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1d
comment Non real complex in metric completions of $\mathbb Q$
@LuisGomezSanchez, sorry for the delay, but I was away from wifi. If $\Phi_n$ is the $n$-th cyclotomic polynomial, whose roots are the primitive $p$-th roots of unity, and if $\xi$ is a root of $\Phi_n$, then $\xi$ may be sent into $\Bbb C$ to an $\exp(k\pi i/n)$, but *which $k$*? Similarly, $\xi$ may be sent into $\Bbb Q_p$ if $n|(p-1)$, but in which of the possible $\phi(n)$ ways? There is simply no way to identify which element of a $p$-adic field $\exp(i\pi/n)$ corresponds to.
1d
comment What does “bounded away from zero” actually mean?
@TonyK, ED would never have written anything so pedestrian. It’s a saying of my great-grandmother’s. Anyhow, it’s fine to say that a set is bounded away from zero, and in your answer below, you’ve handled exactly correctly the unfortunate ambiguity in the expression “the function is bounded away from zero”.
Jun
26
comment What does “bounded away from zero” actually mean?
@TonyK, “spoken words are light as air; written words are everywhere.” In lectures, we get away with all sorts of sloppinesses and infelicities. If it is from a lecture, I grit my teeth and let it pass. In a book, though, it should be edited to be perfectly precise.
Jun
26
comment What does “bounded away from zero” actually mean?
They’re apparently talking about the values of the function, not the domain. Indeed, $1/z\in\{w:|w|>1\}$ on that domain, certainly a set bounded away from zero. The wording is ambiguous, though. Poor writing on the part of your author!
Jun
26
comment Isomorphism of quadratic extensions (of a number field)
Certainly: the statement is simply false.
Jun
26
comment Galois group and traslations by rational numbers.
@JosephCurwen, your suggestion is good, but it’s specific to the point $x=0$ on the line, while the question is invariant under translations of the line. I think that when you smear your suggestion over the whole line, it becomes the hypothesis that $g$ has all roots simple. [ Whoops — not quite right, but I’ll leave what I said anyway. ]
Jun
26
comment $E/F$ is a normal extension iff $E$ is a splitting field for some polynomial $f\in F[X]$.
Sorry, but the only help I could give passes through another (fairly easily provably equivalent) definition of normality, namely that $E$ is normal over $F$ if every $F$-homomorphism of $E$ into an algebraically closed field containing $E$ sends $E$ into (and thus onto) itself.
Jun
25
comment Reference for the p-adic numbers
This is the one I recommend most strongly.
Jun
25
comment $E/F$ is a normal extension iff $E$ is a splitting field for some polynomial $f\in F[X]$.
I don’t see normality coming into your argument. And while we’re at it, you’d better tell us your definition of normality (there are several).
Jun
24
comment Show that $\mathbb{Q}(\sqrt{5}+\sqrt[3]{2})=\mathbb{Q}(\sqrt{5},\sqrt[3]{2})$
This would be my method, too. You get a polynomial of degree $6$, but you must also show that it’s irreducible.
Jun
23
comment Exercise about cyclic field extension
Some ideas (hints) for (a): Isn’t it the case that for each $n\ge1$, there is at most one extension of $k$ of degree $n$? Isn’t there a sequence of integers $1|N_1|N_2|\cdots$ that’s cofinal with the set of degrees of extensions of $k$?
Jun
22
comment Finding the subfields of the cyclotomic field of order $5$
And so we see a relation between the magical mystical Golden Ratio and the magical mystical Pentagram.
Jun
22
comment Polynomials over a finite field
But, @user26857, your $P$ is not of the required form.
Jun
22
comment A constructive method for finding a subfield $\mathbb{K}$ such that $Gal\left(\mathbb{C}\left(x\right)/\mathbb{K}\right)$ is isomorphic to $S_3$.
No, @EwanDelanoy, I believe that this will typically give you a nonnormal extension. What you say about the minimal polynomial is right up to a point: I would say that the minimal polynomial over $\Bbb C(f)$ is $f(X)-f$, as confusing as that looks. Considering your field $\Bbb C(f)$ as $\Bbb C(t)$, the minimal becomes $f(X)-t$.
Jun
21
comment Is there a ring homomorphism $M_2(\mathbb Z)\to \mathbb Z$?
This was the “much better” method I had in mind in my answer.
Jun
21
comment Is there a ring homomorphism $M_2(\mathbb Z)\to \mathbb Z$?
Most people require a ring homomorphism to take the identity element to the identity.
Jun
21
comment Subgroup of $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{Q} / \mathbb{Z}$
What I meant was that the lifts in $\Bbb Q$ of the generators of your subgroup of $\Bbb Q/\Bbb Z$ generate a cyclic subgroup of $\Bbb Q$, as you already know, and the image of a cyclic group is cyclic. Done.
Jun
20
comment Ring of integers of a cyclotomic number field
Do you know the technique of localization?
Jun
20
comment Subgroup of $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{Q} / \mathbb{Z}$
For $\Bbb Q/\Bbb Z$, you almost have it: for a finitely generated subgroup there, just lift your generators back to $\Bbb Q$ via $\pi$, and consult the subgroup they generate.
Jun
19
comment What fraction of a sphere can an external observer see?
Thanks, it’s always nice to see a familiar name in these parts.