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Let $R$ be a UFD and suppose $f(x),g(x)$ are co-prime in $R[X]$. I want to show that $f(x), g(x)$ are then also co-prime in $\operatorname{Frac}(R)[X]$, where $\operatorname{Frac}(R)$ is the field of fractions of $R$.

For me $f, g$ co-prime in $R[X]$ means that $\langle f \rangle + \langle g \rangle = R[X]$. Similarly $f,g$ are co-prime in $\operatorname{Frac}(R)[X]$ iff $\langle f \rangle + \langle g \rangle = \operatorname{Frac}(R)[X]$. So pick some $h(x)$ from $\operatorname{Frac}(R)[X]$. I seek $ a,b$ in $R[X]$ such that $af + bg = h$. Since $ h$ is in $\operatorname{Frac}(R)[X]$ its coefficient are fractions of elements of $R$. Therefore I can clear denominators and obtain $ c ( af + bg) = h’$ where $c$ is just some member of $R$ and $h’$ is in $R[X]$. But $f,g$ are co-prime so we can find $A,B$ such that $Af+Bg=h’$. Hence $a= A/c$ and $b= B/c$ should do the trick. My proof is complete.

I feel there is something wrong with this argument as it seems too simple. My textbook wants me to use Gauss Lemma on products of primitives for this. But my argument seems to do fine without it. Can anyone help me here?

Michael Hardy
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    Why do you believe $,f,g,$ coprime $\Rightarrow (f,g)=(1) $? Informed that this is false can you find a counterexample? You seem to be implicitly assuming that $R[X]$ is a PID, but you are only given that $R[X]$ is a UFD (recall PIDs are precisely the UFDs of dimension $\le 1,,$ i.e. where all nonzero prime ideals are maximal) – Bill Dubuque Apr 04 '20 at 16:17
  • Hmmm... That’s odd. My definition states that two ideals are co-prime iff their sum is the entire ring, I assumed that two elements are then co-prime just when their ideals are? Is this not correct? –  Apr 04 '20 at 16:22
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    @user It just means their greatest common divisor is a unit. For example, $2$ and $x$ are coprime in $\mathbb Z[x] $, but there are no $a, b$ such that $2a+bx=1$. – Matt Samuel Apr 04 '20 at 16:26
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    Generally elements are defined to be coprime iff they have only unit common divisors, i.e. their gcd is (associate to) $1$. In a PID (or Bezout domain) this is equivalent to them generating comaximal ideals (your notion), but that equivalence fails in more general rings. Double check which notion is intended in your textbook (which book?) – Bill Dubuque Apr 04 '20 at 16:31
  • I see. Using your definition the application of Gauss Lemma would make sense and I can see how to prove the statement this way. Is my proof correct if we use my definition? Many thanks. –  Apr 04 '20 at 16:42
  • Also what is your definition of co-primality for ideals? –  Apr 04 '20 at 16:58

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If $f=0$, then $g$ is a unit of $R$, and we are done. Let $f,g\neq 0$. Let $F$ be the field of fractions of $R$. Let $D$ be a common divisor of $f$ and $g$ in $F[x]$. Then $f(x)=A(x)D(x)$ and $g(x)=B(x)D(x)$ for some $A,B\in F[x]$. Taking out the common denominators of $A,D,B$, we have $$ f(x)=\frac{1}{rt}a(x)d(x),\quad g(x)=\frac{1}{st}b(x)d(x), $$ where $a,b,d\in R[x]$ and $r,s,t\in R$, and $tD(x)=d(x)$. Let $C(f)$ be the content of $f$. Then we have $$ rtC(f)f_{1}(x)=C(a)C(d)a_{1}(x)d_{1}(x),\quad stC(g)g_{1}(x)=C(b)C(d)b_{1}(x)d_{1}(x), $$ where $f_{1},g_{1},a_{1},b_{1},d_{1}\in R[x]$ are primitive. Taking the content on the equations, we have $$ rtC(f)u=C(a)C(d), \quad stC(g)v=C(b)C(d), $$ where $u,v\in R$ are units. So $$ f_{1}(x)=ua_{1}(x)d_{1}(x), \quad g_{1}(x)=vb_{1}(x)d_{1}(x). $$ Then $d_{1}$ is a common divisor of $f$ and $g$ in $R[x]$. So $d_{1}$ is a unit of $R$. Since $tD(x)=d(x)=C(d)d_{1}(x)$, $D$ is a unit of $F$. So $f$ and $g$ are coprime in $F[x]$.

Delong
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