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Jun
26
answered Non real complex in metric completions of $\mathbb Q$
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
25
answered Prove that field $Q(x)$ is a field of fractions of ring $F[x]$
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
23
answered $\mathbb{Q}(\sqrt2) $ is isomorphic to $\mathbb{Q}(\sqrt2 +2 )$
Jun
22
awarded  Nice Answer
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
awarded  algebraic-number-theory
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
answered Is there a ring homomorphism $M_2(\mathbb Z)\to \mathbb Z$?
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.