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? If you have any questions about this or need more help, please write back, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.