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I am trying to prove this theorem:

Let $L/k$ be field extension, and let $f(x),g(x)$ be two polynomials in $k[x]\setminus\{0\}$ and $h(x)=\gcd(f(x),g(x))$ in $k[x]$, then prove that the $\gcd(f(x),g(x))$ is also $h(x)$ when $f(x),g(x)$ are considered as elements in $L[x]$.

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2 Answers 2

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We have $f(x)=h(x)f^\ast(x)$ and $g(x)=h(x)g^\ast(x)$, where $f^\ast(x)$ and $g^\ast(x)$ are polynomials in $k[x]$, and $f^\ast(x)$ and $g^\ast(x)$ are relatively prime over $k[x]$.

It follows by Bezout's Theorem that there exist polynomials $a(x)$ and $b(x)$ in $k[x]$ such that $$a(x)f^\ast(x)+b(x)g^\ast(x)=1.$$ Thus $f^\ast(x)$ and $g^\ast(x)$ are relatively prime over $L[x]$, and therefore $h(x)$ is a greatest common divisor of $f(x)$ and $g(x)$ over $L[x]$.

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Hint $ $ gcds in Euclidean domains D persist in extension rings because the gcd may be specified by the solvability of (linear) equations over D, and such solutions always persist in extension rings, i.e. roots in D remain roots in rings $\rm\,R \supset D.\:$ More precisely, the Bezout identity for the gcd yields the following ring-theoretic equational specification for the gcd

$$\rm\ gcd(a,b) = c\ \iff\ a\: \color{#C00}x = c,\ b\:\color{#C00} y = c,\,\ a\:\color{#C00} u + b\: \color{#C00}v = c\ \ has\ roots\ \ \color{#C00}{x,y,u,v}\in D$$

Proof $\ (\Leftarrow)\:$ In any ring $\rm R,\:$ $\rm\:a\: x = c,\ b\: y = c\:$ have roots $\rm\:x,y\in R$ $\iff$ $\rm c\ |\ a,b\:$ in $\rm R.$ Further if $\rm\:c = a\: u + b\: v\:$ has roots $\rm\:u,v\in R\:$ then $\rm\:d\ |\ a,b$ $\:\Rightarrow\:$ $\rm\:d\ |\ a\:u+b\:v = c\:$ in $\rm\: R.\:$ Hence we infer $\rm\:c = gcd(a,b)\:$ in $\rm\: R,\:$ being a common divisor divisible by every common divisor. $\ (\Rightarrow)\ $ If $\rm\:c = gcd(a,b)\:$ in D then the Bezout identity implies the existence of such roots $\rm\:u,v\in D.\ $ QED

Rings with such linearly representable gcds are known as Bezout rings. As above, gcds in such rings always persist in extension rings. In particular, coprime elements remain coprime in extension rings (with same $1$). This need not be true without such Bezout linear representations of the gcd. For example, $\rm\:\gcd(2,x) = 1\:$ in $\rm\:\mathbb Z[x]\:$ but the gcd is the nonunit $\:2\:$ in $\rm\:\mathbb Z[x/2]\subset \mathbb Q[x]$.

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