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of degrees and respectively, we already know how to determine whether and have a common factor: we simply preform Euclid's algorithm on and (or, to say the same thing in a rather silly way, we compute a Gröbner basis for the ideal ). If the coefficients of or were changed, however, we would have to start over with Euclid's algorithm, i.e. we can't just preform Euclid's Algorithm on the general polynomials of degree and . It might be useful then if it were possible to write down (for fixed and ) a polynomial in the coefficients of and which is zero precisely when and have a common factor.
In order to do this, we look again at a question we've studied before: given polynomials and of degrees and respectively and another polynomial , what are the solutions to the equation
for polynomials and ? Let us first assume that and have no common factors. Then it is possible to solve the equation
and hence we may solve the equation for any polynomial : certainly is a solution. To find all the solutions, we just note that any two solutions must differ from one another by a solution to the equation
This equation, however, is much easier to solve: we can rewrite it as
and since
and
have no factors in common, this means that the solutions to this equation are
,
The fact that if
and
and
have no common factors, then
of course follows from the uniqueness of the factorization of a polynomial into irreducible polynomials, but we can also show it directly. We know from Euclid's Algorithm that we can write
parametrized by an arbitrary polynomial .
If we want to find a solution that minimizes the degree of , we simply note that there are unique polynomials and such that and by the division algorithm. Thus we see that there is a unique solution to the original equation in which . Similarly, it can be shown that there is a unique solution in which . Moreover, if , then these two solutions are the same, since if , then so , and as well.
Proposition Let be polynomials over a field of degrees and , respectively. Then for any of degree less than , there are unique polynomials with and such that
if and only if and have no common factors.
We've just shown that if and are relatively prime, then the above equation has a unique solution. We are left with the “only if” part: we must show and do have a common factor, then either the existence or the uniqueness of the solutions must fail.
In fact, both fail. Existence fails because we can not solve the equation , since any common factor of and must also divide 1. Uniqueness fails even for where a solution exists because if is a common factor of and , then is another solution to with and .
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