question regarding resultant of 2 polynomials defined over finite field

Question

I have been studying "Introduction to finite fields" by Lidl & Neiderreiter and came across the following statement concerning resultant $R(f,g)$ of 2 polynomials $f(x) = a_{0}x^{n} + a_{1}x^{n - 1} + \ldots + a_{n} \in K[x]$ and $g(x) = b_{0}x^{m} + b_{1}x^{m - 1} + \ldots + b_{m} \in K[x]$. The statement says "$R(f,g) = 0$ iff $f$ and $g$ have a common root, which is the same as saying that $f$ and $g$ have a common divisor in $K[x]$ of positive degree". My question is, why is it necessary that common divisor of $f$ and $g$ must be a polynomial in $K[x]$? e.g. if $f$ and $g$ have a single common root $\alpha$ in some extension field, then $x - \alpha$ does not belong to $K[x]$.

Answer 1

Before doing anything else, we may consider where everything is living. As far as I'm aware of, the resultant $R(f,g)$ of two polynomials over a field $K$ is defined as the product of all $(\alpha-\beta)$ where $f(\alpha)=0$ and $g(\beta)=0$, where the roots are taken in an algebraic closure $\overline{K}$ of $K$.

With that definition, the resultant would be $0$, precisely if $f$ and $g$ have a common zero in the algebraic closure of $K$, which isn't entirely clear from the quotation. Your question then boils down to the question why two polynomials $f$ and $g$ having a common zero in some extension field $L/K$ need to have a common divisor in $K[x]$. The answer to this question is given by the concept of minimal polynomial. If $\eta$ is an element of an extension field, the minimal polynomial $\phi\in K[x]$ of $\eta$ is the monic polynomial of lowest degree having $\eta$ as a zero (where one can check that it is unique and irreducible over $K$). As an example, $x^2-2$ is a minimal polynomial of $\sqrt{2}$ over $\mathbb{F}_5$ (over $\mathbb{Q}$ as well, but you were talking about finite fields). The important property is that any polynomial over $K$ having $\eta$ as a root, has the minimal polynomial $\phi$ as a divisor. In the previous example, every polynomial in $\mathbb{F}_5[x]$ having $\sqrt{2}$ as a root, is divisible by $x^2-2$.

In our current case, this means that the polynomials $f$ and $g$ have a common root in $\alpha$ in $\overline{K}$ and so both $f$ and $g$ have the minimal polynomial of $\alpha$ over $K$ as a divisor. In particular, this minimal polynomial lies in $K[x]$.