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Last year Mathlover posted a very good question about Galois theory: specifically, the existence of funtions of roots which map to each other under permutations of those roots. You can see his question here: Number of distinct $f(x_1,x_2,x_3,\ldots,x_n)$ under permutation of the input

The question was answered quite well by Generic Human but I am still unable to close the loop on the question of the resolvent of the fifth degree equation. I know that if A, B, C and D are the roots of the fourth degree, then there are three functions which are the roots of a cubic equation:

$AB+CD = p$

$AC+BD = q$

$AD+BC = r$

I know how to construct the cubic equation whose roots are $p$, $q$, and $r$ and I assume this is what people mean when they talk about the resolvent of the quartic being a third-degree equation.

I also know that people say the resolvent of the quintic is a sixth-degree equation. What I don't know how to do is construct six functions in $A$, $B$, $C$, $D$ and $E$ which, by analogy with my $p$, $q$ and $r$ for the 4th-degree, are mapped to each other by permutations of the five roots.

I've almost convinced myself that no such functions exist, but then people say there is something called the "resolvent", which I'm assuming must be analogous to what I know for the case of the fourth-degree. So is there or isn't there...can anyone write out those six functions? (Or at least just one of them from which I can infer the other five by analogy?)

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Marty Green
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1 Answers1

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I initially intended this to post it as a comment, but it seems to get a bit bigger to do that. The answer is positive. In fact, there are lots of such resolvents. For example,

Runge's resolvent:

$$F(x_1,x_2,x_3,x_4,x_5,x_6)=(x_1-x_2)^2(x_2-x_3)^2(x_3-x_4)(x_4-x_5)^2(x_5-x_1)^2+(x_1-x_3)^2(x_3-x_5)^2(x_5-x_2)^2(x_2-x_4)^2(x_4-x_1)^2$$

Dummit's resolvent:

$$F(x_1,x_2,x_3,x_4,x_5,x_6)=x_1^2x_2x_5+x_1^2x_3x_4+x_2^2x_1x_3+x_2^2x_4x_5+x_3^2x_2x_4+x_4^2x_1x_2+x_4^2x_3x_5+x_5^2x_1x_4+x_5^2x_2x_3$$

Malfatti's resolvent:

$$F(x_1,x_2,x_3,x_4,x_5,x_6) = (x_1x_2 + x_2x_3 + x_3x_4 + x_4x_5 + x_5x_1 - x_1x_3 - x_3x_5 - x_5x_2 - x_2x_4 - x_4x_1)^2$$

Conjugates for all of them can be obtained by the permutations $(1, 2, 3), (1, 3, 2), (1, 2), (2, 3)$ and $(1, 3)$

Balarka Sen
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  • 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? – Marty Green Dec 26 '13 at 15:50
  • And the fact that I spent two days looking for one and couldn't find any...would that be a proof? – Marty Green Dec 26 '13 at 16:09
  • I searched with my dumb brain. oh my god you're just a kid! – Marty Green Dec 26 '13 at 16:59
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    You might want to check out my blog on this topic at http://marty-green.blogspot.ca/2013/12/in-which-i-get-help-from-unexpected.html – Marty Green Dec 26 '13 at 17:34
  • I suppose it must be obvious but the comments have been edited so I appear to be talking to myself. I don't know if Balarka removed himself or someone else did. – Marty Green Aug 15 '16 at 06:52
  • @MartyGreen Ah, yes, I did (I am sorry about that); I removed many of my older comments, including these, during the previous years' winterbash. There was apparently a hat for removing year-old comments or so. Feel free to remove your comments too if you think they look odd off-context. – Balarka Sen Aug 15 '16 at 09:33
  • Thanks, Balarka, But I already have a hat. ;^) – Marty Green Aug 15 '16 at 12:04
  • @MartyGreen But do you have one like this? – Balarka Sen Aug 15 '16 at 14:44