3h
comment Roots Of An Inseparable Polynomial.
Right. (Well, except you should say $f$ only has one root, not $x$.)
3h
comment Roots Of An Inseparable Polynomial.
@J.DavidTaylor Well, $\Bbb Z_p$ is used by many for what others (including me) call $\Bbb F_p$. More are aware of the finite field of order $p$ than are aware of the $p$-adic integers. Of course there is no ambiguity in this question: it must be the finite field of order $p$.
3h
comment Roots Of An Inseparable Polynomial.
Um, $p$ equals $0$ in $\Bbb F_p$, and of course the roots are not powers of $0$. Use my hint to completely factor $x^p-t$ using only one root $\sqrt[p]{t}$.
3h
comment Roots Of An Inseparable Polynomial.
Hint: $(a+b)^p=a^p+b^p$ in characteristic $p$.
13h
comment Is $\left\{ e^{ \frac{2\pi i }{n}}: n\in \mathbb{N}\right\}$ compact in complex plane?
Yes that is correct. If closed and bounded is your definition of compact then you are done, but if not then you might want to namedrop the Heine-Borel theorem.
17h
comment General way to solve equations of congruent classes?
A trick for large prime powers is Hensel lifting.
2d
comment Show that $D_n$ is a subgroup of Perm($\mathbb{C}$).
I would use $S_{\Bbb C}$ or ${\rm Perm}(\Bbb C)$ or any standard notation to denote the group of permutations of the set $\Bbb C$. Clearly your $D_n$ is a subset, you just need to show it's closed under composition and inverses. Did you try? If you don't know that multiplying by a modulus $1$ complex number is a rotation of the plane, then may I ask what kind of experience you have with complex numbers? Do you have any thoughts, work, ideas to show? Have you ever done any exercises where you have to prove a specific subset of a group is a subgroup before?
2d
comment Show that $D_n$ is a subgroup of Perm($\mathbb{C}$).
Multiplying by $e^{2\pi ir/n}$ adds $2\pi r/n$ to the phase of $z$. Use polar coordinates to see this. (This kind of fact is usually discussed quite early in any course on complex variables.) Now, what are you trying to show it's a subgroup of again? Clearly it's not a subgroup of $(\Bbb C,+)$ or $(\Bbb C,\times)$, so what operation are you thinking of having on $\Bbb C$ here?
Oct
22
comment Evaluate the angle between two curves at their intersection: $y=x^2+1, x^2+y^2=1$
Both equations have a corresponding graph. These graphs intersect at a unique point. What is the angle between them, or equivalently the angle between their tangent lines at that point?
Oct
21
comment Vandermond identity corollary $\sum_{v=0}^{n}\frac{(2n)!}{(v!)^2(n-v)!^2}={2n \choose n}^2$
The summands are $\displaystyle\binom{2n}{n}\cdot\binom{n}{v}^2$.
Oct
21
comment How to prove a subset is an ideal
You are overusing the letter $a$. You're supposed to prove that if $b\in L(a)$ (notice the new letter) and $r\in R$ then $rb\in L(a)$ also. In other words, you need to prove $ba=0\Rightarrow rba=0$. That last bit should be obvious to you. (Notice how I replaced $b\in L(a)$ with $ba=0$: the latter is what it means for $b$ to be in $L(a)$. Part of doing the basic textbook exercises is being able to unpackage the meaning of things. That's important.)
Oct
21
comment Using divisibility and greatest common divisor for a proof
Lemma: if $v\mid ab$ and $\gcd(v,b)=1$ then $v\mid a$. Application: $v\mid \frac{t}{u}u$ and $\gcd(v,u)=1\implies v\mid\frac{t}{u}$.
Oct
21
comment Riemann hypothesis: An query about the primes
The prime counting function has an explicit formula in terms of the zeros of the Riemann zeta function. The imaginary parts control the local oscillatory behaviour and the real parts control the long-term growth rate.
Oct
21
comment A Contradiction of Riemann Zeta Residues
Why should it imply $\zeta(-1)^2=\zeta(-3)$? You're the one making the claim; back it up if you can. Which you can't. Because it doesn't imply that. And that about wraps things up here. (Also, you are misusing the term "residues" - do you know what a residue is?)
Oct
20
comment How would I solve this congruence?
The group of units is cyclic. Fixing just means picking one, i.e. "let $g$ be a generator" and then write $r,a,b$ as a power of $g$. Then you are effectively working on a linear equation in the additive group $\Bbb Z/(p-1)\Bbb Z$.
Oct
20
comment How would I solve this congruence?
Fix a generator and rewrite the congruence using it; then it will be a linear problem.
Oct
19
comment Suppose two transitive G-sets X and Y are isomorphic as G-sets. Show that the two corresponding actions have the same kernel.
Thoughts, ideas, work?
Oct
17
awarded  Necromancer
Oct
16
comment Definition of Representation in terms of Group Action
The Wikipedia quote is correct, and is using the normal definition of a group action. A representation is a special case of a group action, so WP is right to say a representation defines a group action. Your claim "according to Wikipedia, for finite groups an equivalent definition is an action of $G$ on $V$" is wrong though: just because a representation defines ("defines" here being interchangeable with "induces" or "is") a group action does not mean that just any group action defines a representation; it must be linear.
Oct
16
comment Definition of Representation in terms of Group Action
@goatman2743 A barebones group action on $V$ would be $G\to{\rm Perm}(V)$ where $V$ is the vector space. The group of autobijections/permutations ${\rm Perm}(V)$ of the underlying set of $V$ is much bigger than ${\rm GL}(V)$, which is by definition the group of linear transformations of $V$ (it is called the "general linear group" if you want to look it up). For instance, a two-dimensional real representation would look like $G\to{\rm GL}_2(\Bbb R)$, but the so-called affine group ${\rm Aff}(\Bbb R^2)$ acts on $\Bbb R^2$ by some maps that are not additive (e.g. translations).