Polynomial Version for Fermat Last Theorem over $\mathbb{Z}_p$

Question

How to prove the polynomial version for Fermat Last Theorem over $\mathbb{Z}_p$, that is, prove there exists no non-constant coprime solution $f(x), g(x), h(x) \in \mathbb{Z}_p[x]$ satisfies the following equation $$f(x)^n + g(x)^n = h(x)^n.$$

I have already proven the polynomial version for Fermat Last Theorem in $\mathbb{C}[x]$. It can be used if necessary. Any hint will be helpful. Thanks in advance.

Answer 1

Let's derive it as a corollary to the Mason-Stothers theorem, which states

Let $F$ a field and $f, g, h \in F[x]$ coprime with $f + g = h$ and at least one of $f', g', h'$ nonzero. Then $$\max\{\deg(f),\deg(g),\deg(h) \}\leq \deg\big(\text{rad} (fgh)\big)-1 $$ where $\text{rad}(f) \equiv \prod\limits_{\text{irred } p \mid f } p$

In fields of characteristic $0$, like $\mathbb{C}$, having nonvanishing derivative is equivalent to being nonconstant. But in fields of characteristic $p > 0$, this is not the case. However, things aren't very much more complicated, for our purposes: if $F$ has characteristic $p > 0$, then the kernel of the derivative $F[x] \rightarrow F[x]$ is precisely $F[x^p]$. We can use this fact to reduce the proof for characteristic $p>0$ to the proof for characteristic $0$.

Now suppose we are in a field $F$ of characteristic $p > 0$. If we have $f^n + g^n = h^n$, $\gcd(n,p) = 1$, then $(f^n)' = nf'f^{n-1} =0$ iff $f' = 0$ iff $f \in F[x^p]$, and similarly for $(g^n)'$ and $(h^n)'$.

We thus observe that

There exists $m \in N$ such that, setting $\bar{f} = f(x^{1/p^m}), \bar{g} = g(x^{1/p^m}), \bar{h} = h(x^{1/p^m})$, we have that $\bar{f}, \bar{g}, \bar{h} \in F[x]$ are coprime, $\bar{f}^n + \bar{g}^n = \bar{h}^n$, and one of $\bar{f}, \bar{g}, \bar{h}$ has nonvanishing derivative.

Indeed if all of $f^n, g^n, h^n$ have vanishing derivatives, then $f,g,h \in F[x^p]$, and setting $\bar{f} = f(x^{1/p}), \bar{g} = g(x^{1/p}), \bar{h} = h(x^{1/p})$ (which are certainly in $F[x]$) we get a relation $\bar{f}^n + \bar{g}^n = \bar{h}^n$. If all of $\bar{f}, \bar{g}, \bar{h}$ have vanishing derivatives, we can repeat the process, on so and.... you can confirm that it terminates after finitely many steps with the desired polynomials.

The above reduction gives us a relation $$\bar{f}^n + \bar{g}^n = \bar{h}^n$$ which satisfies all the conditions of Mason-Stothers as stated above. The argument is now the same as in the case of $F$ having characteristic $0$. The bounds $$\deg\big(\text{rad}(\bar{f}^n \bar{g}^n \bar{h}^n)\big) - 1 < \deg(\bar{f}) + \deg(\bar{g}) + \deg(\bar{h})$$ and

$$\max\{ \deg( \bar{f}^n), \deg( \bar{g}^n), \deg( \bar{h}^n ) \} \geq \frac{n}{3}\big(\deg( \bar{f}) + \deg( \bar{g}) + \deg( \bar{h} ) \big) $$

are immediate. When $n > 2$ this blatantly contradicts the conclusion of Mason-Stothers, yielding the "Fermat's last theorem" for polynomials over fields of arbitrary characteristic.