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JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

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.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Number of roots of polynomials

Hi Jane;

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)

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