GCD of Polynomials does not depend on Base Field

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Theorem

Let $E / F$ be a field extension.

Let $P, Q \in F \sqbrk X$ be polynomials.

Let:

$\gcd \set {P, Q} = R$ in $F \sqbrk X$

$\gcd \set {P, Q} = S$ in $E \sqbrk X$.

Then $R = S$.

In particular, $S \in F \sqbrk X$.

Proof

By definition of greatest common divisor:

$R \divides S$ in $E \sqbrk X$

By Polynomial Forms over Field is Euclidean Domain, there exist $A, B \in F \sqbrk X$ such that:

$A P + B Q = R$

Because $S \divides P, Q$:

$S \divides R$ in $E \sqbrk X$

By $R \divides S$ and $S \divides R$:

$R = S$

$\blacksquare$