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15h
comment Only finitely many reduced cubics with given discriminant?
After half a glass of wine, I think I disagree. Doesn’t the experimental evidence seem to show that in any “reasonable” arithmetic family of elliptic curves, the rank is zero half the time and $1$ half the time?
20h
comment Field homomorphism induces an isomorphism between their prime subfields
@GerryMyerson, in my upbringing, a homomorphism between rings with unit must take unit to unit. This jibes well with the geometric interpretation of ring morphisms.
20h
comment General way to find actions of automorphisms of the group of $x^3-2$ over $\mathbb{Q}$
I like this a lot, ’cause it’s exactly what I would have said (except that you’ve probably said it better than me).
21h
comment intersection of plane elliptic curve with coordinate hyperplanes
Oops! Very careless of me, and I’ve deleted that comment. The truth is that nothing whatever can be said about the order of such a point. It can be $2$, $3$, or anything. Examples: $y^2=x^3-x$ and $y^2=x^3+1$ for the $2$ and the $3$. And need I say that you should expect that a point of that form would be of infinite order?
1d
comment Is $\mathbb{Q}(\alpha_i^2)$ an intermediate field of $\mathbf{K}/\mathbb{Q}$, where $K$ is the splitting field of $x^3-2$ over $\mathbb{Q}$?
My critique: excellent answer, and efficient. Keep making contributions as good!
1d
comment Can the “$\forall x\in X $” be moved in this statement? “$\Gamma$ satisfiable $\implies \exists v:v(\alpha)=1 \forall \alpha \in \Gamma$”.
Quantifiers don’t commute. I think anybody would be satisfied with commas. In all these things I must defer to someone like @MauroALLEGRANZA, who knows the field.
1d
comment Can the “$\forall x\in X $” be moved in this statement? “$\Gamma$ satisfiable $\implies \exists v:v(\alpha)=1 \forall \alpha \in \Gamma$”.
Indeed, it’s always confusing and often ambiguous to have a quantifier at the end. Much better to put all the quantifiers at the beginning, in the right order.
2d
comment How to find irreducible polynomials in a given ring or field?
When I was teaching Algebra, I would partition the elements of a commutative ring $R$ thus: $\{0\}$; nonzero zero-divisors: units; reducible elements, namely things writable as the product of two nonunit nonzero-divisors; and everything else, which are the irreducibles. If you take a field $k$ and form $k[x]/(f)$ where $f$ is reducible, then there are no irreducibles nor reducibles: everything is either a zero-divisor or a unit. This is not obvious, would need a proof.
2d
comment Is it possible to follow another way to perform this calculation steps?
If you’re really talking computer-science notation, your first two displayed equations are not true: if $a=12$ and $b=13$ and $n=5$, then the left-hand side of your first equation is $1$, while the right-hand side is $6$. You have to apply “mod” once more on the right-hand side.
2d
comment Why if $\alpha$ is a root of $x^3-5x^2+2$, then $\mathcal O_{\Bbb Q[\alpha]}=\mathbb Z[\alpha]$?
Maybe it uses methods not familiar to OP, but it often helps to remember that the discriminant is the product of the local discs.
Sep
2
comment A quick Galois Theory question
In agreement with @JyrkiLahtonen’s comments, I think that your argument is perfect and applies to any Galois extension, not just the degree-$d$ extension of $\Bbb F_q$. But with the extra information that the smaller field is finite, you can simply, as he says, list the conjugates.
Sep
1
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.
Aug
31
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.
Aug
30
comment Fixed Field of $\sigma, \tau$
Added a couple of paragraphs to show the polynomial.
Aug
30
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.
Aug
30
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.
Aug
30
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
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
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.