GCD of polynomials over a field and its extension

Question

Let $E$ be an extension of a field $F$.

Let $f(x),g(x)\in F[x]$ with $d(x)$ their GCD in $F[x]$.

Claim: $d(x)$ is GCD of $f(x)$ and $g(x)$ in $E[x]$.

Proof. Case 1. $d(x)=1$. Thus $f(x),g(x)$ are coprime in $F[x]$, and so we can write $$1=a(x)f(x) + b(x)g(x)$$ for some $a(x),b(x)\in F[x]$, and since this expression is valid in $E[x]$ too, so $f(x)$ and $g(x)$ are coprime in $E[x]$.

Case 2. Suppose $d(x)$ is not unit in $F[x]$.

Then $f(x)/d(x)$ and $g(x)/d(x)$ are polynomials in $F[x]$ with GCD $1$, so their GCD is $1$ in $E[x]$. Hence $d(x)$ is GCD of $f(x)$ and $g(x)$ in $E[x]$.

Q. Is the assertion and proof correct?

Suppose the gcd is $d(x)=x^2+1$ over the reals. Then over the complex numbers, $x^2+1$ splits. — Wuestenfux, Aug 28 '19 at 10:18
I might simply argue that you can calculate the gcd with Euclid's algorithm. But that algorithm runs through exactly the same steps in both $F[x]$ and $E[x]$. Claim follows. — Jyrki Lahtonen, Aug 28 '19 at 11:48

score 1 · Accepted Answer · answered Aug 28 '19 at 11:47

Yes, this is the case. I'm not entirely sure about your argument, but consider the following.

Let $h \in F[x]$ be the GCD of $f,g \in F[x]$ and $h' \in E[x]$ the GCD of $f,g \in E[x]$.

$h \vert h'$: Since $h \in E[x]$ and $h \vert f,g$ we have $h \vert h'$.

$h' \vert h$: Since $F[x]$ is a PID we have $(f,g)_{F[x]} = (h)_{F[x]}$. Hence $h = af + bg$ for some $a,b \in F[x]$. Now since $a,b \in E[x]$ we have $h \in (f,g)_{E[x]} = (h')_{E[x]}$ and thus $h' \vert h$.

Hence $h'$ and $h$ are associated.

GCD of polynomials over a field and its extension

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