# Proving that $\gcd(f(x),g(x))=1$ at a ring and sub-ring [duplicate]

$$F$$ 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!

## marked as duplicate by Bill Dubuque polynomials StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); 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.

• – lhf Jul 10 at 16:26

## 2 Answers

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.

• Can you explain me please the last sentence? - f/d and g are coprime in E[X] for E=F and E=K respectively. – CS1 Jul 10 at 16:59
• I hope you asw what I wrote above... :-) – CS1 Jul 19 at 17:17
• @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? – k.stm Jul 19 at 17:19
• @CS1 Two elements of a ring are coprime if their greatest common divisor is $1$. – k.stm Jul 22 at 14:26
• @CS1 “$f$ divided by $d$”, that is the unique polynomial $e ∈ E[X]$ such that $f = de$. For example $\frac{X^2 - 1}{X - 1} = X + 1$. – k.stm Jul 22 at 14:29

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 $$F[x]$$.

• 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? – CS1 Jul 10 at 16:29
• $a(x)$ 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 $1$. – Robert Shore Jul 10 at 16:31