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For two polynomials, there exist a method to find their monic GCD by a variation of Euclid's algoritm, is there any method exist for finding the exact nonmonic factor between two polynomials?


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What kind of polynomials? – Qiaochu Yuan May 2 '12 at 22:32
I think you misunderstood that Wikipedia page. The Euclidean algorithm method finds a monic GCD $d$ of any two polynomials $f$ and $g$ (over a field). Whether or not $f$ and $g$ are monic is irrelevant. – Robert Israel May 2 '12 at 22:33
If two polynomials have a common monic polynomial factor $c(x)=x^n+\cdots$, then $k\cdot c(x)=k\cdot x^n+\cdots$ is also a common factor, for all $k\in\mathbb{R}$ (at least, assuming you're working with real [or even complex] coefficients)... – Isaac May 2 '12 at 23:21
up vote 3 down vote accepted

If you do not allow the full rational field, you expect to lose the Euclidean property. That is, $\mathbb Z[x]$ is not a PID or a Euclidean domain.

A Euclidean domain $R$ is when we have some sort of degree function, Rotman calls it $\partial,$ with $\partial: R \setminus \{0\} \rightarrow \mathbb N,$ such that $\partial(f) \leq \partial(fg) $ for all nonzero $f,g \in R,$ and as soon as $f \neq 0$ we can always write $$ g = q f + r,$$ with either $r=0$ or $\partial(r) < \partial(f).$

Anyway, in $\mathbb Z[x]$ we take $\partial$ to be the degree of a polynomial. Suppose we then take $g=x$ and $f=2.$ We are asking about writing $$ x = q \; 2 + r $$ with either $r=0$ or $\partial(r) < \partial(f) = \partial(2) =0.$ The latter is impossible, so we need $r=0$ and $x = 2 q$ for some polynomial $q.$ This is fine in $\mathbb Q[x],$ but the evident solution $q = x/2$ is not a legal element of $\mathbb Z[x].$ We could be more formal about proving there is no $q,$ but in the end $\mathbb Z[x]$ is not a Euclidean domain.

Meanwhile, in case you wonder about some exotic degree map that makes $\mathbb Z[x]$ into a Euclidean domain, be aware that $\mathbb Z[x]$ is not a PID either, which rules out that strategy. Indeed, the ideal generated by $(2,x)$ is not principal.

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