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In the second page of "Algeriac Curve" by Fulton, he stated

If $R$ is a UFD with $K=\operatorname{Frac}(R)$, then by Gauss Lemma for UFD, any irreducible $F\in R[x]$ remains irreducible when considered in $K[x]$. It follows that if $F$ and $G$ are polynomials in $R[x]$ with no common factors in $R[x]$, they have no common factors in $K[x]$.

I understand the first sentence just fine, but I have no idea why the second sentence follows from the first. I tried to prove the second sentence using proof by contradiction, but If we let $H$ be the common factor of $F$ and $G$ in $K[x]$, then we can't say much about $H$ since $H$ is not in $R[x]$, and even if we multiply it with all denominators, so that $aH\in R[x]$ for some $a\in R$, we still can not claim $aH$ is irreducible in $R[x]$ to get any meaningful result.

So my question is: Can you prove the second sentence from the first one? Thank you in advance and in all forms.

Edit 1: By two polynomial $F,G\in R[x] $ having no common factor, I think Fylton means that there are no polynomials $H, P,L\in R[x]$ such that $F=HP,G=PL$ with all of them being non-comstant. Please let me know if that’s a wrong interpretation. And I am willing to accept any answer that interpret what Fulton mean correctly, even if one didn't show their interpretation is equivalent to mine, as long as they point out if mine is correct or not.

John Frank
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  • Hint: in an extension of UFDs: irreds(= primes) persist $!\iff!$ prime factorizations persist $!\iff!$ gcds persist. This is likely a dupe but I don't have time to search now. $\ \ $ – Bill Dubuque Apr 23 '25 at 19:09

1 Answers1

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I suppose what Fulton means is this. If $R$ is a UFD then so is $R[X]$. Because $F$ and $G$ have no common factor in $R[X]$, we can write $F$ $=$ $u\Pi_{\,i=1}^{\,m}F_i^{a_i}$ and $G$ $=$ $v\Pi_{\,j=1}^{\,n}G_j^{b_j}$, where $u,v$ are in $R$ (and don't have a common factor), and the $F_i$ and $G_j$ are non-constant irreducibles in $R[X]$ that are not associated to one another. The factors $F_i$ and $G_j$ remain irreducible in $K[X]$ and no two are associated in $K[X]$. Thus $F$ and $G$ have no irreducible factor in common in $K[X]$.

Matthé van der Lee
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  • Is it true that in your answer for each fixed $i$, if $F_i=a_nX^n+\cdots +a_1X+a_0$, then the greatest common divisor (as an element of $R$) of ${a_n,\dots ,a_0}$ must be unit in $R$? – John Frank Apr 24 '25 at 02:17
  • If that's the case, I think my previous interpretation of what Fulton means is wrong. – John Frank Apr 24 '25 at 02:17
  • Thanks for your answer anyway! If you could elaborate on why $F_i,G_j$ are not associated with $K[x]$, I would really appreciate it. – John Frank Apr 24 '25 at 02:21
  • Continuing my last comment, I think I have figured it out. They are not associated with each other in $K[x]$ because if they have different degrees, then clearly, one can not divide the other; otherwise, the one divided becomes reducible. If they are of the same degree, they are not associated because $K$ is a field: all nonzero elements of $K$ are units. Am I correct here? – John Frank Apr 24 '25 at 02:26
  • The units of $K[X]$ are the elements of $K^\times$. So if $H=a_nX^n+\cdots +a_1X+a_0$ and $P=b_mX^m+\cdots +b_1X+b_0$ are in $R[X]$ and they are associated in $K[X]$, then $m=n$ and there are $0\ne u,v\in R$ such that $uH=vP$ in $R[X]$. So if the GCD of the $a_i$ in $R$ is 1, and same for the $b_i$, then the GCD of the coefficients of $uH$ is $u$, while for $vP$ it is $v$. So $u=tv$ for some unit $t$ of $R$. But then $P=tH$ is associated to $H$ in $R[X]$. – Matthé van der Lee Apr 24 '25 at 07:37