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Suppose $f$ and $g$ are are two polynomials with complex coefficents (i.e $f,g \in \mathbb{C}[x]$). Let $m$ be the order of $f$ and let $n$ be the order of $g$.

Are there some general conditions where

$fg= \alpha x^{n+m}$

for some non-zero $\alpha \in \mathbb{C}$

Mykie
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    Unless f and g are monomials themselves, I can't see how else this will hold. – J. M. ain't a mathematician Aug 26 '10 at 14:51
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    TIP It's not a good idea to accept an answer 10 minutes after posting your question. Being accepted, this will cause the software to give the post much less exposure, so you will have much less a chance to receive other good answers. And, generally, it's highly unlikely that you'll get the best answer in 10 minutes. Generally you shouldn't accept an answer for at least a few days if not more if you want to have the best possible chance of learning from the collective expertise here. – Bill Dubuque Aug 26 '10 at 15:09
  • Sorry I will keep that in mind. I was somewhat eager to close the question since the answer occured to me a few minutes after I posted it. – Mykie Aug 26 '10 at 15:23

4 Answers4

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Polynomials over $\mathbb{C}$ (in fact, over any field) are a Unique Factorization Domain (see http://en.wikipedia.org/wiki/Unique_factorization_domain); since $x$ is an irreducible, the only way for that to happen is for $f=ax^m$ and $g=bx^n$, with $ab=\alpha$.

(If you don't want to bring in the sledgehammer of unique factorization, you can just do it explicitly: look at the lowest nonzero term in $f$ and the lowest nonzero term in $g$; their product will be the lowest nonzero term in $fg$, hence must be of degree $m+n$. Since the degree of the lowest nonzero term of $f$ is at most $m$ and the one of $g$ is at most $n$, you have that they must be exactly of degree $m$ and $n$, respectively, and you get the result)

Arturo Magidin
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  • Here one needs only the weak statement that products of primes factor uniquely - not the strong statement that products of irreducibles factor uniquely ($\Rightarrow$ UFD when factorizations are finite). But this weaker statement is true in any domain. Indeed, the well-known one-line proof for $\mathbb Z$ generalizes immediately. That $x$ is prime here is immediate - see my answer. – Bill Dubuque Aug 29 '10 at 17:58
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We don't need the strong property of UFD. If $\rm D$ is a domain $\rm D$ then $\rm x$ is prime in $\rm D[x]$ (by $\rm D[x]/x \cong D$ a domain), and products of primes factor uniquely in every domain (same simple proof as in $\Bbb Z$). In particular, the only factorizations of the prime power $\rm x^i$ are $\rm \,x^j x^k,\ i = j+k\ $ (up to associates as usual). This fails over non-domains, e.g. $\,\rm x = (2x+3)(3x+2) \in \mathbb Z/6[x].$

Bill Dubuque
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  • Beware that over non-domains there are not even standard definitions of basic divisibility notions, e.g. "associate" and "irreducible" bifurcate into a few inequivalent notions. See here for literature. – Bill Dubuque May 12 '20 at 17:01
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The answer just occured to me. The roots of $f$ and $g$ must be be at 0.

Mykie
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Yes, the intuitively evident ones: all other terms in $f$ and $g$ must vanish. To see this, note that the product of the constant terms of $f$ and $g$ equals the constant term of $fg$, which is zero, whence at least one of these polynomials is multiple of $x$. Without any loss of generality assume it is $f$. Then $$fg = x\, \left(\frac{f}{x}\right) g,$$

implying $(f/x) g$ is a multiple of $x^{n+m-1}$. By induction this reduces us to the case $n+m=0$, which is trivial (because $f$ and $g$ then have no other terms). QED.

whuber
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  • Without evaluation: $x$ is prime in $D[x]\iff D$ is a domain, so in a domain $,x\mid fg\iff x\mid f,$ or $,x\mid g.\ $ Your argument essentially repeats the standard proof of uniqueness of factorization of prime products for the special case of a power of the prime $x$ (cf. my answer). – Bill Dubuque May 14 '20 at 16:11
  • @Gone That's right. I hope you don't view that as problematic :-). (NB: (i) you posted your answer later. (ii) I upvoted yours.) – whuber May 14 '20 at 16:24
  • The point of my comment is to help readers understand how the evaluation (coef) argument is a special case of uniqueness of prime factorizations. It has nothing at all to do with the other matters you mention. This is rarely mentioned in textbooks and often overlooked by students, so it is worth extra emphasis (which it might possibly get due to the recent edits). – Bill Dubuque May 14 '20 at 16:34