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Let $p(x)\in K[x]$ and $q(x)\in K[x]$, $F$ is an extension of field $K$. Will $gcd(p(x),q(x))$ over $K[x]$ equal to $gcd(p(x),q(x))$ over $F$.

My guess is yes. I couldn't find an elegant way to prove it.

Jack
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    Yes. When you say you don't know an elegant way to prove it, do you mean you know a way but it isn't elegant, or actually you don't know a way at all? Concretely, think about how you compute the gcd up to a nonzero scaling factor: by Euclid's algorithm. If you start with two polynomials in $K[x]$, even if you view them in $F[x]$ all the calculations for Euclid's algorithm on them in $F[x]$ still take place in $K[x]$, so the $K[x]$-gcd is also the $F[x]$-gcd when the two polynomials are in $K[x]$. – KCd Apr 27 '17 at 02:08

3 Answers3

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$\gcd(p,q)$ can be computed with the Euclidean algorithm, which operates on the coefficients of $p$ and $q$ and so never leaves $K$.

lhf
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If $pK[x]+qK[x]=dK[x]$, then clearly $pF[x]+qF[x]=dF[x]$.

MooS
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Here is an elegant way.

Let $d$ be the GCD over $\mathbb{K}[x]$ and $D$ be the GCD over $\mathbb{F}[x]$. Since: $$d(x)=a(x)p(x)+b(x)q(x)$$ and $D$ divides $p$ and $q$, it is clear that $D$ divides $d$.

On the other hand, in the big field $\mathbb{F}[x]$, $d$ is a polynomial which divides $p$ and $q$ so it will divide the GCD of $p$ and $q$. Hence $d$ divides $D$.

hardmath
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user128787
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