Thread: f|g wy?

Let be polynomials with rational coefficients, and is irreducible. Suppose further that there exists a complex number such that . Prove that

I do not know how to prove. and I've thought it for many times....Would you help me out?

consider the map from Q[x] to Q(α) given by: p(x)-->p(α). this is a ring homomorphism.

furthermore, since α is algebraic over Q (it satisfies f(x) in Q[x]), Q[a] = Q(α) (that is, Q[α] is actually a FIELD).

now the image of this homomorphism contains Q (we can take as the pre-image of q in Q the constant polynomial q) and α (which has pre-image x), and is therefore surjective.

by the fundamental isomorphism theorem (for rings), Q(α) = Q[x]/I, for some ideal I of Q[x] (namely, the kernel of our homomorphism).

now Q[x] is a principal ideal domain, so I is generated by some polynomial, h(x) in Q[x].

since f(x) is in the kernel (since f(α) = 0), we have f(x) = h(x)k(x), for some polynomial k(x) in Q[x].

but f(x) is irreducible, so k(x) must be a unit (that is, an element of Q*). but then f(x) = qh(x), so h(x) = (1/q)f(x).

this means that (h(x)) = (f(x)).

now g(α) is also in the kernel, which is generated by f(x), so g(x) = f(x)s(x), for some polynomial s(x) in Q[x], that is: f|g.

Thank you very much. I need to spend some time writting your words in the language of only linear algebra...

divisibility is not a linear algebra concept, as it deals with multiplication, and a vector space need not have one. the set of polynomials over a field does indeed form a vector space, but it is also a RING, this is important.

divisibility is indeed a linear algebra concept. And this problem is a problem from linear alegebra in a entrance examination for graduate.

shall i ask you how? what does it mean for one vector to divide another?

i'm not sure what you MEAN by "the language of linear algebra". if you explain it to me, perhaps i can put what i said above in different terms.