Let $K:F$ be a field extension. Let $p, q\in F[x_1, \ldots, x_n]$ be polynomials with greatest common divisor $d\in F[x_1, \ldots, x_n]$.
Question. Is it necessary that the g.c.d. of $p$ and $q$ in $K[x_1, \ldots, x_n]$ is also $d$?
Is GCD Preserved Under Extension of a Polynomial Ring?
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polynomials
ring-theory
unique-factorization-domains
1 Answers
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The gcd is the same in both fields (up to multiplication by some nonzero scalar).
First, a couple of reductions.
It is enough to assume that $f,g$ are coprime over $F$, since otherwise you can divide by their gcd over $F$ (which by definition divides the gcd over $K$). From now on I will denote by $d=d(x_1,...,x_n)$ their gcd over $K$.
It is enough to assume that both $f,g$ are irreducible over $F$ (and nonequivalent - $f\neq c g$ for some $c\in F$) , since if they have a common divisor over $K$, then they have irreducible factors that have common divisor.
Suppose first that $n=1$. In this case $F[x]$ is PID so you can write $1=f(x)a(x)+g(x)b(x)$ over $F$. Since $d$ divides both $f$ and $g$ it must also divide 1 and you are done.
We now prove for general $n$. Since they are divisible (over $K$) by a nonconstant polynomial $d$, we may assume wlog that $x_n$ appears in $d$ nontrivially and therefore in $f,g$ also. We now think of these polynomial as polynomials over $F(x_1,...,x_{n-1})[x_n]$ so that by Gauss' lemma they are also irreducible there. If they become equivalent irreducible, then $f=g\cdot h(x_1,...,x_{n-1})$ where $h$ is a rational function and after normalizing we can write $$h_1(x_1,...,x_{n-1})f(x_1,...,x_n)=h_2(x_1,...,x_{n-1})g(x_1,...,x_n)$$ as a product in $F[x_1,...,x_n]$. Since both $f,g$ are irreducible and contain $x_n$ non-trivially, we get by contradiction that they are equivalent over $F$. It follows that $f,g$ are coprime irreducible polynomials over $F(x_1,...,x_{n-1})[x_n]$. Using the $n=1$ case we get that $f,g$ are coprime as polynomials in $K(x_1,...,x_{n-1})[x_n]$. If they are not coprime over $K[x_1,...,x_n]$, then the gcd must be in $K[x_1,...,x_{n-1}]$, but we assumed that $x_n$ appears non trivially in $d$ - contradiction. It then follows that $d$ is constant, i.e. $f$ and $g$ are also coprime over $K$.