GCD in polynomial rings with coefficients in a field extension

Question

Let $E/F $ be a field extension and $f,g$ $\in$ $F[x]$ (the polynomial ring with coefficients in $F$ ). Let's denote with $\gcd_F(f,g)$ the greatest common divisor of $f$ and $g$ in $F[x]$. Is it true that $\gcd_F(f,g)$ $=$ $\gcd_E(f,g)$ ?

Answer 1

Hint: You compute the gcd with the Euclidean algorithm, right?

Answer 2

Here is a slightly different perspective to the posted answers: As Bill Dubuque observes in his answer, by dividing $f$ and $g$ through by their GCD, you can reduce to the case where $f$ and $g$ are coprime (i.e. have GCD equal to $1$), and then you have to show that they remain coprime in $E[x]$.

One can certainly deduce this using the Euclidean algorithm (more precisely, one can find $h(x)$ and $k(x) \in F[x]$ such that $f h + g k = 1$, and this equation persists in $E[x]$, showing that $f$ and $g$ are coprime in that ring as well), but one can also deduce this from the fact that coprime polynomials have no common zero in $\overline{F}$ (the algebraic closure of $F$), hence have no common zero in $\overline{E}$ (the algebraic closure of $E$) either, and hence remain coprime in $E[x]$.

Also: see this question and its answers for another discussion of the persistence of coprimality under field extensions. (The preceding paragraph is essentialy a summary of those answers.)

Answer 3

Hint $\ $ It follows rather trivially from the fact that properties that are definable by the existence of solutions to ring equations necessarily persist in extension rings. Thus the property of being a common divisor clearly persists in extension domains, and the property of being a greatest common divisor also persists since it follows from the fact the the gcd is representable as linear combination (Bezout identity), and this identity necessarily persists in extensions.

In detail: gcds in a PID D persist in extension rings because the gcd may be specified by the solvability of (linear) equations over D, and such solutions always persist in extension rings, i.e. roots in D remain roots in rings $\rm\,R \supset D.\:$ More precisely, the Bezout identity for the gcd yields the following ring-theoretic existential $(\exists_1)$ equational specification for the gcd

$$\begin{eqnarray} \rm\gcd(a,b) = c &\iff&\rm (a,b) = (c)\\ &\iff&\rm a = c\:\color{#C00}x ,\ b= c\:\color{#C00} y ,\,\ a\:\color{#C00} u + b\: \color{#C00}v = c\ \ has\ roots\ \ \color{#C00}{x,y,u,v}\in D\end{eqnarray}$$

Proof $\ (\Leftarrow)\:$ In any ring $\rm R,\:$ $\rm\:a\: x = c,\ b\: y = c\:$ have roots $\rm\:x,y\in R$ $\iff$ $\rm c\ |\ a,b\:$ in $\rm R.$ Further if $\rm\:c = a\: u + b\: v\:$ has roots $\rm\:u,v\in R\:$ then $\rm\:d\ |\ a,b$ $\:\Rightarrow\:$ $\rm\:d\ |\ a\:u+b\:v = c\:$ in $\rm\: R.\:$ Hence we infer $\rm\:c = gcd(a,b)\:$ in $\rm\: R,\:$ being a common divisor divisible by every common divisor. $\ (\Rightarrow)\ $ If $\rm\:c = gcd(a,b)\:$ in D then the Bezout identity implies the existence of such roots $\rm\:u,v\in D. $

Remark $ $ While this is true for all purely existential $(\exists_1)$ properties it does not hold true for more general properties. E.g. universal $(\forall)$ properities (laws) generally do not persist, e.g. an extension ring of a commutative ring need not be commutative. Such persistence properties are studied at length when one studies model theory = universal algebra + logic.

Answer 4

Let $d=\gcd_F(f,g)$. Then $$dE[x]=(dF[x])E[x]=((f,g)F[x])E[x]=(f,g)E[x].$$Thus $d=\gcd_E(f,g).$