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$F$ is a field and $K$ is a sub-field. $K\subseteq F$.
$f(x),g(x)\in K[x]$.

How to prove that:
If in $K[x]$, $\gcd(f(x),g(x))=1$ also at $F[x],\; \gcd(f(x),g(x))=1$?

Can you give me a hint?
Because I even don't know how to begin this proof :-(

Thank you!

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marked as duplicate by Bill Dubuque polynomials Jul 10 at 16:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Hint. Show that $f$ and $g$ are coprime in $F[X]$ if and only if they are coprime in $K[X]$.

For this, observe the following:

  • Any common divisor of $f$ and $g$ in $K[X]$ is also a common divisor for $f$ and $g$ in $F[X]$.
  • For $r, s ∈ K[X]$, if $rf + sg = 1$ in $K[X]$, then $rf + sg = 1$ in $F[X]$.

More generally, you can conclude from this that the greatest common divisor of polynomials $f, g ∈ K[X]$ in $K[X]$ is still the greatest common divisor of $f$ and $g$ in $E[X]$. For this, use that, for $d ∈ E[X]$, you have $d = \gcd( f, g)$ if and only if $f/d$ and $g$ are coprime in $E[X]$, for $E = F$ and $E = K$ respectively.

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  • $\begingroup$ Can you explain me please the last sentence? - f/d and g are coprime in E[X] for E=F and E=K respectively. $\endgroup$ – CS1 Jul 10 at 16:59
  • $\begingroup$ I hope you asw what I wrote above... :-) $\endgroup$ – CS1 Jul 19 at 17:17
  • $\begingroup$ @CS1 I didn’t. Or I did, but then forgot about it. I’m not sure what’s unclear about that phrase, though. Can you specify your question? $\endgroup$ – k.stm Jul 19 at 17:19
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    $\begingroup$ @CS1 Two elements of a ring are coprime if their greatest common divisor is $1$. $\endgroup$ – k.stm Jul 22 at 14:26
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    $\begingroup$ @CS1 “$f$ divided by $d$”, that is the unique polynomial $e ∈ E[X]$ such that $f = de$. For example $\frac{X^2 - 1}{X - 1} = X + 1$. $\endgroup$ – k.stm Jul 22 at 14:29
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If $\gcd(f(x), g(x)) =1$ in $K[x]$, then $\exists a(x), b(x) \in K[x]$ with $a(x)f(x)+b(x)g(x)=1.~~ K \subseteq F \Rightarrow a(x), b(x) \in F[x] \Rightarrow \gcd(f(x), g(x)) =1$ in $F[x]$.

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  • $\begingroup$ How do you know the last sentence? $K \subseteq F \Rightarrow a(x), b(x) \in F[x] \Rightarrow \gcd(f(x), g(x)) =1$ Can you explain little bit more please? $\endgroup$ – CS1 Jul 10 at 16:29
  • $\begingroup$ $a(x)$ and $b(x)$ are polynomials with coefficients in $K$, which is a subfield of $F$, so they are also polynomials with coefficients in $F$. That means that an $F$-linear combination of $f(x)$ and $g(x)$ is equal to $1$, which in turn means that their $\gcd$ in $F$ is $1$. $\endgroup$ – Robert Shore Jul 10 at 16:31

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