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### What Makes Polynomials Relatively Prime?

```Date: 11/20/2007 at 19:32:24
From: Chris
Subject: Relatively prime polynomials.

Why are polynomials whose only common factors are constants considered
'relatively prime'?  Why are the common constants not considered?

For example, 3x + 6  and 3x^2 + 12  are considered relatively prime
even though they have a common constant factor of 3.

```

```
Date: 11/21/2007 at 12:13:33
From: Doctor Vogler
Subject: Re: Relatively prime polynomials.

Hi Chris,

Thanks for writing to Dr. Math.  That's a good question.  It reminds
me of a computer program called "magma" which can give you both
answers (gcd = 1 or gcd = 3) depending on how you define your
polynomial.  And it is correct.  It depends on what kind of polynomial
you are talking about.

Just like the integers, polynomials form a ring, which means that you
can add and subtract polynomials, and you can multiply polynomials and
still get a polynomial, but you can't always divide two polynomials
(because this doesn't always result in a polynomial).  This is true
whether you allow the coefficients to belong to some field (like
rational numbers, or real numbers, where you *can* always divide by
nonzero numbers) or to another ring (like the integers).

A ring has some units, which are numbers that have inverses (like -1
is a unit in the integers, because its inverse 1/-1 is also an
integer).  All units in a polynomial ring are constant polynomials,
polynomials of degree 0.  When you speak of "unique factorization" in
a ring, there are primes and then there is a leftover unit.  For
example, if you factor the polynomial

3x^2 - (1/2)x - 1/2,

you get

3(x - 1/2)(x + 1/3).

Or you might prefer to say

(x - 1/2)(3x + 1),

but who's to say which is "more factored"?  These are polynomials over
the field of rational numbers, and all rational numbers have inverses,
so 3 is a unit (with inverse 1/3).  So whether you consider 3x + 1 or
x + 1/3 to be the preferred representation of the prime factor, you
get the unit 3 leftover.

But what if we considered the polynomial

6x^2 - x - 1 ?

This is just 2 times the previous polynomial, so I might say that it
factors as

6(x - 1/2)(x + 1/3),

and this would be correct.  This would be treating it as a polynomial
with rational coefficients, which it is.  But it is also a polynomial
with integer coefficients, and if you factor it as a polynomial in the
ring of polynomials with integer coefficients, then you can't get

6(x - 1/2)(x + 1/3),

because these factors are not polynomials with integer coefficients.
But you can factor it as

(2x - 1)(3x + 1),

and you can factor it as

-(1 - 2x)(3x + 1)

or

(1 - 2x)(-3x - 1).

But -1 is the only unit that we can move around.  So now let's look
back at your example.  You have the two polynomials

3x + 6  and  3x^2 + 12

If we treat these as polynomials with rational coefficients (or real
coefficients, or complex coefficients), then they have no common
factors except units, so you can take a linear combination of them
which is 1, such as

(1/12 - (1/24)x)(3x + 6) + (1/24)(3x^2 + 12) = 1.

But if you treat these as polynomials with integer coefficients, then
3 is no longer a unit, so there is no way that you can take a linear
combination of them and get 1, because if f(x) and g(x) are
polynomials with integer coefficients, then

f(x)(3x + 6) + g(x)(3x^2 + 12)

with have all of its coefficients divisible by 3.  Their gcd is, in
fact, 3.

So you see that the whole question is whether you treat these
polynomials as polynomials with rational coefficients or as
polynomials with integer coefficients.  It changes how the
factorizations are done, and it changes the meaning of "relatively
prime."  Your polynomials are relatively prime in one ring, but not in
the other.  Does that make sense?

back, and I will try to offer further suggestions.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Modern Algebra
College Number Theory

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