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Number of roots of polynomials
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 of can 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 won’t 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 .
Let be 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 (Gauss’s lemma), our proof is complete.
If is 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. That’s because is not an integral domain and so we cannot construct its field of fractions.
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Re: Number of roots of polynomials
Last edited by bobbym (2009-08-29 10:19:02)
In mathematics, you don't understand things. You just get used to them.
No explanations, just DGA.
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