Nonsingular cubic curve, quotient of $d(x/z)$ and $y/z$ is differential which is regular everywhere. Let $C \subset \mathbb{P}_2$ be the cubic curve defined by$$y^2z = x(x-z)(x-\lambda z)$$with $\lambda \in \mathbb{C} - \{0,1\}$. Let $p = [0, 0, 1]$, $q = [1, 0, 1]$, $r = [\lambda, 0, 1]$, and $s = [0, 1, 0]$. Let $D$ be the divisor $p + q + r - 3s$. Show that $D$ is the divisor of both $d(x/z)$ and $y/z$, so that their quotient is a differential which is regular everywhere.
 A: We may restrict to appropriate affine coordinate patches to check zeros and poles. Indeed, away from $z = 0$, taking the affinization $z=1$, the curve is defined by $$y^2 = x(x-1)(x-\lambda).$$Then it is clear $y/z$ indeed has zeros of degree $1$ at$$p = (0, 0),\text{ }q = (1, 0), \text{ }r = (\lambda, 0),$$and that these are the only zeros of the function on this affine coordinate patch. There are clearly no zeros on the line $z = 0$, so these are all the zeros of $y/z$ on the entire curve. Correspondingly, the only pole can be when $z = 0$, which uniquely specifies the point $s$. Since it is a principal divisor, this shows that the divisor associated to $y/z$ is indeed $p + q + r - 3s$.
On the other hand, $d(x/z)$ is a bit of a hassle, since we will need to compute explicitly its representation in the corresponding local rings. To compute the local rings, we can work in appropriate affinizations, since the local rings are determined only by the local algebraic structure. We find that the uniformizing parameter of the local ring at $p$, $q$, $r$ can all be given by $y$, since the maximal ideals are given by$$(x, y),\text{ }(x-1, y),\text{ }(x - \lambda, y),$$respectively, and we have the relation$$y^2 = x(x-1)(x-\lambda),$$where $2$ of the $3$ factors on the right-hand side are units in each DVR. Then we find through implicit differentiation that$${{dx}\over{dy}} = {{2y}\over{3x^2 - 2(1 + \lambda)x + \lambda}},$$so we obtain valuation $1$ at each of those points.
When $z = 0$, there is a pole since $x/z$ itself has a pole. We claim that besides these $4$ points, there are no other zeros or poles. Indeed, we need only check the rest of the $z = 1$ affine coordinate patch. Indeed, if we localize at$$(y \pm \sqrt{a(a-1)(a-\lambda)}, x - a),\text{ }a \neq 0,\, 1,\,\lambda,$$we have $y \neq 0$, so$$y^2 - a(a-1)(a-\lambda) = x(x-1)(x-\lambda) - a(a-1)(a-\lambda).$$Say, without loss of generality, that our localization point is $y - \sqrt{a(a-1)(a-\lambda)}$; the same argument works if we flip the signs. We have$$y - \sqrt{a(a-1)(a-\lambda)} = {{x(x-1)(x-\lambda) - a(a-1)(a-\lambda)}\over{y + \sqrt{a(a-1)(a-\lambda)}}}.$$The denominator is a unit since $a \neq 0$, $1$, $\lambda$, so our maximal ideal is actually just$$(x(x-1)(x-\lambda) - a(a-1)(a-\lambda), x-a) = (x-a),$$so $(x-a)$ is our uniformizing parameter, and hence$$d\left({x\over{z}}\right) = dx = d(x-a)$$has no zero or pole at any point on the affine patch $z = 1$ away from $p$, $q$, $r$.
Hence, since the degree of a canonical divisor is$$d(d-3) = 0,$$we again obtain $p + q + r - 3s$ as the divisor.
So we have that the quotient of $d(x/z)$ and $y/z$ has divisor $0$, hence it is a differential which is regular everywhere.
