Davenport's theorem general case (Algebra by S Lang).

Question

This is in regards to question 13 from chapter 4 of Lang. I got the case when $f,g$ are coprime, and for the general case I got the point where if $A$ is the gcd of $f$ and $g$, and $f'$, $g'$ are such, that, $Af'=f$, and $Ag'=g$, deg $f' \leq $ 2deg $((f')^3-(g')^2) - 2 \implies $ deg $f$ + 3 deg $A \leq $ 2deg $( f^2f' - g^2) - 2$. But I cannot show that deg$(f^2f'-g^2) \leq$ deg$(f^3-g^2)+ 3$deg$A$. Is the way I'm going about it wrong, I cannot see how to apply the hint Lang gave either.

Edit for Clarification The question being asked is to prove that deg$(f) \leq$ deg$(f^3 - g^2) - 2$, when $f,g$ are polynomials over an algebraically closed field. Question

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. — Community, Mar 12 '23 at 07:41
Davenport's theorem follows immediately from Mason's $abc$ theorem for polynomials - which has a very short elementary proof. — Bill Dubuque, Mar 12 '23 at 08:54

score 0 · Answer 1 · answered Jul 18 '24 at 08:51

If you write $d = \gcd(f,g)$ then $f = d \cdot \overline f$ and $g = d \cdot \overline g$ where now $(\overline f, \overline g) = 1$, you need: $\deg f \leq 2 \deg h - 2$, where $h = f^3 - g^2 = d^3\hat f^3 - d^2\hat g^2 = d^2(d\hat f^3-\hat g^2)$. So equivalently you need that: $\deg d + \deg \hat f \leq 2\left(\deg d^2 + \deg (d\hat f^3 - \hat g^2)\right) - 2$.

This should point you in the right direction. For a final hint, remember that you can freely choose $A$ and $B$. Perhaps a particular choice of $A$ would help in trying to reach an inequality of the above form.

Davenport's theorem general case (Algebra by S Lang).

1 Answers1