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Jan
27
awarded  Nice Question
Jan
8
awarded  Nice Question
Dec
6
comment Vague definitions of ramified, split and inert in a quadratic field
Do you assume that $n$ is squarefree in the calculation of the discriminant?
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