# Mason's Theorem Proof. From Algebra (S. Lang)

I have a question about the last step in the proof of Mason's Theorem. I will write it.

Mason's Theorem. If $$a, b$$ and $$c$$ are relatively prime polynomials such that $$a + b = c$$, then $$\max \{ \deg(a) , \deg(b) , \deg(c)\} \leq N_0(abc) -1,$$ where $$N_0(f)$$ is the number of distinct factors of $$f$$.

Proof: Let $$\displaystyle a(t) = \prod_{i=1}^{s_a} (t-\alpha_i)^{u_i},~ b(t) = \prod_{i=1}^{s_b}(t-\beta_i)^{v_i}, ~c(t) = \prod_{i=1}^{s_c}(t-\gamma_i)^{w_i}$$

$$N_0(abc) = s_a + s_b + s_c$$. Let $$\displaystyle f = \frac{a}{c},~ g= \frac{b}{c}$$. Hence $$f + g = 1 \Rightarrow f' + g' = 0 \Rightarrow f' = - g'$$.

$$\displaystyle\frac{f'}{f} = \sum_{i=1}^{s_a} \frac{u_i}{t-\alpha_i} - \sum_{i=1}^{s_c} \frac{w_i}{t-\gamma_i}, ~~ \frac{g'}{g} =\sum_{i=1}^{s_b} \frac{v_i}{t-\beta_i} - \sum_{i=1}^{s_c} \frac{w_i}{t-\gamma_i}$$

$$\displaystyle \frac{b}{a} = \frac{g}{f} =-\frac{f'/f}{g'/g}=-\frac{M f'/f}{M g'/g},\tag{1}$$

where $$M =\prod_{i=1}^{s_a}(t-\alpha_i) \prod_{i=1}^{s_b}(t-\beta_i)\prod_{i=1}^{s_c}(t-\gamma_i)$$.

So the denominator and nominator in $$(1)$$ has the degree at most $$s_a + s_b + s_c -1$$. Then we conclude the inequality of the theorem.

I understand all the proof except the last thing.

• Perhaps you want see What is the abc conjecture?, by UConn Math, from YouTube, around 33'. It isn't neccesary a response of this comment, it is only you want see it.
– user243301
May 9, 2016 at 17:00

Assume that polynomials have coefficients in a field of characteristic $$0$$. From $$b/a=-(f^{'}/f)/(g^{'}/g)$$ we observe that $$\deg(f^{'}/f)=-\infty$$ if $$b=0$$ and $$\deg(f^{'}/f)=-1$$ if $$a$$ equals to zero (I edit here, because $$f$$ is defined as $$a/c$$ and $$\deg(f^{'})=\deg(f)-1$$).

Hence $$\deg(M f^{'}/{f})\leq n_{0}(abc)-1$$

thus from $$-a(M\cdot f^{'}/{f})=b(M\cdot g^{'}/{g})$$, we deduce that $$a$$ divides $$M\cdot \frac{g^{'}}{g}$$ since $$\gcd(a,b)=1$$, this and previous divisibility relation implies $$\deg(a)\leq n_{0}(abc)-1$$.

A similar argument works for $$\deg(b)\leq n_{0}(abc)-1$$.

Obviously from $$a+b=c$$ we have $$\deg(c)\leq\max(\deg(a),\deg(b))$$. Putting together these results we obtain $$\max(\deg(a,b,c))\leq n_{0}(abc)-1$$.

References (I haven't an open access, I took the details from this, I hope that it is possible):

Jeffrey Paul Wheeler' Thesis, The abc conjecture.