Graduate student mostly dabbling in algebraic geometry.

2d
answered How many zeros does $g(z)= z^4+iz^3 +1$ have in the first quadrant?
Jul
1
comment Rotman's exercise 2.8 "$S_n$ cannot be imbedded in $A_{n+1}$"
Though it seemed like a good idea at 03:00 AM last night, I see now that my proof doesn't work. But perhaps it can be saved. I'll give it some more thought.
Jul
1
comment Galois groups of quintics
My (belated) thanks for your very clear answer, it has been a great help. However, the case of a quintic $f\in\Bbb{Z}[X]$ with Galois group of order $10$ is left unanswered. I have found that $$f=X^5+X^4-5X^3-4X^2+3X+1,$$ with discriminant $401^2$ does the trick, but I can only show it by means of 'magic'. Is there a method to find such polynomials like the method you describe for abelian groups?
Jul
1
revised Convergence of $\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$
Corrected the argument for the case $z=\pm n$.
Jul
1
comment Convergence of $\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$
@DanielFischer Thanks for the clarification! All these sequences and their converence are things pretty far outside my field (I haven't deal with such things since basic undergraduate calculus, years ago), so I went with the naïve idea that a product converges whenever it contains a factor $0$. But if I'm not mistaken the exact same argument will hold if $z=n$. I'll double-check and edit soon.
Jul
1
revised Convergence of $\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$
Added a second, more complete answer.
Jul
1
comment Convergence of $\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$
The disk of converges of that series is (at most) $|z|<(2\pi^{-2})^{-1}$, as we certainly cannot have convergence for $\log(0)$. So your question about $z=1$ is a good one; it shows that my answer leaves a big part of the problem open. I will revise my answer.
Jul
1
comment Convergence of $\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$
@Dr. MV I think there's a typo there. And it is currently 03:39 AM here, so I'll look at what i wrote again tomorrow.
Jul
1
comment Convergence of $\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$
The series converge on all of $\Bbb{C}$ except the $z\in\Bbb{C}$ for which we get $\log(0)$, so except for $$z=\pm(\pi\sqrt{2})^{-1}i.$$ So yes, if $z=1$ it converges to $\log(1+2\pi^2)$.
Jul
1
revised Rotman's exercise 2.8 "$S_n$ cannot be imbedded in $A_{n+1}$"
added 75 characters in body
Jul
1
answered Rotman's exercise 2.8 "$S_n$ cannot be imbedded in $A_{n+1}$"
Jun
30
asked Counterexample to a variation on "The politician theorem".
Jun
30
awarded Nice Question
Jun
30
revised Convergence of $\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$
added 107 characters in body
Jun
30
answered Convergence of $\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)$
Jun
30
comment Aut$(G)\cong \Bbb{Z}_8$
I expanded my answer to settle the question for finitely generated abelian groups, and in effect to all abelian groups. I might rephrase my answer later to make this clearer.
Jun
30
revised Aut$(G)\cong \Bbb{Z}_8$
Expanded answer
Jun
30
answered Aut$(G)\cong \Bbb{Z}_8$
Jun
30
awarded Enlightened
Jun
30
awarded Nice Answer
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