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Galois groups and the symmetries of polynomials

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POSTED BY: Paul Abbott
4 Replies

Fascinating!

This is an application of group theory that chemists like me are rarely exposed to. We can take a molecule and its 3D geometry and work out which point group symmetry it possesses, or its chemical graph and the corresponding permutation-inversion group.

This post give me some ideas for other ways to explain symmetry in chemistry.

POSTED BY: Robert Nachbar

Very useful for error correction code as well.

POSTED BY: Shenghui Yang

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POSTED BY: EDITORIAL BOARD

Hi Paul,

thank you very much for this nice notebook. This really makes the article in the quanta magazine twice as valuable.

I only want to add one more sentence, which makes the difference between the two polynomials more transparent, hopefully.

For your function f(x) ( in the article it is denoted with g(x)) the expression of the roots $(a-b)(b-c)(c-a)=\pm 7$ is rational and therefore breaks half of the symmetries.

For the other function g(x), the expression of the roots $(r-s)(s-t)(t-s)\notin {\mathbb Q}$ is irrational and therefore it cannot be used to reduce the number of symmetries. $g(x)$ ends up with the full symmetric group as Galois group.

Finally, when you look at the radical expressions for the roots you can convince yourself, that the solutions for $f(x)$ are simpler than for $g(x)$. While the solutions for $g(x)$ contain nested roots the solutions for $f(x)$ don't.

POSTED BY: Johannes Martin
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