Galois Theory ProofsDate: 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/ |
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