This can be proved using the equivalent of the division algorithm for polynomials over a field.
A simple proof of this using induction on the degree ofcan be found on p.28 of D.J.H. Garling, A Course in Galois Theory (1986), Cambridge University Press.
Note that this applies to polynomials over a field only. This wont work with rings in general, e.g.and in . However, if is an integral domain, we can apply the division algorithm to polynomials over its field of fractions .
Letbe a polynomial of degree in . Let be a root of Considering as a polynomial in , we can write where and . Then .
Thus. If is another root of distinct from , write where and . Then as .
Thus. Continuing this way, we find that if are all the distinct roots of , then we have for some . It follows that cannot have more than roots in since has degree and so cannot be factorized into more than linear factors in . And since a root of polynomial in is also a root of a polynomial in (Gausss lemma), our proof is complete.
Ifis not an integral domain, it may be possible for an th-degree polynomial in to have more than roots. For instance, as a polynomial over , the quadratic polynomial actually has 4 roots, namely 1, 3, 5, 7. Thats because is not an integral domain and so we cannot construct its field of fractions.
Thanks for the reason why x^2 + 7=0 mod 8 can have 4 solutions. I didn't know that.
Last edited by bobbym (2009-08-28 12:19:02)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.