Cohomology of sheaf of differentials on a hyperelliptic curve. This is 21.4.C in Vakil's Foundations of Algebraic Geometry.
Let $C$ be a hyperelliptic curve over a field $k$ (algebraically closed, char not 2), with $r$ branch points ($r \geq 4$ and is even). In Proposition 19.5.2, a cover with two affine open sets is given : $\operatorname{Spec}(k[x, y] / (y^2 - f(x)))$ and  $\operatorname{Spec}(k[u, v]/(v^2 - g(u)))$, where $g$ is $f$ with coefficients reversed, neither $g$ nor $f$ have any repeated roots, and $0$ is not a root of either $f$ nor $g$. The transition function on the overlap is : $(k[x, y] /(y^2 - f(x)))_x \rightarrow (k[u, v] /(v^2 - g(u)))_u$ by $x \rightarrow 1/u, y \rightarrow v / u^{r/2}$.
The task is to find $h^1(C, \Omega_{C/k})$, and that $x^{-1}dx$ is a generator. From Serre duality we know this is 1 dimensional. I'm stuck on the generator part.
My approach is below.
Let $A = k[x, y] / (y^2 - f(x)),B = k[u, v] / (v^2 - g(u)) $. Then, $\Omega_{A/k} = A^2 / (-f'(x), 2y)$, and $\Omega_{B/k} = B^2 / (-g'(u), 2v)$. I then computed $du = d(1/x) = \frac{-1}{x^2} dx $ and $dv =d(y/x^{r/2}) = (dy x^{r/2} -  \frac{r}{2}x^{r/2 - 1}y  dx) /  x^r = x^{-r/2} dy - \frac{r}{2}x^{-r/2 - 1}y dx$
For simplicity, I'll write $f dx + g dy$ as $(f, g)$ , so we have $(-1/x^2, 0) $ and $(-\frac{r}{2} x^{-r/2-1} y, x^{-r/2})$.
Now I try to compute $h^1(C, \Omega_{C/k})$ using Cech cohomology using the two affine covers. I want to compute homology at the "two open sets" part. Everything on $\Omega_{A_x/k}$ is sent to 0. I just need to compute the range of $\Omega_{A/k} \oplus \Omega_{B/k}\rightarrow \Omega_{A_x/k}$, and show that $(\frac{1}{x}, 0)$ is not in it.
I'm not sure how to do this part. I think I see a way: the only way for there to be a difference of 1 in the coefficients is to use $dv$, but that introduces an extra $y$ that we can't get rid of. How do I make this precise?
 A: It helps to first describe $\Omega_{A_x/k}$. Choose $\alpha(x),\beta(x)\in k[x]$ such that
\begin{equation}
 \det\begin{pmatrix}
   \alpha(x)&-f'(x)\\\beta(x)&2f(x)
  \end{pmatrix}=1. 
\end{equation}
Set $e_0,e_1,e_2\in \Omega_{k[x,y]/k}$ as follows
\begin{equation}
  e_1:=\mathrm{d}x,\quad (e_2,e_3):=\big(y\mathrm{d}x,\mathrm{d}y\big)\begin{pmatrix}
   \alpha(x)&-f'(x)\\\beta(x)&2f(x)
  \end{pmatrix}. 
\end{equation}
Then it is easy to see in $\Omega_{k[x,y]/k}$ we have
\begin{equation}
 k[x]y\mathrm{d}x\oplus k[x]\mathrm{d}y=k[x]e_2\oplus k[x]e_3. 
\end{equation}
Then we have
\begin{equation}
 \begin{split}
  \Omega_{A_x/k}\overset{A_x}{\cong}\quad&\frac{A_x\mathrm{d}x\oplus A_x\mathrm{d}y}{2y\mathrm{d}y-f'(x)\mathrm{d}x}\\
  \overset{k[x,x^{-1}]}{\cong}\quad&\frac{k[x,x^{-1}]\mathrm{d}x\oplus k[x,x^{-1}]y\mathrm{d}x\oplus k[x,x^{-1}]\mathrm{d}y\oplus k[x,x^{-1}]y\mathrm{d}y}{-f'(x)\mathrm{d}x+0+0+2y\mathrm{d}y,\;0-f'(x)y\mathrm{d}x+2f(x)\mathrm{d}y+0}\\
  \overset{k[x,x^{-1}]}{\cong}\quad&\frac{k[x,x^{-1}]\mathrm{d}x\oplus k[x,x^{-1}]y\mathrm{d}x\oplus k[x,x^{-1}]\mathrm{d}y}{0-f'(x)y\mathrm{d}x+2f(x)\mathrm{d}y+0}\\
  \overset{k[x,x^{-1}]}{\cong}\quad&\frac{k[x,x^{-1}]e_1\oplus k[x,x^{-1}]e_2\oplus k[x,x^{-1}]e_3}{e_3}\\
  \overset{k[x,x^{-1}]}{\cong}\quad&k[x,x^{-1}]e_1\oplus k[x,x^{-1}]e_2
 \end{split}
\end{equation}
where isomorphisms are between $k[x,x^{-1}]$-modules.
Then similarly we describe $\Omega_{A/k},\Omega_{B/k}$. Notice that modulo $e_3$ we have
\begin{equation}
 \begin{split}
  \mathrm{d}v=&-\frac{r}{2}x^{-r/2-1}y\mathrm{d}x+x^{-r/2}\mathrm{d}y\\
  =&\big(y\mathrm{d}x,\mathrm{d}y\big)\begin{pmatrix}
   -\frac{r}{2}x^{-r/2-1}\\x^{-r/2}
  \end{pmatrix}\\
  =&(e_2,e_3)\begin{pmatrix}
   2f(x)&f'(x)\\-\beta(x)&\alpha(x)
  \end{pmatrix}\begin{pmatrix}
   -\frac{r}{2}x^{-r/2-1}\\x^{-r/2}
  \end{pmatrix}\\
  =&x^{r/2}g'(u)e_2+\cdots\\
  v\mathrm{d}u=&2x^{r/2}\big(u^2g(u)\big)e_2+\cdots. 
 \end{split}
\end{equation}
For $\Omega_{A/k}$ and $\Omega_{B/k}$ as modules over $k[x]$ or $k[x^{-1}]$, we have
\begin{equation}
 \begin{split}
  \Omega_{A/k}\cong&k[x]\mathrm{d}x+k[x]y\mathrm{d}x+k[x]\mathrm{d}y+k[x]y\mathrm{d}y\\
  =&k[x]e_0+k[x]e_2\\
  \Omega_{B/k}\cong&k[u]\mathrm{d}u+k[u]v\mathrm{d}u+k[u]\mathrm{d}v+k[u]v\mathrm{d}v\\
  =&k[u]\mathrm{d}u+k[u]v\mathrm{d}u+k[u]\mathrm{d}v\\
  =&k[u]\mathrm{d}u+x^{r/2}\Big(k[u]\big(u^2g(u)\big)e_2+k[u]g'(u)e_2\Big)+\cdots\\
  =&k[u]\mathrm{d}u+x^{r/2}\langle u^2g(u),g'(u)\rangle e_2+\cdots\\
  \cong&x^{-2}k[x^{-1}]e_1+x^{r/2}\langle u^2g(u),g'(u)\rangle e_2+\cdots
 \end{split}
\end{equation}
where $\langle u^2g(u),g'(u)\rangle\subset k[u]$ is the ideal generated by $u^2g(u),g'(u)$. Since $g,g'$ are coprime, we have
\begin{equation}
u^2k[u]\subset\langle u^2g(u),g'(u)\rangle\subset k[u]
\end{equation}
and then modulo $e_3$, we have
\begin{equation}
x^{-2}k[x^{-1}]e_1+x^{r/2-2}k[x^{-1}]e_2\subset\Omega_{B/k}\subset x^{-2}k[x^{-1}]e_1+x^{r/2}k[x^{-1}]e_2+\cdots. 
\end{equation}
In particular, since $r\geq 4$, we have $k[x^{-1}]e_2\subset \Omega_{B/k}$.
We do not consider $e_3$ since its image in $\Omega_{A_x/k}$ is zero. Note that the summation is not direct.
With these calculations, it is easy to see the image of $\Omega_{A/k}\times\Omega_{B/k}\to \Omega_{A_x/k}$ is the following vector subspace
\begin{equation}
 \begin{split}
  \mathrm{im}\Big(\Omega_{A/k}\times\Omega_{B/k}\to \Omega_{A_x/k}\Big)=&\big(k[x]+x^{-2}k[x^{-1}]\big)e_1\oplus \Big(k[x]+x^{r/2}k[x^{-1}]\Big)e_2\\
  =&\Big(\bigoplus_{m\ne -1}kx^m\Big)e_1\oplus k[x,x^{-1}]e_2. 
 \end{split}
\end{equation}
