24plzz help
There are 2 irreducible polynomials f and g with rational coeff.Now let there be 2 complex numbers a and b s.t. f (a) = g (b) = 0. Prove that if a+b is rational number, then f and g must have same deg.
-
UP 0 DOWN 0 0 5
5 Answers
341This has to do with the concept of minimal polynomials.
[http://en.wikipedia.org/wiki/Minimal_polynomial_(field_theory)]
Let a = p+iq and b = r-iq where p,q,r are real. Let p+r = s where s is rational
Let f(x) = a_nx^n +a_{n-1}x^{n-1}+...+a_0.
and g(x) = b_mx^m +b_{m-1}x^{m-1}+...+b_0.
WLOG, say n>m.
Now f(x) is the minimal polynomial over Q with a as its root. Similarly g(x) is the minimal polynomial for b
g(b) = 0
i.e.\sum_{k=0}^m b_k(r-iq)^k = 0
\Rightarrow \sum_{k=0}^m b_k(s -p -iq)^k = 0 \Rightarrow \Rightarrow \sum_{k=0}^m b_k [s-(p+iq)]^k = 0
\Rightarrow \sum_{k=0}^m b_k(s -a)^k = 0
Thus a is a root of the rational irreducible polynomial \sum_{k=0}^m b_k(s -x)^k of degree m<n
Thus we have exhibited a polynomial over Q of degree<n that has a = p+iq as a root contradicting the minimality of f(x).
[By the way this is not JEE math, so if you sourced this from some JEE material God help us all!]
24thanx a lot sir...
and no this is not from any JEE source..[1]
So no need to disturb dear God[1]