多项式恒等式$f(x)^3 - f(x)h(x)^2 = g(x)^2h(x)$

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若存在多项式$f(x),g(x),h(x)\inC[x]$使恒等式$f(x)^3 - f(x)h(x)^2= g(x)^2h(x)$成立，且$g(x)$不恒为0，则$\frac{f(x)}{g(x)},\frac{h(x)}{g(x)}$都是常数。


Henry Swanson在MSE写道：

So I  showed that it would give rise to an integer solution to $pq(p + q)(p - q) = r^2$. The pathway to get there is really neat, but long, so unless someone asks, I'll omit it. Using a vaguely geometric argument from Fermat, I showed there are no integer solutions.

But this was a very 1) lengthy 2) tricky-to-motivate 3) bizarre proof, and it would have been much easier if I could have proved that elliptic curves do not admit a rational parameterization.

他使用了Fermat的无穷递降法。他因为证明很冗长，所以没有写出。但具体怎样做呢？

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仿照Corollary 1.8.的证明：
设$s(x)=\gcd(f(x),h(x))$，设$f(x)=u(x)s(x),h(x)=v(x)s(x)$，则 $u(x)$与$v(x)$是互质的多项式。
\[f(x)h(x)(f(x)^2-h(x)^2)=(g(x)h(x))^2\]
设$w(x)=\frac{g(x)v(x)}{s(x)}$，则
\[\label1\tag1\implies u(x)v(x)(u(x)-v(x))(u(x)+v(x))=w(x)^2\]
设$(u,v,w)$是\eqref{1}的解，使得$\max(\deg u, \deg v)$最小。
继续仿照1.6.的证明：
因为$u(x)$与$v(x)$互质，所以左边4个因子是互质的，右边是完全平方，所以左边4个因子都是完全平方，即$u(x)$与$v(x)$与$u(x)+v(x)$与$u(x)-v(x)$都是完全平方。设
\[
u=a^2,\quad v=b^2,\quad u-v=c^2,\quad u+v=d^2
\]
则
\[(a+b)(a-b)(a+ib)(a-ib)=c^2d^2\]
\[\implies(a+b)(ia-ib)(a+b+ia-ib)(a+b-(ia-ib))=i(1+i)(1-i)c^2d^2\]
\[\implies(a+b)(ia-ib)(a+b+ia-ib)(a+b-(ia-ib))=(\sqrt{2i}cd)^2\]
设$u'=a+b,v'=ia-ib,w'=\sqrt{2i}cd$，这说明$(u',v',w')$也是\eqref{1}的解！
但$\max(\deg u',\deg v')\le\max(\deg a,\deg b)=\frac{\max(\deg u, \deg v)}2$，
根据假设的“$(u,v,w)$是使得$\max(\deg u, \deg v)$最小的\eqref{1}的解”，$\max(\deg u',\deg v')\ge\max(\deg u, \deg v)$，因此$$\frac{\max(\deg u, \deg v)}2\ge\max(\deg u, \deg v)$$
因此$\max(\deg u, \deg v)=0$，因此$u$和$v$是常数，代入$f(x)=u\cdot s(x),h(x)=v\cdot s(x)$得，$g(x)=\sqrt{\frac uv(u^2-v^2)}s(x)$，所以$\frac{f(x)}{g(x)},\frac{h(x)}{g(x)}$都是常数。

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Isomorphic function fields of projective curves, bijection of points.

It can be shown that two algebraic varieties are birationally equivalent if and only if their function fields are isomorphic.

Undergraduate Algebraic Geometry第46页：