GCD of Polynomials does not depend on Base Field

Theorem
Let $E/F$ be a field extension.

Let $P, Q \in F \left[{X}\right]$ be polynomials.

Let:
 * $\gcd \left({P, Q}\right) = R$ in $F \left[{X}\right]$
 * $\gcd \left({P, Q}\right) = S$ in $E \left[{X}\right]$.

Then $R = S$.

In particular, $S \in F \left[{X}\right]$.

Proof
By definition of greatest common divisor:
 * $R \mathrel \backslash S$ in $E \left[{X}\right]$

By Polynomial Forms over Field is Euclidean Domain, there exist $A, B\in F \left[{X}\right]$ such that:
 * $A P + B Q = R$

Because $S \mathrel \backslash P, Q$:
 * $S \mathrel \backslash R$ in $E \left[{X}\right]$

By $R \mathrel \backslash S$ and $S \mathrel \backslash R$:
 * $R = S$