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16h
answered Difficulty understanding how an element of a quotient ring/field can be represented a certain way…
16h
comment Difficulty understanding how an element of a quotient ring/field can be represented a certain way…
Because $\alpha^2$ and $2$ are in the same coset of the ideal. You test whether this is true by asking whether the difference $\alpha^2-2$ is in the ideal. And it is.
1d
comment Galois invariants of the Tate module of an elliptic curve over a number field
Indeed, bad reduction is one of my many blind spots and areas of ignorance. I was definitely thinking of good reduction.
2d
comment Fixed Field of $\sigma, \tau$
Added a couple of paragraphs to show the polynomial.
2d
revised Fixed Field of $\sigma, \tau$
Went into much more detail, at OP’s request.
2d
comment Prime ideals decomposition
You’ll need to give more information, especially explaining how much you know, and how far you’ve gotten so far.
2d
comment **Location** of shortest distance between two skew lines in 3D?
To tell the truth, I would be much more simple-minded about this problem than to use a formula like that. I would parametrize the two lines linearly, like $\ell_1: (kt+a, mt+b, nt+c)$, similarly for the other line, and (using different parameters $s$ and $t$), write out the square of the distance between the $s$-point on the first line and the $t$-point on the second. You get a quadratic expression in $s$ and $t$, which you can easily minimize. This gives you the values of $s$ (point on first line) and $t$ (point on second line). Easy as that.
2d
comment Fixed Field of $\sigma, \tau$
You have a group, call it $G$, generated by $\sigma$ and $\tau$. It’s necessary to check that this group has order $6$. It follows that the fixed field $F$ had $[K\colon F]=6$. I told you how to get an element $u$ of $F$. You only need to show that this element generates all of $F$, not a proper subfield. One way to do this is find a sextic polynomial over $k(u)$ that has $x$ as a root. This shows that $[k(x)\colon k(u)]\le 6$. But what you already know shows that $[k(x)\colon k(u)]\ge 6$. Thus $F=k(u)$. If you want to see the sextic polynomial of which $x$ is a root, let me know.
Aug
29
comment Fixed Field of $\sigma, \tau$
Awright, without applying any geometric intuition or knowledge, take s plain linear, such as $x-3$, and apply to it all six elements of the group. Now take the product of these six. You’re taking the norm of $x-3$, from $k(x)$ down to the fixed field. There you are, end of story.
Aug
29
revised Roots of $\Phi_{31}(x)$ as roots of unity.
corrected an embarrassing typo
Aug
29
comment Show that $K=k(u)$ iff $u=\frac{ax+b}{cx+d}$ where $a,b,c,d \in k$ with $ad-bc \neq 0$
I think you really have it already. Certainly, if $u=(ax+b)/(cx+d)$, then $k(u)\subset k(x)$, right? Now use what you have already shown.
Aug
29
answered Fixed Field of $\sigma, \tau$
Aug
28
answered Galois invariants of the Tate module of an elliptic curve over a number field
Aug
28
comment Galois invariants of the Tate module of an elliptic curve over a number field
Let me give an unsatisfactory answer below, which would easily be annihilated by a short answer from an expert, I’m sure.
Aug
28
comment Ramification in $\mathbb Q(i,\sqrt[4]\pi)/\mathbb Q(i)$
Yes, I tried going from the quadratic extensions to the quartic ones, and did not succeed. Others might, however; and I’m not going to be so rash as to claim that the method I showed you is the best or shortest!
Aug
28
comment Galois invariants of the Tate module of an elliptic curve over a number field
Good reduction at $\ell$?
Aug
27
comment Prove that if $f$ and $g$ are analytic at $w$, then so is $fg$
Exactly, since analyticity is just differentiability in the complex sense.
Aug
27
comment Equilateral triangle
Have you drawn a picture?
Aug
27
comment Ramification in $\mathbb Q(i,\sqrt[4]\pi)/\mathbb Q(i)$
Just an addendum: over $\Bbb Q(i)$, you get the fifth roots of unity by adjoining a fourth root of $-15-20i$, so (modulo $8\mathfrak m$) this is what’s needed for a fourth-root extension in which $(1+i)$ remains prime.
Aug
27
comment Notation: $\varphi$ and $\phi$
In total agreement, especially on your last paragraph. I think I’m a coauthor on a paper that uses $\pi$ and $\varpi$, but few recognize the latter as a pi.