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Optional lab: a complete abc theorem for polynomials

This is a proved analogy, not a proof of the integer abc conjecture or of any IUT step. It is worth studying because it makes the conflict between size and distinct prime factors visible in a setting with a simple proof. See also Keith Conrad's exercises on the Mason--Stothers theorem.

Let \(k\) be an algebraically closed field of characteristic zero. Suppose nonzero, pairwise coprime polynomials \(A,B,C\in k[T]\) satisfy \(A+B=C\) and are not all constant. Define \(\operatorname{rad}(ABC)\) as the product of the distinct monic irreducible factors of \(ABC\). The Mason--Stothers theorem says

\[ \max(\deg A,\deg B,\deg C) \leq \deg\operatorname{rad}(ABC)-1. \tag{polynomial abc} \]

Over an algebraically closed field, these irreducible factors are the distinct roots \(T-\alpha\). For example, \(T^m+1=C\) with \(A=T^m,B=1\) has many distinct roots in \(C\); its radical accounts for them even though the power \(T^m\) contributes just one distinct root.

Proof, with every divisibility step exposed

  1. Set \(W=A'B-AB'\), where primes denote formal derivatives. Because the characteristic is zero, \(W=0\) would mean \((A/B)'=0\), hence \(A/B\) is constant. Since \(A,B\) are coprime, both would be constant, contradicting the hypothesis. Thus \(W\neq0\).
  2. If \(\alpha\) is a root of \(A\) of multiplicity \(m\), then \(B(\alpha)\neq0\). In \(A'B-AB'\), each term is divisible by \((T-\alpha)^{m-1}\). The same argument applies to roots of \(B\).
  3. Because \(B=C-A\), we can also write \(W=A'C-AC'\). At a root of \(C\) of multiplicity \(m\), \(A(\alpha)\neq0\); again \((T-\alpha)^{m-1}\) divides \(W\).
  4. The root sets of \(A,B,C\) are disjoint by coprimality. Multiply the factors just obtained, without double-counting: \(ABC/\operatorname{rad}(ABC)\mid W\). Therefore \(\deg A+\deg B+\deg C-\deg\operatorname{rad}(ABC)\leq\deg W\leq\deg A+\deg B-1\). Cancelling the first two degrees proves \(\deg C\leq\deg\operatorname{rad}(ABC)-1\).
  5. Rewrite the same equation as \(A=C-B\) or \(B=C-A\) and repeat the argument to bound \(\deg A\) and \(\deg B\). This proves the claimed maximum bound.

For a sharp example, \(T+1=T+1\) with \(A=T,B=1,C=T+1\) has \(\deg\operatorname{rad}(ABC)=2\), so the bound reads \(1\leq1\).

What this proof uses that integers do not give us

The derivative lowers the multiplicity of each repeated polynomial root by at most one, and its degree is strictly less than the degree of the original polynomial. The same \(W\) sees roots of all three polynomials by rewriting \(A+B=C\). These two facts force the bound. For ordinary integers, no directly corresponding derivative produces this divisibility-and-size argument. Replacing \(\deg\) by \(\log\), and distinct roots by distinct rational primes, describes a conjectural analogy, not a proof; the integer formulation additionally allows an exponent \(1+\varepsilon\) and a constant \(K_\varepsilon\).

Why characteristic zero matters. In characteristic \(p>0\), take \(A=T^p,\ B=1,\ C=(T+1)^p\). Their product has radical \(T(T+1)\), with degree 2, while \(\deg C=p>1\). The derivatives of \(A\) and \(C\) vanish, so \(W=0\) and the proof cannot begin. Exact hypotheses, not just an attractive diagram, are decisive.

Reader exercise: For \(A=T^2,B=1,C=T^2+1\) over \(\mathbb C\), compute \(W=2T\) and check that \(\deg\operatorname{rad}(ABC)=3\). Verify each inequality in step 4. Then go back to the integer examples in 01 and identify precisely which line of this proof cannot be copied.