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#1 2009-08-27 08:28:15

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. smile

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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#2 2009-08-28 11:49:08

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 108,456

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)


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.

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