$f$ and $g$ have a common divisor with degree at least 1 in $R[X]$ if and only if $\operatorname{Res}_X(f,g)=0$ Let's say $f(X)= a_0X^m+...+ a_m$ and $g(X)=b_0X^n+...+b_n$ two polynomials in $R[X]$, with $a_0, b_0 \neq 0$. I want to prove that the following statements are the same:
(1) $f$ and $g$ have a common divisor with degree at least 1 in $R[X]$
(2) $\operatorname{Res}_X(f,g)=0$
I thought you should switch to the fraction field of $R$. We also don't know if $R$ is algebraically closed, so we can only make a statement about the common divisor.
I tried the following but It doesn't work out:
$(1)\to (2)$:
If $f,g$ have a common divisor, let's say $q \in R[x]$ we can rewrite $f,g$ as $f(X)=q^a f_1(X)$ an d $g(X)=q^ag_1(X)$, with $a>0$ and $g_1, f_1 \in R[x]$ ... I really don't know how to move on.
$(2) \to (1)$:
Let's say $\operatorname{Res}_X(f,g)=0$, then $f,g$ have a common root, say $c$. So we can rewrite $f(X)=(X-c)f_1(X)$ and $g(X)=(X-c)g_1(X)$ you can see that the common divisor is $X-c$. This has $\deg (X-c) =1$.
 A: You're correct that we want to work over the fraction field of $R$. So let $K=\operatorname{Frac} R$, and let $P_d\subset K[x]$ be the vector space of polynomials of degree less than $d$. Letting $d=\deg f$ and $e=\deg g$, we have that $\operatorname{Res}_x(f,g)$ is the determinant of the map $P_e\times P_d\to P_{d+e}$ by $(p,q)\mapsto fp+gq$.
In the forward direction, suppose $f$ and $g$ have some common factor $h$ of positive degree. Then $fg+pq\in (h)$ for all $p,q$, and $(h)\cap P_{d+e}$ is a proper subspace of $P_{d+e}$. Therefore our map $P_d\times P_e\to P_{d+e}$ isn't surjective, and must have determinant zero. So $\operatorname{Res}_x(f,g)=0$.
In the reverse direction, suppose the resultant is zero. Then there's a nontrivial solution to $fp+gq=0$, which means we can write $fp=-gq$ for $p\in P_e$ and $g\in P_d$ with $p,q\neq 0$. As $K[x]$ is a UFD and $\deg p<\deg g$, there must be some irreducible factor of $g$ which divides $f$. By Gauss' lemma, such a factor of $g$ in $K[x]$ is (up to a unit) a factor of  $g$ in $R[x]$, and we're done.

Let me also point out that your proposed proof that 2 implies 1 has an issue: $K$ need not be algebraically closed, so there's no guarantee that you can choose such a $c$.
