Differentials on the Fermat curve $x^3 + y^3 = z^3$ Consider the Fermat curve $x^n + y^n = z^n$ in $\mathbb{P}^2$, where we'll assume we work over a field of characteristic $0$. I am able to infer from its Hilbert polynomial that this curve has genus $g = (n-1)(n-2) / 2$. In particular, if $n = 3$, then $g = 1$, and through Serre duality, one infers for this case that the sheaf $\Omega^1_{C}$ of $1$-forms is isomorphic to $\mathscr{O}_C$. I would like to confirm this by hand, but I do not know how to approach this.
Take the three usual affine opens $U_i = \operatorname{Spec} k[x_{0/i},x_{1/i},x_{2/i}] / (x_{i/i} - 1)$ on $\mathbb{P}^2$. We have
$$\begin{split}
\Omega^1_{C}\big|_{U_0} &= \big\langle dx_{1/0},dx_{2/0} : 3x_{1/0}^2 \,dx_{1/0} = 3x_{2/0}^2\,dx_{2/0}\big\rangle, \\
\Omega^1_{C}\big|_{U_1} &= \big\langle dx_{0/1},dx_{2/1} : 3x_{0/1}^2 \,dx_{0/1} = 3x_{2/1}^2\,dx_{2/1}\big\rangle, \\
\Omega^1_{C}\big|_{U_2} &= \big\langle dx_{0/2},dx_{1/2} : 3x_{0/2}^2 \,dx_{0/2} = -3x_{1/2}^2\,dx_{1/2}\big\rangle.
\end{split}$$
This should somehow define a line bundle, and I am aware that a section such as $dx_{1/0}$ on $U_0$ should transform to $-(1/x_{0/1}^2)\,dx_{0/1}$ on $U_1$. Is it possible to see from these equations that $\Omega^1_C \cong \mathscr{O}_C$?
 A: Here's a way to cheat to get the answer. Since the Fermat curve $C: U^3 + V^3 = W^3$ has genus $1$ and a rational point ($[1:0:1]$, for instance), then it is an elliptic curve, hence has a short Weierstrass model $E: y^2 = f(x)$. It is well-known that $\omega := \frac{dx}{y}$ is a nonvanshing regular differential on $E$, so we can pullback $\omega$ along the isomorphism $\varphi: C \overset{\sim}{\to} E$ to obtain a nonvanishing regular differential $\eta := \varphi^* \omega$ on $C$.
So let's find an isomorphism $\varphi$. Let $F = U^3 + V^3 - W^3$. We make the following changes of variable.
\begin{align*}
  \begin{pmatrix}
    U'\\ V'\\ W'
  \end{pmatrix}
  &=
  \begin{pmatrix}
    1/3 & 0 & 0\\
    0 & 1/18 & -2Z\\
    0 & 1/18 & 2Z
  \end{pmatrix}
  \begin{pmatrix}
    U\\ V\\ W
  \end{pmatrix}
\end{align*}
Then $27 F(U', V', W') = U^3 - V^2 W - 432 W^3$. Letting $E: Y^2 Z = X^3 - 432 Z^3$, then we have an isomorphism
\begin{align*}
  \varphi: C &\to E\\
  [U:V:W] &\mapsto [X:Y:Z] = \left[3U:9(V+W):-\frac{1}{4}(V-W)\right]
\end{align*}
with inverse $[X:Y:Z] \mapsto [U:V:W] = \left[\frac{1}{3}X : \frac{1}{18}Y - 2Z :  \frac{1}{18}Y + 2Z\right]$. To pull back $\omega$, let's write things in the affine coordinates $u = U/W, v = V/W$ and $x = X/Z, y = Y/Z$ on the affine opens where $W \neq 0$ and $Z \neq 0$. On these patches we have
\begin{align*}
  \varphi: (u,v) \mapsto (x,y) = \left(\frac{-12u}{v-1}, \frac{-36(v+1)}{v-1}\right)
\end{align*}
Letting $\eta = \varphi^* \omega$, then
\begin{align*}
  \eta &= \varphi^* \omega = \varphi^*\left(\frac{dx}{y}\right) = \frac{d\left(-12\frac{u}{v-1}\right)}{\frac{-36(v+1)}{v-1}}
\end{align*}
which after some simplification becomes
\begin{align*}
  \eta = \frac{1}{3(v+1)} du - \frac{u}{3(v^2-1)} dv \, .
\end{align*}
We know that $\eta$ has no zeroes or poles since $\omega$ doesn't, but we can also check this explicitly. For instance, on the distinguished affine open where $u \neq 0$ we have
\begin{align*}
  du = -\frac{v^2}{u^2} dv
\end{align*}
and after some calculation (using the relation $u^3 + v^3 = 1$) one can show that $\eta = \frac{dv}{3u^2}$. Since $\frac{1}{3u^2}$ has no zeroes or poles on this patch, that means $\eta$ doesn't either. Similarly when $v \neq 0$ then $\eta = -\frac{1}{3v^2} du$ so there are no zeroes or poles on this patch, either. That shows there are no zeroes or poles on $U_2$ and one should be able to check on $U_0$ and $U_1$, as well.
As discussed in the comments, this shows that $\Omega_C$ has a nonvanishing global section, which implies that $\Omega_C \cong \mathscr{O}_C$.
