Let $f(x), g(x), h(x)$ be elements of $F[x]$, where $f(x)$ and $g(x)$ are relatively prime. If $f(x) \mid h(x)$ and $g(x)\mid h(x)$ prove that $f(x)g(x)\mid h(x)$.
Here $F$ is a field.
This is a practice question I need to know for my finals.
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What exactly is F(x)? – Alex Provost Apr 06 '13 at 23:46
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@silencer F(x) is a field – Kj Tada Apr 06 '13 at 23:51
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1@KjTada : You're omitting context. You must have seen this question posed in a context in which it was stated that $f(x)$, $g(x)$, and $h(x)$ are polynomials, or whatever it is that they are, and that $F$ is a field, or whatever it was intended to be. You omitted that information. – Michael Hardy Apr 06 '13 at 23:58
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@YACP found the error. – Kj Tada Apr 07 '13 at 00:17
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@MichaelHardy Yes f(x),g(x), h(x) are polynomials. F is a field. – Kj Tada Apr 07 '13 at 00:20
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Then $F[x]$ is an euclidean domain, so what you state follows. The proof of this fact is almost line by line the same as for integers. – vonbrand Apr 07 '13 at 01:15
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That $\rm\:gcd(f,g) = 1\:\Rightarrow\:lcm(f,g) = fg\:$ can be proved the same as for integers, namely:
Here is a proof of $\rm\ f,g\,|\,h\:\Rightarrow\:fg\,|\,hd,\ \ d = gcd(f,g)\, [= af+bg\ $ by Bezout]
$\rm\begin{eqnarray} \rm\quad f,g\mid h\, &\Rightarrow&\,\rm fg\mid hf,hg &\Rightarrow&\,\rm fg\mid ahf\, +\, bhg \ =\ h(af\!+\!bg)\, = hd\quad [Bezout\ form] \\ \rm\quad f,g\mid h\, &\Rightarrow&\,\rm fg\mid hf,hg &\Rightarrow&\,\rm fg\ |\ \gcd(hf, hg)\, =\: h \gcd(f,g) = hd\quad [GCD\ form] \\ \end{eqnarray}$
Remark $\ \ $ More generally $\rm\ \ gcd(f,g)\, lcm(f,g) = fg,\ $ in any domain where the lcm exists
Theorem $\rm\quad gcd(a,b)\, =\, ab/lcm(a,b)\ \ $ if $\ \ \rm lcm(a,b) \;$ exists, and $\rm\ ab\ne 0$
Proof $\rm\quad d\mid a,b\!\iff\! a,b\mid ab/d \!\iff\! lcm(a,b)\mid ab/d \iff d\mid ab/lcm(a,b)$
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We are working in a some polynomial, which is a UFD, so we can use Euclid's lemma. To say that $f\mid h$ and $g\mid h$ is to say that there exist elements $f',g'$ such that $h = ff'$,$h=gg'$, so that $ff' = gg'$. Now $f,g$ are coprime, so that $g\mid f'$; thus there exists $k$ such that $f' = gk$ (this is where we apply Euclid's lemma), and we have $h = fgk$, as required.
Alex Provost
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I really like your explanation and it seems like it extends to polynomials as f(x),g(x) and h(x) in my question were polynomials. Could you explain how you concluded this " $f,g$ are coprime, so that $g∣f'$ ′" – Kj Tada Apr 07 '13 at 00:40
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@mathGems, what would be a correct proof in this instance. I am just finding your proof a little hard to grasp. This question is from a first course in Abstract Algebra – Kj Tada Apr 07 '13 at 00:51
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@Silencer The proof does not invoke any special properties of the ring $,F[x].:$ So if the proof were valid, it would be true in any ring. But the result is not true in any ring. Probably you are implicitly assuming some properties of $,R = F[x].,$ Perhaps you assume $,R,$ is a UFD? Or perhaps you are assuming Euclid's Lemma? (a consequence of $,R,$ being a Euclidean domain). One can only guess, since the proof gives no justification for the key claim $,g\mid f'$ (which, in fact, is equivalent to the sought result). – Math Gems Apr 07 '13 at 06:51
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@KjTada What results do you know about $,F[x],?$ Do you know it has a division and Euclidean algorithm for the gcd? Euclid's Lemma? Bezout's identity for the gcd? That $,F[x],$ is a UFD (unique factorization domain)? – Math Gems Apr 07 '13 at 06:59
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@MathGems Yes, I was implicitly using the fact that we were working in a polynomial ring (and using Euclid's lemma!), despite my (misleading) early claim of working in some general ring. Sorry for the confusion. – Alex Provost Apr 07 '13 at 13:14