let $\zeta_0$ be any eleventh root of unity unequal to $1$, and thus primitive. Then any field automorphism at all sends $\zeta_0$ to a power of itself, and the extension $\Bbb Q(\zeta_0)\supset\Bbb Q$ is normal.