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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 $\gcd_F(f,g)$ $=$ $\gcd_E(f,g)$ ?

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up vote 3 down vote accepted

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.)

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

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Thank you...the euclidean algorithm depends only on $f$ and $g$ and therefore yields the same solution in $E[x]$. – Emilio Ferrucci Nov 22 '11 at 18:03
@EmilioFerrucci: Correct. Well done. – Jyrki Lahtonen Nov 22 '11 at 19:19

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.

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.

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Thank you for the hint. The coefficients of the Bezout identity are the same in $E$ and $F$ since in both cases they are obtained with the euclidean algorithm. Still, without the euclidean algorithm (which isn't necessary to define the gcd in a ring) I wouldn't know how to procede. Did you have something else in mind? I still find it quite interesting that no matter how "large" the extension $E|F$ is, there is no improvement on the gcd of $f$ and $g$. – Emilio Ferrucci Nov 22 '11 at 18:21

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