# f|g wy?

• Nov 5th 2012, 08:06 PM
f|g wy?
Let $f(x),g(x)$ be polynomials with rational coefficients, and $f(x)$ is irreducible. Suppose further that there exists a complex number $\alpha$ such that $f(\alpha)=g(\alpha)=0$. Prove that $f(x)|g(x).$

I do not know how to prove. and I've thought it for many times....Would you help me out?
• Nov 5th 2012, 09:06 PM
Deveno
Re: f|g wy?
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.
• Nov 5th 2012, 11:24 PM
Re: f|g wy?
Thank you very much. I need to spend some time writting your words in the language of only linear algebra...
• Nov 6th 2012, 07:56 AM
Deveno
Re: f|g wy?
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.
• Nov 6th 2012, 03:50 PM