9h
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?
1d
comment Vandermond identity corollary $\sum_{v=0}^{n}\frac{(2n)!}{(v!)^2(n-v)!^2}={2n \choose n}^2$
The summands are $\displaystyle\binom{2n}{n}\cdot\binom{n}{v}^2$.
1d
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.)
1d
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 $\gcd(v,u)=1\implies v\mid\frac{t}{u}$.
1d
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.
1d
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?)
1d
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 $\Bbb Z/(p-1)\Bbb Z$.
1d
comment How would I solve this congruence?
Fix a generator and rewrite the congruence using it; then it will be a linear problem.
2d
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).
Oct
16
comment Definition of Representation in terms of Group Action
A group action on a vector space does not need to be linear. In order to define a representation as a group action, you must specify that it is a linear action.
Oct
16
comment Why isn’t f(6Z + n) = 9Z + 2n a homomorphism from Z/6 to Z/9, even though g(n) = 2n is a homomorphism from Z to Z?
$n+6\Bbb Z\mapsto 2n+6\Bbb Z$ is a group homomorphism $\Bbb Z/6\Bbb Z\to\Bbb Z/6\Bbb Z$. In fact, $[n]\mapsto[2n]$ is a group homomorphism $\Bbb Z/m\Bbb Z\to\Bbb Z/m\Bbb Z$ for any $m$. It is not, however, a homomorphism of unital rings (and neither is $n\mapsto 2n$ a ring endomorphism of $\Bbb Z$). Of course, if you actually did mean $f(6\Bbb Z+n)=\color{Red}{9}\Bbb Z+2n$ and that's not just a typo, then that $f$ is not even a map $\Bbb Z/6\Bbb Z\to\Bbb Z/6\Bbb Z$... Do you intend to fix/clarify your question?
Oct
7
comment What is the meaning of this notation in algebraic geometry (from /): $k\left[x_{1},\ldots,x_{r}\right]\mathbf{/\left(f_{1},\ldots,f_{r}\right)}$?
en.wikipedia.org/wiki/…
Oct
7
comment Show $\sum\limits_{b=0}^{p-1}\left(\frac{b}{p}\right) = 0$
Or one could use symmetry: argue it's invariant under multiplication by $-1$. (Since $\left(\frac{\cdot}{p}\right)$ is a group homomorphism on $\Bbb F_p^\times$, and you can pick $t$ for which $\left(\frac{t}{p}\right)=1$.)
Oct
7
comment Let $\alpha$ and $\beta$ be disjoint cycles. Prove for every positive integer n, $(\alpha\beta)^n=\alpha^n\beta^n$
But how does your conclusion follow from your observation? I would first prove commutativity: $\alpha\beta=\beta\alpha$.
Oct
5
comment How do I prove that if gcd(n,m) divides a-b, then $x\equiv a \pmod n$ and $x\equiv b \pmod m $ has a solution?
Break the problem up so that you do have relative primality.
Oct
4
comment Normal subgroup of a free product : how does $f_1(h)f_2(h)^{-1}$ generate a normal subgroup of $G_1 * G_2$?
@ggfgfg First off, $f_1(h)f_2(h)^{-1}$ is an element (depending on $h$). What you really need to be talking about is the subset $\{f_1(h)f_2(h)^{-1}:h\in H\}$. But that is not a subgroup, and nobody claims it is. It does generate a subgroup, $\langle f_1(h)f_2(h)^{-1}:h\in H\rangle$, but that is not guaranteed to be normal, and again nobody claims it is. Rather, we are constructing "the normal subgroup generated by" the subset $\{f_1(h)f_2(h)^{-1}:h\in H\rangle$.
Oct
4
comment Normal subgroup of a free product : how does $f_1(h)f_2(h)^{-1}$ generate a normal subgroup of $G_1 * G_2$?
It's not saying the subgroup they generate is normal. Terminology: The subgroup generated by a subset is the unique subgroup minimal among those containing that subset; similarly, the normal subgroup generated by a subset is the unique subgroup which is minimal among normal subgroups containing that subset.