# The abc Theorem of Polynomials

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

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

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

This implies

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

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

It follows that

Also observe that, just as in calculus, if $f$ is a constant function, then $D(f)=0$. Now we write

Then it follows that

Now we can be back to the proof.

*Proof (continued).* Since $K$ is algebraically closed,

We see, for example

Therefore

Likewise

Combining both, we obtain

Next, multiplying $f’/f$ and $g’/g$ by

which has degree $n_0(abc)$ (since $(a,b,c)=1$, these three polynomials share no root). Both $N_0f’/f$ and $N_0g’/g$ are polynomials of degrees at most $n_0(abc)-1$ (this is because $\deg h’=\deg h-1$ for non-constant $h \in K[X]$, while $f$ and $g$ are non-constant (why?); we assume $\operatorname{char} K=0$ for this reason).

Next we observe the degrees of $a,b$ and $c$. Since $a+b=c$, we actually have $\deg c \le \max\{\deg a,\deg b\}$. Therefore $\max\{\deg a,\deg b,\deg c\}=\max\{\deg a,\deg b\}$. From the relation

and the assumption that $(a,b)=1$, one can find polynomial $h \in K[X]$ such that

Taking the degrees of both sides, we see

This proves the theorem. $\square$

# Applications

We present some applications of this theorem.

Corollary 1 (Fermat’s theorem for polynomials).Let $a(X),b(X)$ and $c(X)$ be relatively prime polynomials in $K[X]$ such that not all of them are constant, and such thatThen $n \le 2$.

Alternatively one can argue the curve $x^n+y^n=1$ on $K(X)$.

*Proof.* Since $a,b$ and $c$ are relatively prime, we also have $a^n$, $b^n$ and $c^n$ to be relatively prime. By Mason-Stothers theorem,

Replacing $a$ by $b$ and $c$, we see

It follows that

In this case $n<3$. $\square$

Corollary 2 (Davenport’s inequality).Let $f,g \in K[X]$ be non-constant polynomials such that $f^3-g^2 \ne 0$. Then

One may discuss cases separately on whether $f$ and $g$ are coprime, and try to apply Mason-Stothers theorem respectively, and many documents only record the proof of coprime case, which is a shame. The case when $f$ and $g$ are not coprime can be a nightmare. Instead, for sake of accessibility, we offer the elegant proof given by Stothers, starting with a lemma about the degree of the difference of two polynomials.

Lemma 1.Suppose $p,q \in K[X]$ are two distinct non-constant polynomials, then

*Proof.* Let $k(f)$ be the leading coefficient of a polynomial $f$. If $\deg p \ne \deg q$ or $k(p) \ne k(q)$, then $\deg(p-q)\ge \deg p \ge \deg p - n_0(p)-n_0(q)+1$ because $n_0(p) \ge 1$ and $n_0(q) \ge 1$.

Next suppose $\deg p = \deg q$ and $k(p)=k(q)$. If $(p,q)=1$, then by Mason-Stothers,

Otherwise, suppose $(p,q)=r$. Then $p/r$ and $q/r$ are coprime. Again by Mason-Stothers,

Therefore

On the other hand,

Combining all these inequalities, we obtain what we want. $\square$

*Proof (of corollary 2).* Put $\deg{f}=m$ and $\deg{g}=n$. If $3m \ne 2n$, then

because $m \ge 1$. Next we assume that $3m=2n$, or in other word, $m=2r$ and $n=3r$. By lemma 1, we can write

This proves the inequality. $\square$

One may also generalise the case to $f^m-g^n$. But we put down some more important remarks. First of all, Mason-Stothers is originally a generalisation of Davenport’s inequality (by Stothers). I personally do not think any mortal can find the original paper of Davenport’s inequality, but on [Shioda 04] there is a reproduced proof using linear algebra (lemma 3.1).

For more geometrical interpretation, one may be interested in [Zannier 95], where Riemann’s existence theorem is also discussed.

In Stothers’s paper [Stothers 81], the author discussed the condition where the equality holds. If you look carefully you will realise his theorem 1.1 is exactly the Mason-Stothers theorem.

# References / Further Reading

- [Davenport 65] H. Davenport,
*On $f^3(t)-g^2(t)$*, 1965. (can someone find a digital copy of this paper?) - [Ma 84] R. C. Mason,
*Diophantine Equations over Function Fields*, 1984. - [Shioda 04] Tetsuji Shioda,
*The abc-theorem, Davenport’s inequality and elliptic surfaces*, 2004 (https://www2.rikkyo.ac.jp/web/shioda/papers/esdstadd.pdf) - [Stothers 81] W. W. Stothers,
*POLYNOMIAL IDENTITIES AND HAUPTMODULN*, 1981. (https://doi.org/10.1093/qmath/32.3.349) - [Zannier 95] Umberto Zannier (Venezia),
*On Davenport’s bound for the degree of $f^3-g^2$ and Riemann’s Existence Theorem*, 1995. (https://eudml.org/doc/206763)

The abc Theorem of Polynomials

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