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bio website marty-green.blogspot.com
location Canada
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visits member for 3 years, 3 months
seen Aug 2 at 14:49

Jul
30
answered Roots for quintic equations
Jul
11
awarded  Enlightened
Jul
11
awarded  Nice Answer
Jul
2
awarded  Curious
Jun
6
accepted Resolvent of the Quintic…Functions of the roots
Jun
6
revised Intuitive reasoning why are quintics unsolvable
point to follow-up discussion
May
13
awarded  Yearling
Apr
30
answered Galois group of $x^3 - 2 $ over $\mathbb Q$
Feb
19
awarded  Nice Question
Jan
22
awarded  Talkative
Dec
26
comment Resolvent of the Quintic…Functions of the roots
You might want to check out my blog on this topic at marty-green.blogspot.ca/2013/12/…
Dec
26
comment Resolvent of the Quintic…Functions of the roots
I searched with my dumb brain. oh my god you're just a kid!
Dec
26
comment Resolvent of the Quintic…Functions of the roots
And the fact that I spent two days looking for one and couldn't find any...would that be a proof?
Dec
26
answered Number of distinct $f(x_1,x_2,x_3,\ldots,x_n)$ under permutation of the input
Dec
26
comment Resolvent of the Quintic…Functions of the roots
Thank you for this very helpful answer. Can it be that Dummit's resolvent was only discovered in the last 20 years? (Dummit is a prof at U of Vermont.) I calculated the conjugates as you suggested, and then spent the last two days searching for permutations which map all the conjugates to themselves. I've just about concluded that there are no such non-trivial permutations...is that right? For example, each 5-cycle maps one of the conjugates to itself, but shuffles the others around...right?
Dec
25
awarded  Benefactor
Dec
24
awarded  Promoter
Dec
21
asked Resolvent of the Quintic…Functions of the roots
Dec
21
comment Number of distinct $f(x_1,x_2,x_3,\ldots,x_n)$ under permutation of the input
I added an update to this question but I don't know what happened to it...its not there anymore.
Dec
20
comment Number of distinct $f(x_1,x_2,x_3,\ldots,x_n)$ under permutation of the input
@GenericHuman: excellent answer which I am not smart enough to completely understand. I hope you will look at my new update to this question and respond further.