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Let $k$ be a field and let $f\in k[x]$. Let's say that there are some $g, h\in k[x]$ such that $\gcd(g, h)=1$, $g|f$ and $h|f$. I want to show that $gh|f$.

Since $g|f$ and $h|f$, we may write $f=gu$ and $f=hv$ for some $u, v\in k[x]$. Now, $\gcd(g, h)=1$ implies that there are some $\alpha, \beta \in k[x]$ such that $\alpha g+\beta h=1$.

We may now write that $gu=hv\implies \alpha gu=\alpha hv \implies (1-\beta h)u=\alpha h v\implies u=h(\alpha v +\beta u)$. Thus, $f=gh(\alpha v + \beta u)$, so $gh |f$.

Is this correct? I remember hearing this fact in high school, but in my abstract algebra course from this year the lecturer never mentioned it, even though it would have made some things easier. This is why I tried to prove this myself.

user26857
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TheZone
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  • Your proof is correct. Well done! – cos_dm_math21 Jun 08 '21 at 13:51
  • Thank you both! @DietrichBurde yes, that's true - I posted this mostly as a sanity check (some proofs from the lectures could have been easily simplified by using this simple argument and it seemed quite strange not to see it used) – TheZone Jun 08 '21 at 13:57
  • IBy Euclid's Lemma: $,\color{#c00}{(g,h)=1,\ g\mid h}(f/h)\Rightarrow g\mid f/h\Rightarrow gh\mid f.\ $ See the linked dupes for proofs of Euclid's Lemma. All of these closely related properties are well-known and their proofs are already given here many times, so it is best to delete dupes like this to avoid cluttering search results. – Bill Dubuque Jun 08 '21 at 22:35
  • @DietrichBurde That's a bit misleading. It is not "factorial" (UFD) that enables the Bezout based proof but rather Euclidean. For non-Euclidean UFDs (or GCD domains), e,g, $,\Bbb Z[x],$ or $,\Bbb Q[x,y]),,$ there need be no Bezout equation for the gcd so we need to resort to other methods (cf. the dupes). – Bill Dubuque Jun 08 '21 at 22:39

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