Re: Elementary Galois theory

From: Michael Barr (barr_at_barrs.org)
Date: 06/22/04 

Date: 21 Jun 2004 18:35:07 -0700

sigoldberg1@yahoo.com (Seth Goldberg) wrote in message news:<decd2256.0406211454.1dc9ef7b@posting.google.com>...
> In a recent issue of the notices, Lars Garding at
>
> http://www.ams.org/notices/200403/rev-garding.pdf
>
> mentions that one reason that the the roots of the general rational
> equation of degree 5 or higher cannot be described in terms of the
> coefficients by means of repeated rational operations and extraction
> of roots is that there are n nth roots, so that in general there is
> no way to specify a particular root of a given equation from the much
> larger number of roots which will be generated from a given sequence
> of operations. He uses the example that
> (2+sqr(3))^(1/3)+(3+sqr(2))^(1/3) has 36 possible values.
>
> Is this the only reason for the unsolvability of the general such
> equation? In other words, does Galois' theorem with solvable groups
> etc. actually just translate into this, or is there something else
> involved?

There is most assuredly something else involved. Here is one way to
emphasize that point. Let Q be the rationals and A denote the
algebraic numbers. Let B be the subfield of A generated by all
possible root extractions. So all 16 values of the above expression
added to any of the seven values of the 7th root of i, multiplied by
... would all be in B. Then Galois showed that B is a proper subfield
of A.

What he did was to define (what we call) the Galois group of an
equation (any element of A is one of the roots of a polynomial
equation with rational coefficients) to be the set of all permutations
of the root of the equation that extend to an automorphism of A. Then
he showed (or we show, anyway, I am not sure exactly how Galois viewed
this) that that group has a property called solvability iff the
polynomial has all its roots in B. Finally, "most" equations of
degree 5 the full symmetric group on 5 elements as their Galois group
and that group is pretty easily shown not to be solvable.

Incidentally, there is a subfield of B consisting of all elements
whose Galois group has order a power of 2. These are exactly the
numbers constructible using a ruler and compass construction. Now
cube root is there unless the cube root is already rational. Since
the cube root of 2 is not rational (same proof as for the square
root), you cannot duplicate a cube with ruler and compass. A slightly
more complicated argument shows that you cannot construct a 20 degree
angle so that a 60 angle cannot be trisected.

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