Reputation
23,716
Next privilege 25,000 Rep.
Access to site analytics
Badges
2 25 53
Newest
 Necromancer
Impact
~294k people reached

4h
comment Prove $n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0
Maybe you meant $n^3 -7n+3$?
20h
comment Solve $e^{4z} +e^{3z} + e^{2z} + e^z + 1 = 0$.
Just substitute $e^z=u$, find that $u$ is a primitive $5$-th root of unity, and take it from there.
20h
comment Cartesian State Vectors → Keplerian Orbit Elements
Still not enough information. Maybe the $T$ is for transpose, turning a row vector into a column vector? This is definitely not my meat, but please don’t hesitate to e-mail me.
20h
comment Show that the elements of the form $1+\zeta + \zeta^2 + \dots + \zeta^m$, with $1 \leq m \leq p-2$, are invertible in $\mathbb{Z}[\zeta]$
Oh, conjugate elements always have the same norm. And $\zeta^{m+1}$ and $\zeta$ have the same minimal polynomial, so are conjugate.
20h
comment When is the limit of an infinite product equal to the infinite product of the limit?
I suppose that for $n$ large enough, your individual fraction factors are less than $3/4$, so that your product is dominated by $\lim_n(3/4)^n$, going to zero. Not necessary to bother with logarithms.
20h
comment Cartesian State Vectors → Keplerian Orbit Elements
I am around, but what kind of mathematical object is $T$ in this case? I can’t think of an interpretation of it that makes the expression grammatical.
20h
answered What does this notation mean on this derivative question?
20h
comment Cartesian State Vectors → Keplerian Orbit Elements
Here, the length of a vector $r$ is written $\Vert r\Vert$, so that you’re dividing a vector by its (scalar) length, to get a unit vector in the same direction. It’s a common way of getting the direction of a vector, throwing away its magnitude.
20h
answered Show that the elements of the form $1+\zeta + \zeta^2 + \dots + \zeta^m$, with $1 \leq m \leq p-2$, are invertible in $\mathbb{Z}[\zeta]$
1d
answered Is the map $\phi: \mathbb{P^2\rightarrow \mathbb{P^1}}$ defined by $\phi(x,y,z)=(x,y)$ birational?
1d
comment Is the map $\phi: \mathbb{P^2\rightarrow \mathbb{P^1}}$ defined by $\phi(x,y,z)=(x,y)$ birational?
No, $Y=Z$ is closed and not open. The set where $Y\ne Z$ is open, and maybe you were thinking of that?
1d
comment Is the map $\phi: \mathbb{P^2\rightarrow \mathbb{P^1}}$ defined by $\phi(x,y,z)=(x,y)$ birational?
But $(x,y,y)$ has the last two coordinates equal, in other words lies in the line $Y=Z$, whereas $(x,y,z)$ does not necessarily lie on this line. That is, $(1,2,3)\mapsto(1,2)\mapsto(1,2,2)$, for a specific example. Your $\phi^{-1}$ is not an inverse.
1d
comment Is the map $\phi: \mathbb{P^2\rightarrow \mathbb{P^1}}$ defined by $\phi(x,y,z)=(x,y)$ birational?
Your inverse is not an inverse of the given $\phi$, right? You’d need to find one.
1d
answered How do you say in a correct way using some technique?
1d
comment A nontrivial solution to the quadratic form $x^2 - xy + y^2$ over the finite field $𝔽_p$ with $p ≡ 1 \pmod3$ a prime
You beat me to it by a couple of minutes.
1d
answered A nontrivial solution to the quadratic form $x^2 - xy + y^2$ over the finite field $𝔽_p$ with $p ≡ 1 \pmod3$ a prime
1d
comment Does the determinant give you the index over $\mathcal{O}_k$ as well as over $\mathbb{Z}$?
Yes, I guess that was my point.
1d
comment Upper bound on exact power of wild prime that divides the different
Yes, @JyrkiLahtonen, I think the local case of the following extension of $\Bbb F_p((t))$ also shows this interesting difference from the mixed-characteristic case. Let $\pi$ be a root of $X^p+t^nX+t$. I suppose it shows how there are separable elements arbitrarily close to the inseparable element $t^{1/p}$.
1d
comment An automorphism of the field of $p$-adic numbers
Maybe it’s late to be making a comment here, but I love this argument. I proved the triviality of the automorphism group of $\Bbb Q_p$ for myself by a far more elaborate argument, and this one really appeals to me. But it seems to me that talking about the cubehood of $1+p^2x^3$ should work just as well, and not need any special arguments, since it’s fine at $3$.
2d
comment Does the determinant give you the index over $\mathcal{O}_k$ as well as over $\mathbb{Z}$?
Are you aware of the theorem in algebraic number theory that the index of $Ra$ in $R$ is $|\mathbf N(a)|\,$? And then, locally, one ordinarily defines $v_{\mathfrak P}(a)$ exactly to be $v_p(\mathbf N(a))$. In the latter case, I’m thinking of the complete situation, where one then shows that this $v_{\mathfrak P}$ is a valuation by a Hensel’s-Lemma argument.