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

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

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 function, then $D(f)=0$. Now we write

$$f(X)=c\prod(X-\alpha_i)^{m_i}.$$

Then it follows that

$$\begin{aligned} f'/f=L(f)&=L(c)+\sum L\left((X-\alpha_i)^{m_i} \right) \\ &= m_i\sum L(X-\alpha_i) \\ &= \sum\frac{m_i}{X-\alpha_i}. \end{aligned}$$

Now we can be back to the proof.

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

$$a(X)=c_1\prod(X-\alpha_i)^{m_i}, \quad b(X)=c_2\prod(X-\beta_j)^{n_j}, \quad c(X)=c_3\prod(X-\gamma_k)^{r_k}.$$

We see, for example

$$f(X)=c_1 c_3^{-1}\prod(X-\alpha_i)^{m_i}\prod(X-\gamma_k)^{-r_k}.$$

Therefore

$$f'/f=\sum\frac{m_i}{X-\alpha_i}-\sum\frac{r_k}{X-\gamma_k}.$$

Likewise

$$g'/g=\sum\frac{n_j}{X-\beta_j}-\sum\frac{r_k}{X-\gamma_k}.$$

Combining both, we obtain

$$\frac{b}{a}= -\frac{\sum\frac{m_i}{X-\alpha_i}-\sum\frac{r_k}{X-\gamma_k}}{\sum\frac{n_j}{X-\beta_j}-\sum\frac{r_k}{X-\gamma_k}}.$$

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

$$N_0(X)=\prod(X-\alpha_i)\prod(X-\beta_j)\prod(X-\gamma_k),$$

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

$$\frac{b}{a}=-\frac{N_0f'/f}{N_0g'/g},$$

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

$$bh=-N_0f'/f,\quad ah = N_0g'/g.$$

Taking the degrees of both sides, we see

$$\begin{aligned} \deg b \le \deg N_0f'/f \le n_0(abc)-1, \\ \deg a \le \deg N_0g'/g \le n_0(abc)-1. \end{aligned}$$

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 that

$$a(X)^n+b(X)^n=c(X)^n.$$

Then $n \le 2$.

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,

$$\begin{aligned} \deg a^n = n\deg a &\le n_0(a^nb^nc^n)-1 \\ &= n_0(abc)-1 \\ &\le \deg(abc)-1 \\ &= \deg a + \deg b + \deg c - 1. \end{aligned}$$

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

$$\begin{cases} n\deg a \le \deg a + \deg b + \deg c - 1 \\ n\deg b \le \deg a + \deg b + \deg c - 1 \\ n\deg c \le \deg a + \deg b + \deg c - 1 \end{cases}$$

It follows that

$$n(\deg a + \deg b + \deg c) \le 3(\deg a + \deg b + \deg c) - 1.$$

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

$$\deg (f^3-g^2) \ge \frac{1}{2}\deg f + 1.$$

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

$$\deg(p-q) \ge \deg p - n_0(p)-n_0(q)+1.$$

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,

$$\begin{aligned} \deg p &\le n_0(pq(p-q))-1 \\ &=n_0(p)+n_0(q)+n_0(p-q)-1 \\ &\le n_0(p)+n_0(q)+\deg(p-q)-1. \end{aligned}$$

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

$$\begin{aligned} \deg(p/r) &\le n_0\left\{\frac{p}{r}\cdot\frac{q}{r}\cdot\frac{p-q}{r}\right\}-1 \\ &\le n_0(p/r)+n_0(q/r)+\deg((p-q)/r)-1 \end{aligned}$$

Therefore

$$n_0(p/r)+n_0(q/r) \ge \deg(p/r)-\deg((p-q)/r)+1=\deg(p)+\deg(p-q)+1$$

On the other hand,

$$n_0(p)+n_0(q) \ge n_0(p/r)+n_0(q/r)+n_0(r) \ge n_0(p/r)+n_0(q/r).$$

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

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

$$\deg(f^3-g^2) \ge \deg(f^3)=3m\ge \frac{1}{2}m+1$$

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

$$\begin{aligned} \deg(f^3-g^2) & \ge \deg(f^3)-n_0(f^3)-n_0(g^2)+1 \\ & \ge 6r - 2r - 3r + 1 \\ & \ge r+1 \\ & = \frac{1}{2}\deg f + 1. \end{aligned}$$

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)