Can a ring extension leads to new gcd?

Question

Q. Given $f,g$ coprime in ring $R$, i.e. $\exists h,k \in R$, s.t. $fh+gk=1$. Is it possible that, after somehow extend $R$ to ring $S$ we have $e,f_1,g_1 \in S$, $f=ef_1$, $g=eg_1$ s.t. $e\ne1$?

I don't know if there is such an "extension" in ring, comparing to "field extension".

Background: I'm reading David R. Finston and Patrick J. Morandi's book Abstract Algebra: Structure and Application and in section 7.3 page 112 there is an exercise

Exercise 5. Let $L$ be any extension field of $F$ . A pair of polynomials $f$ and $g$ in $F[x]$ can also be viewed as lying in $L[x]$. Use Proposition 5.5 to see that if $d$ and $e$ are the gcds of $f$ and $g$ computed over $F$ and $L$ respectively, then each one divides the other. Thus $gcd(f,g)$ is well defined in the sense that it is the same polynomial in $F[x]$ whether $f$ and $g$ are viewed in $F[x]$ or $L[x]$.

I agree with all that's said in the book. However, if what I asked in Q is possible, what prevent this to happen in Exercise 5 setting?Is it due to $F$ is a field? or due to $F[x], L[x]$ are PID?

Answer 1

$h$ and $k$ are still in $S$, so $1=fh+gk=ef_1h+eg_1k=(f_1h+g_1k)e$ which says $e$ divides $1$, so $e$ is a unit.

Answer 2

Let $R$ be a ring and $S$ containing $R$ as a subring. Assume that for $f,g\in R$, there exists $h,k$ such that $fh+gk=1$.

This equality can be viewed in $S$, so $f,g$ are also coprime in $S$ (I assume that for you, "coprime" means that there is a Bézout equality, which is not the right definition, but whatever...)

To answer your original question: let $e\in S$ dividing $f$ and $g$ is $S$. Then $e$ divides $fh+gk=1$, meaning that $e$ is invertible (in your question you should replace "$e\neq 1$" by "$e$ is invertible")

A more interesting question is to ask if the gcd can change when you extend the ring $R$ (and i suspect that it is really the question you meant to ask). The answer is NO if $R$ is a PID, or even a Bézout ring (an integral domain for which every finitely generated ideal is principal). Indeed, in this case, then a gcd $d$ satisfies $aR+bR=dR$. Then $aS+bS=dS$. But it is easy to see that $d$ is then a gcd of $a$ and $b$ in $S$.

If $R$ is not a Bézout domain, nasty things may happen. Inspired by Gerry Myerson's example in the comments, here is a detailed example.

Take $R=\mathbb{Z}[i\sqrt{6}]$ and $S=\mathbb{Z}[\sqrt{2},j]$. The ring $S$ is a PID, so any pair of elements has a gcd.

Now take $a=2, b=i\sqrt{6}$. Then a gcd of $a$ and $b$ is $1$: indeed, if $d $is a common divisor in $R$, then $\vert d\vert^2$ must divide $4$ and $6$ in $\mathbb{Z}$. Hence $\vert d\vert^2=1$ or $2$. But here is no element in $R$ whose square of modulus is $2$ ($x^2+6y^2=2,x,y\in\mathbb{Z}$ has no solutions). Hence $ \vert d\vert^2=d\bar{d}=1$, so $d$ is a unit (since $\bar{d}\in R$).

However, $\sqrt{2}$ is a common divisor of $a$ and $b$ in $S$, which is not a unit (the potential inverse is $\dfrac{\sqrt{2}}{2}$, which easily does not lie in $S$)