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Jul
1
comment connection between absolute irreducibility and smooth+geometrically connected
So the condition neededto get smoothness is that the branch loci are disjoint?
Jul
1
comment connection between absolute irreducibility and smooth+geometrically connected
Mohan, I forgot to say that the fields are linearly disjoint. Fixed the question
Jul
1
revised connection between absolute irreducibility and smooth+geometrically connected
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Jun
29
asked connection between absolute irreducibility and smooth+geometrically connected
Mar
11
awarded  Yearling
Dec
10
awarded  Excavator
Oct
31
awarded  Notable Question
Sep
9
comment Morse functions dense in a trigonometric polynomial space
With polynomials, one proves that for any $f(x)$, and for almost all scalar $a$, $f(x)+a x$ is Morse. Maybe you can try to imitate it here? (I think that with polynomials it is allowed that x is a tuple of variables, a is a tuple of scalars and ax is the scalar product, but I am not sure)
Aug
14
comment What is the non-trivial, general solution of these equal ratios?
I would suggest you editing the question...
Aug
14
comment What is the non-trivial, general solution of these equal ratios?
What about $a=0$, $b=7$, $c=7$? Do you need all solutions?
Aug
13
answered Help Determining Degree of a Field Extension
Aug
13
revised Question about calculating the degree of a finite field extension.
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Aug
13
answered Question about calculating the degree of a finite field extension.
Jul
30
comment N is a normal subgroup of G if $aNa^{-1} \subset N $ for all $a ∈ G$. Prove that in that case, $aNa^{-1} = N $.
Weirdness is a matter of whether you are used to it or not. In any rate you can write $N^{g^{-1}}$.
Jul
30
comment N is a normal subgroup of G if $aNa^{-1} \subset N $ for all $a ∈ G$. Prove that in that case, $aNa^{-1} = N $.
@blue It does matter: $N^g=g^{-1} N g$, since it is a right action. If you want a left action of $G$ on the conjugates of $N$, then you take the other one $^gN=N^{g^{-1}} = g N g^{-1}$.
Jul
30
comment $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$, $\theta$ must be in radians. But $x$ can be in degree for $\lim_{x\to\pm\infty}\frac{\sin x}{x}=0$?
I find it a bit strange, from the point of view of meta-mathematics, to consider $\sin x/x$ where $x$ is in degrees. Because, intuitively, $\sin x$ is length, so the units of $\sin x/x$ will be length/degrees, very strange to me. However, if $x$ is in radians, then $x$ is also length (the length of the arc) and so $\sin x/x$ is ratio between to length, and this does make sense to me...
Jul
29
answered Galois extension preserves irreducibility
Jul
28
comment Number of zero digits in factorials
This heuristic reasoning can be made into a theorem if we knew that $\log_{10}(n!)$ is equi-distributed modulo $1$. The latter seems to follow from van der Corput's inequalities, I don't know how yet, but see <a href="mathoverflow.net/questions/159393/…; related question in mathoverflow.
Jul
27
awarded  Necromancer
Jul
27
answered $F/K$ algebraic and every nonconstant polynomial in $K[X]$ has a root in $F$ implies $F$ is algebraically closed.