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### Galois Theory Proofs

```Date: 02/22/2003 at 17:00:23
From: Mikey
Subject: Sum of algebraic numbers is an algebraic number

Prove that a sum of 2 algebraic numbers is also an algebraic number.

I tried to find a formula to find the polynomial, but it seems
impossible.
```

```
Date: 02/23/2003 at 08:56:42
From: Doctor Jacques
Subject: Re: Sum of algebraic numbers is an algebraic number

Hi Mikey,

This is one of the first things one learns in Galois theory.

As I'm not sure of your background, I'll try to make up an elementary
(but correct) proof, and I'll outline the standard proof after that.

Let x and y be two algebraic numbers. This means that they are roots
of polynomials with rational coefficients; by dividing by the first
coefficient, we can assume that that coefficient is 1:

x^m + a_(m-1)*x^(m-1)..... + a_0 = 0
y^n + b_(n-1)*y^(n-1)..... + b_0 = 0

We can also assume that these polynomials have been chosen to have
the least possible degree.

Note that we can write these equations in another form:

(1)    x^m = -a_(m-1)*x^(m-1) .... - a_0

This means that we can express ANY rational polynomial in x in terms
of 1, x, ..., x^(m-1), with rational coefficients. This is obvious
for exponents less than m, and, whenever you have a power of x >= m,
you can replace it using the equation above to get a lower exponent,
and you can continue this until all exponents are < m.  As you start
with rational coefficients, and all the substitutions only involve
rational coefficients, the coefficients of the final polynomial will
be rational.

Let u = x + y. We want to find a polynomial:

f(u) = c_k*u^k + c_(k-1)*u^(k-1) + .... + c_0

such that f(u) = 0.

We will show that such a polynomial exists, and that its degree is at
most mn.

If we choose k = mn in f(u), we have k+1 coefficients c_i to
determine.

If we replace u by x+y in f(u), we get a polynomial in x and y. By
using equation (1) above, we can write this polynomial as a sum of
terms of the form:

(2)  d_ij*x^i*y^j

with

0 <= i <= m
0 <= j <= n

and the d_ij are linear expressions in the c_i, with rational
coefficients.

We will have f(u) = 0 if we can make all d_ij equal to 0.

For each pair (i,j), we have an equation like:

d_ij = e_0*c_0 + .... + e_k*c_k = 0

i.e. a homogeneous linear equation in c_0 ... c_k, with rational
coefficients. There are k = mn such equations, and there are k+1
unknowns (the c_i). Therefore, the system has always a non-zero
solution in c_i, and this gives the desired polynomial.

Let us illustrate this with an example (I'll use a very simple one,
since the algebra quickly gets out of hand).

We will show that sqrt(2) + sqrt(3) is algebraic.

We have the equations:

x^2 - 2 = 0
y^2 - 3 = 0

Using equation (1), we express the powers of x and y in terms of
polynomials of degree 1:

x^2 = 2      y^2 = 3
x^3 = 2x     y^3 = 3y
x^4 = 4      y^4 = 9

We seek a polynomial such that:

f(u) = c_4*u^4 + c_3*u^3 + c_2*u^2 + c_1*u_1 + c_0 = 0

where u = x + y.

We expand the powers of (x + y) using the binomial formula, and reduce
them using the relations above:

(x+y)^4 = x^4 + 4*x^3*y + 6*x^2*y^2 + 4*x*y^3 + y^4
= 4 + 8xy + 36 + 12xy + 9
= 20 xy + 49

(x+y)^3 = x^3 + 3*x^2*y + 3*x*y^2 + y^3
= 2x + 6y + 9x + 3y
= 11x + 9y

(x+y)^2 = x^2 + 2xy + y^2
= 2 + 2xy + 3
= 2xy + 5

(x+y)   = x + y

1       = 1

Our polynomial becomes:

f(u) =
(20*c_4 + 2*c_2)*xy +
(11*c_3 + c_1)*x +
(9*c_3 + c_1)*y +
49*c_4 + 5*c_2 + c_0

By equating all terms to 0, we obtain the system of equations:

c_0     + 5c_2        + 49c_4 = 0
c_1        +9c_3          = 0
c_1        +11c_3         = 0
c_2         + 10c_4 = 0

One solution of this system is:

c_0 = 1, c_1 = c_3 = 0, c_2 = -10, c_4 = 1

(Since the system is homogeneous, the solutions are only defined to
within a constant factor)

This gives the equation:

u^4 -10*u^2 + 1 = 0

and you can check that (sqrt(2) + sqrt(3)) is indeed a root of that
equation.

The standard proof is based on the same logic, but expressed in a
more abstract way. The general idea is:

* The products x^i*y^j, where i < m and j < n, generate a vector
space V over Q. Because of the equation (1), any expression x^i*y^j
(whatever the exponents) belongs to V. As V is generated by mn
vectors, its dimension is at most mn.

* By using the binomial theorem, we see that all the powers of x + y
belong to V.

* As the dimension of V is at most mn, any set of mn+1 elements is
linearily dependent. The powers (x+y)^i, for i = 0..mn, are such a
set.

* The conclusion is that these powers are linearly dependent. A
linear combination of these powers is just a polynomial in u with
rational coefficients, and therefore, there exists such a
polynomial f with f(u) = 0.

Please feel free to write back if you want to discuss this further.

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

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