Let $K$ be an algebraically closed field of characteristic $0$. Instead of studying the polynomial ring $K[X]$ as a whole, we pay a little more attention to each polynomial. A reasonable thing to do is to count the number of distinct zeros. We define

$$n_0(f)=\text{number of distinct roots of $f$.}$$

For example, If $f(X)=(X-1)^{100}$, we have $n_0(f)=1$. It seems we are diving into calculus but actually there is still a lot of algebra.

The abc of Polynomials

Theorem 1 (Mason-Stothers). Let $a(X),b(X),c(X) \in K[X]$ be polynomials such that $(a,b,c)=1$ and $a+b=c$. Then

$$\max\{\deg a,\deg b,\deg c\} \le n_0(abc)-1.$$

Proof. Putting $f=a/c$ and $g=b/c$, we have

$$f+g=1.$$

This implies

$$f'+g'=\frac{f'}{f}f+\frac{g'}{g}g=0 \implies \frac{g}{f}=\frac{b}{a}=-\frac{f'/f}{g'/g}.$$

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We interrupt the proof here for some good reasons. Rational functions of the form $f’/f$ remind us of the chain rule applied to $\log{x}$. In the context of calculus, we have $\left(\log{f(x)}\right)’=f’/f$. On the ring $K[x]$, we define $D:K[x] \to K[x]$ to be the formal derivative morphism. Then this endomorphism extends to $K(x)$ by

$$D(f/g)=\frac{gDf-fDg}{g^2}.$$

On $K(x)^\ast$ (read: the multiplicative group of the rational function field $K(x)$), we define the logarithm derivative

$$L(f)=\frac{Df}{f}.$$

It follows that

$$L(fg)=\frac{D(fg)}{fg}=\frac{fDg+gDf}{fg}=L(f)+L(g).$$

Also observe that, just as in calculus, if $f$ is a constant...