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3h
comment Inverse of a constant function on an open set
This has nothing to do with complex analysis or Riemann surfaces. It's purely to do with sets and functions. Do you know what an inverse function is?
7h
comment Is this a valid way of solving modular equations?
This is valid, but it's often more convenient to stick to modular arithmetic than pulling it back to familiar arithmetic. Since $5\equiv-1$ mod $6$, your equation is $-x\equiv 1$ mod $6$, yielding $x\equiv -1$ mod $6$. Note how this mimics the usual operations of arithmetic, but all done in modular arithmetic.
9h
comment Finding the fixed Field of $\sigma \in Aut(\mathbb{R}(t)/\mathbb{R})$
@Lubin $t+t^3$ and $t-t^3$ are both odd, and $\frac{t+t^3}{t-t^3}=\frac{1-t^2}{1+t^2}\in\Bbb R(t^2)$.
10h
comment $ \sum_{n=1}^{\infty}\frac{1}{(an^2+b)^k} $ equals $q_0+q_1\pi+\dots+q_k\pi^k$ for nonzero rational coefficients
So you want every $q_0,\cdots,q_k$ to be nonzero? Weird question.
10h
comment $ \sum_{n=1}^{\infty}\frac{1}{(an^2+b)^k} $ equals $q_0+q_1\pi+\dots+q_k\pi^k$ for nonzero rational coefficients
As a user said in a deleted answer, $a=1,b=0$ works since it gives $\zeta(2k)$.
11h
revised Show that $\mathbb{C} \otimes_\mathbb{Z} \mathbb{C} \cong \mathbb{C} \otimes_\mathbb{Q} \mathbb{C}$
added 22 characters in body
11h
comment Show that $\mathbb{C} \otimes_\mathbb{Z} \mathbb{C} \cong \mathbb{C} \otimes_\mathbb{Q} \mathbb{C}$
@user114539 Even though when I wrote the parenthetical I was only thinking of the isomorphism as being of $D$-algebras, one can also interpret it as true of $F$-algebras too - the way to make scalars in $F$ act on $A\otimes_DB$ should be obvious (at least after knowing $ra\otimes b=a\otimes rb$ for all $r\in F$ from this answer).
1d
comment What is $Hom((S^1)^k , (S^1)^n)$?
Right. Anyway, are you sure the exercise doesn't specify continuous group homomorphisms?
1d
comment What is $Hom((S^1)^k , (S^1)^n)$?
$\hom(G,H)$ behaves somewhat like multiplying $G$ and $H$ - it distributes over finite direct sums (in both arguments), the same way multiplication distributes over addition. So we have $$\hom\left(\bigoplus_i G_i,\bigoplus_j H_j\right)=\bigoplus_i\bigoplus_j\hom(G_i,H_j),$$ which is exactly analogous to $$\left(\sum_i a_i\right)\left(\sum_j b_j\right)=\sum_i\sum_j a_ib_j. $$
1d
comment A property of finite field of order $2^n$
And $3\nmid 2^n-1$ because $n$ is odd. Right.
1d
comment Prove $f(A_1 \cap A_2 \cap A_3 \cap \dotsb \cap A_n)=f(A_1) \cap f(A_2) \cap f(A_3) \cap \dotsb \cap f(A_n)$
What do those things mean? (I know what they mean - but it's important that you unpackage the definitions and meanings. Sometimes proving a claim is as simple as knowing explicitly what the claim is saying.)
1d
answered In $\mathbb Q_p$, proving every open ball is the disjoint union of more than one open ball
2d
comment What is the mathematical distinction between closed and open sets?
I am not sure I understand your comment Ms. Tank.
2d
comment What is the mathematical distinction between closed and open sets?
Closed sets contain all their boundary points, open sets contain none of theirs. In a metric space, closed sets can be so sparse they contain no metric balls at all, unlike open sets which have "enough space" that there is a ball around every point. As for facts that use the adjectives open/closed - they're almost surely using the adjectives for a reason. One wouldn't e.g. state a fact for open sets if it were also true for closed sets for the same reasons. Literally pick any such claim and consider it an exercise to see what goes wrong if you alter its adjectives "open" and "closed."
2d
answered Are binomial series multiplicative in their bases?
2d
comment Do there exist nontrivial quotient groups of arbitrary finite order?
Not only is every natural number the size of a quotient group, but every group is isomorphic to a quotient group.
2d
comment What is the $\lim_{n\rightarrow \infty }(1+\frac{1}{n})^{n^n}$
Saying "for sufficiently large $n$" saves us the trouble of proving e.g. monotonicity of $(1+\frac{1}{n})^n$.
2d
comment Confused about transcendental numbers
@Ghassan As Steven just said, there is no such thing as "super transcendental."
2d
revised DVR, power series expansion.
deleted 208 characters in body
2d
comment Is $\mathbb{Q}(\pi) \cong \mathbb{Q}[[x]]/ \langle \sin(x) \rangle$?
Ah, derp. Nevermind! :-)