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3h
comment Relation between runge domain and polynomial convexity
Sorry, but my contribution stops here: I don't wish to read that article.
11h
comment Relation between runge domain and polynomial convexity
More links: L.Kaup-B.Kaup , Shabat
11h
comment Relation between runge domain and polynomial convexity
No, no you are not arrogant at all . You have adopted the right atitude: defend your position and if you see that something is indeed wrong, say so. By the way there are excellent books on the subject, written by real masters: Grauert-Fritzsche, L.Kaup-B.Kaup, Shabat . You would certainly profit from them.
11h
comment Relation between runge domain and polynomial convexity
I mean exactly what I wrote. The point is that neither you nor the linked article define what a polynomially convex domain is but only what a polynomially convex compact set is. For polynomially convex domains you should require that compact subsets $K\subset \Omega$ have compact polynomial hulls $\hat K$, but not that $K=\hat K$. By the way, with your definition no nonempty domain in the universe is polynomially convex: I challenge you to find just one (even in $\mathbb C$) if you cling to your false definition !!
12h
answered Relation between runge domain and polynomial convexity
13h
comment Relation between runge domain and polynomial convexity
Your definition of polynomially convex is not correct: the strict inequality sign must be replaced by $\leq$ and you should require that $\hat K$ be compact but not that $\hat K=K$.
1d
comment Is the degree of an infinite algebraic extensions always countable?
@XG: yes, I understood that but I don't know the answer.
1d
comment Prove that for any $f_1,f_2,…f_k \in I$, there exists a point $x_0 \in [a,b]$ such that $f_1(x_0)=…=f_k(x_0)=0$.
+1. The same proof works for any compact space in place of $I$. Nice, literate pseudonym by the way.
1d
comment Is the degree of an infinite algebraic extensions always countable?
@X.G. I don't know for complete valued fields.
1d
comment Is the degree of an infinite algebraic extensions always countable?
@Asvin: yes, here is a proof for infinite fields (the algebraic closure of a finite field is denumerable, so that your result is true in that case too). Some anonymous coward downvoted my answer, but my challenge to find the least inaccuracy in it was of course not taken up.
1d
answered Is the degree of an infinite algebraic extensions always countable?
1d
awarded  galois-theory
1d
comment what is the decomposition of $(x^2+y^2+z^2)(x+y+z)(x+y-z)(-x+y+z)(x-y+z)-8x^2y^2z^2$?
Yes, that's right. The point is that results given by software are indeed practically always correct but are sometimes given in a form that is not what a mathematician expects, due to the use of general powerful algorithms . So it may sometimes be advantageous to slightly modify the presentation of the result.
1d
comment what is the decomposition of $(x^2+y^2+z^2)(x+y+z)(x+y-z)(-x+y+z)(x-y+z)-8x^2y^2z^2$?
We humans prefer the more symmetric $\left( {x}^{2}+{y}^{2}-{z}^{ 2} \right) \left( {y}^{2}+{z}^{2}-{x}^{ 2} \right) \left( {z}^{2}+{x}^{2}-{y}^{ 2} \right) $
1d
comment Number of elements in Hom$(S_n,\mathbb{C})$
Dear Alex, Lemma 4 of this handout proves that the the commutator subgroup of $S_n$ is $A_n$, so that the abelianization $S_n^{ab}=S_n/[S_n,S_n]$ is the two-element group. This allows you to conclude since $\text{Hom}(S_n, \mathbb C^*)=\text{Hom}(S_n^{ab}, \mathbb C^*)$ [Recall that for an arbitrary group $G$ and an abelian group $A$ we have $\text{Hom}(G,A)=\text{Hom}(G^{ab},A)$]
2d
comment Number of elements in Hom$(S_n,\mathbb{C})$
+1: Well said, Alex!
2d
comment Is a differentiable function on $(-2, 4)$ always integrable on $[-2, 4]$?
Why don't you you answer the question with the given domain ?
2d
comment Finite covering by finitely generated B-algebras
Read the Masters: EGA I, Chap.I, Proposition (6.3.3), page 145.
2d
comment Example of a divisor of a function
@user42912: the order of $z(x)= \frac{x}{1-x}$ at $x=1$ is $-1$, not $1$.This is because you can write $z(x)=(-x)\cdot (x-1)^{-1}$ and $-x$ is invertible in a neighbourhood of $x=1$, so it has order $0$ while $(x-1)^{-1}$ has order $-1$ (remember that the order of a product is the sum of the orders of the factors).
May
20
comment Example of a divisor of a function
The definition of degree of a rational function can be found in Exercise 6 of the Miscellaneous Exercises to Chapter 13 in Artin's Algebra (page 535) or on page 243 of Jacobson's Basic Algebra I, Exercise 11 to section 4.5.