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Two Integers and a Third Degree Polynomial: Square in Z? A Galois Theory Proof

Date: 07/28/2010 at 08:51:40
From: shams
Subject: show-4a^3-27b^2  square in Z for infintely many a,b integ

This question comes from page 325 of a book by Serge Lang, where it is
exercise 24.

Prove that there are infinitely many non-zero integers a,b such that ...

   -4a^3 - 27b^2
   
... is square in Z. Here, Z stands for ring of integers.

I tried to solve this by contradiction. That is, I assumed that there are
finitely many non-zero integers a,b such that -4a^3 - 27b^2 is square in
Z. But then how do I proceed to get a contradiction?

I know basic algebra, groups, rings, fields, etc. 



Date: 07/30/2010 at 06:28:57
From: Doctor Jacques
Subject: Re:  show-4a^3-27b^2  square in Z for infintely many a,b integ

Hi Shams,

There is a simple answer: if we write ...

   d(a,b) = -4a^3 - 27b^2

... then we have

   d(ak^2, bk^3) = d(a,b)*k^6

This shows that, if d(a,b) is a square in Z, and k is any integer, then
d(ak^2, bk^3) is also a square in Z. As there is at least one solution --
d(-3,1) = 81 -- there are infinitely many solutions.

But this exercise appears in Lang's chapter on Galois theory, so there may
be a deeper answer.

d(a,b) is the discriminant of the polynomial:

   f(x) = x^3 + ax + b       [1]

And this discriminant is a square whenever the Galois group of f(x) is
cyclic of order 3. In particular, if we compare this exercise with the
previous one in the book, it would seem that this one asks us to prove
that there are infinitely many such polynomials.

Although the previous method allows you to find an infinite set of
solutions, these solutions are essentially equivalent, because the
corresponding extensions of Q are the same.

It would be much more interesting to prove that there are infinitely many
algebraic extensions of Q, the Galois group of which is C3.

We need to use the fact that there are infinitely many primes of the form
3k + 1. The proof of this relies on a very difficult theorem of Dirichlet
about primes in arithmetic progression; see, for example:

    http://mathworld.wolfram.com/DirichletsTheorem.html 

(I think Euler found a simpler proof for this particular case, but I don't
know it).

Now, if p is a prime of the form 3k + 1, then the Galois group of the
cyclotomic polynomial of order p is cyclic of order (p - 1) = 3k.

Any cyclic group G of order 3k contains a cyclic (normal) subgroup H of
order k. The fixed field of H is a Galois extension (because H is normal)
of degree 3, the Galois group of which is G/H = C3.

If f(x) is the minimal of any irrational element of that extension, then
the discriminant of f(x) will be a square in Q. This almost solves the
problem, except for the following issues:

(1) f(x) may not be of the form [1]; it may contain a term in x^2.

(2) the coefficients of f(x) may not be integers, although they will
    be rational.

To address these issues, we need to choose an appropriate element of the
field extension.

Consider first issue (1). Assume that x is in the cubic extension in
question, and that the minimal polynomial of x is:

   f(x) = x^3 + ax^2 + bx + c

If a ≠ 0, then we can replace x by y - a/3, which will eliminate the term
in y^2, and not change the discriminant.

Let us now consider issue (2). If the coefficients are not integers, then
x is not an algebraic integer, so we have x = y/k, where y is an algebraic
integer and k is a plain integer (an element of Z). Now, y is in the same
field extension, and its minimal polynomial g(y) will have integer
coefficients. And it has no y^2 term if f(x) has no x^2 term. As the
Galois group of the extension is C3, the discriminant of g(y) will be a
rational square, and, as g(y) has integer coefficients, its discriminant
will be the square of an integer.

To illustrate the technique, let us consider the cyclotomic extension K of
order 13. The Galois group is U(12), the multiplicative group of integers
modulo 12. This group is cyclic of order 12, and a generator is given by
the automorphism ...

   s(x) = x^2,

... where x is a primitive 13th root of unity.

This group contains a subgroup of order 4, generated by s^3 
(s^3(x) = x^8). To find the corresponding subfield, we look at the orbit
of x under s^3; this orbit is:

   {x, x^8, x^12, x^5}

This implies that the element ...

   y = x + x^8 + x^12 + x^5

... is in the fixed field of s^3, and generates a cyclic extension of
degree 3. With a lot of work (or perhaps some software), we find that the
minimal polynomial of y is:

   f(y) = y^3 + y^2 - 4y + 1

To get rid of the term in y^2, we let y = z - 1/3. The minimal polynomial
of z is:

   g(z) = z^3 - (117/27)z + 65/27

As the denominators show, z is not an algebraic integer. We have 
N(z) = -65/27, which suggests that u = 3z may be an algebraic integer.
Indeed, the minimal polynomial of y is ...

   h(u) = u^3 - 39u + 65,

... which has integer coefficients. 

Now check that the discriminant of h is ...

   d(-39,65) = 123201 = 351^2

... and, starting from these values of a and b, you can again generate an
infinity of solutions, using the technique explained at the beginning.
These solutions correspond to the minimal polynomials of integer multiples
of u.

Please feel free to write back if you require further assistance.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Modern Algebra

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