prove that the sum of two algebraic integer is an algebraic integer
If a , b are two algebraic numbers, then Q(a)=Q[a]/(p(a)) for some monic polynomial p∈Q[x] . Since b is algebraic over Q , it is also algebraic over Q(a) . Let q(x) be a minimal polynomial for b over Q(a) . Then Q(a,b)=Q(a)[b]/(q(b)) . Note that since [Q(a,b):Q(a)] is finite (equal to the degree of q ) and [Q(a):Q] is finite (equal to the degree of p ), [Q(a,b):Q] is also finite (equal to the product of the degrees of p and q ).
Note that a+b∈Q(a,b) . Consider the sequence of powers 1,a+b,(a+b)2,… . Since Q(a,b) has finite dimension as a Q -vector space, there will be some n so that 1,a+b,…,(a+b)n is a linearly dependent set over Q . (In particular, n>[Q(a,b):Q] will always work.) Write down a linear combination of these that yields 0:
∑ni=0ci(a+b)i=0,ci∈Q .
Then the polynomial ∑ni=0cixi is a polynomial with coefficients in Q that evaluates to 0 when plugging in a+b for x . This is the definition of algebraic, so a+b is also algebraic.
Comments
Leave a comment