Lior B-S
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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 added 43 characters in body 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. added 355 characters in body 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