GCD of Polynomials does not depend on Base Field

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 $S \divides R$:
 * $R = S$