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Feb
2
reviewed Approve How find this $\sum\limits_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}$
Jan
23
reviewed Approve if $\ f(f(x))= x^2 + 1$ , then $\ f(6)= $?
Jan
23
reviewed Reject and Edit A partial derivative problem related with elasticity of substitution in Advanced Micro
Jan
23
revised A partial derivative problem related with elasticity of substitution in Advanced Micro
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Jan
20
comment Quantitative aspect of Chebotarev Density Theorem
The density of primes in $K$ that are principal is $\frac{1}{h_K}$. This follows from class field theory and Chebotarev.
Jan
14
comment Is there any flaw in the following proposed elementary " proof'' of FLT?
@Isaac: Yes, but you don't know that FLT holds (in the way you stated) even for $z = 1$, as any integral solution $x^n + y^n = z^n$ is exactly the same as a rational solution $(x/z)^n + (y/z)^n = 1$.
Jan
14
comment Is there any flaw in the following proposed elementary " proof'' of FLT?
Yes, but in an inductive proof you have to have a base case. Your base case (which you implicitly assume) is just as hard as FLT in the first place!
Jan
8
comment When is $\mathbb{Z}[\theta]$ a Dedekind domain for an algebraic number $\theta$?
Relevant discussion on MO
Jan
8
answered Units become powers when lifted to unramified extensions?
Jan
7
comment How are unramified extensions of number fields formed?
You are missing the adjective "abelian" throughout your answer. For example, the maximal unramified extension of an arbitrary number field can have degree much larger than $h_K$, although $\mathbf{Q}$ still has no unramified extensions. Also, every abelian extension is contained in a ray class field, but does not need be one.
Jan
7
revised A function of two quadratic forms
edited tags
Dec
7
comment Doubly periodic function is not a rational function
Hint: Look at the poles of $f$.
Dec
7
comment Understanding the proof of this theorem leading up to Dedekind's theorem
Exactly! $\sigma(I) = \{\sigma(\alpha) \, | \, \alpha \in I\}$, so if $p \in \mathbf{Q}$, $\sigma((p)) = \{\sigma(p \cdot \alpha) \, | \, \alpha \in \mathcal{O}_K\} = \{p \cdot \sigma (\alpha) \, | \, \alpha \in \mathcal{O}_K\} = p \mathcal{O}_K = (p)$.
Dec
7
comment Understanding the proof of this theorem leading up to Dedekind's theorem
An ideal $I$ being Galois stable just means that $\sigma(I) = I$ for all $\sigma \in \mathrm{Gal}(K/\mathbf{Q})$.
Dec
7
comment Understanding the proof of this theorem leading up to Dedekind's theorem
That's what the last paragraph is about. $(p)$ is Galois stable, so $x^{p^m} \in (p)$ implies that $\mathrm{Tr}(x^{p^m}) \in (p) = p \mathcal{O}_K$. Since the trace is rational, this must actually be in $p \mathcal{O}_K \cap \mathbf{Q} = p \mathbf{Z}$.
Dec
7
answered Understanding the proof of this theorem leading up to Dedekind's theorem
Dec
6
reviewed Approve A problem related to cyclic group (group theory )
Nov
29
comment Integer solutions of $x^3+y^3=z^3$ using methods of Algebraic Number Theory
@PITTALUGA $\zeta_3 = \frac{1 + \sqrt{-3}}{2}$.
Nov
11
revised Is there a genus-one curve over $\mathbb{Q}$ with no points over any solvable extension?
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Nov
11
answered Is there a genus-one curve over $\mathbb{Q}$ with no points over any solvable extension?