# Is $y^2 =$ quartic in $x$ smooth at infinity?

Let $q(x)\in K(x)$ be a quartic polynomial in x with distinct roots over the algebraically closed field $K$. Consider the curve $C\subset \Bbb P^2$ given by $y^2-q(x)$. Is $C$ smooth?

Well, at least in the open affine $U_{z\neq 0}$, it is because $\textrm{gcd}(q,q')=1$. But at infinity, i.e. $z=0$, the point $[0:1:0]$ is singular because $\frac{\partial \overline{C}}{\partial y}|_{z=0}= 0$, $\frac{\partial \overline{C}}{\partial z}|_{z=0,x=0}= 0$, $\frac{\partial \overline{C}}{\partial x}|_{z=0,x=0}= 0$.

But doesn't this contradict the fact that every curve $C$ of the form $y^2=$ quartic in $x$ with a rational point is an elliptic curve (can be put in Weierstrass form, as in Cassels "Lectures on elliptic curves", chp 8)?

Not every model for a curve is non-singular. Some models may be singular, and others non-singular. For instance, consider the circle $C: x^2+y^2=1$, and now consider this poor model $C': (x^2+y^2-1)^2=0$. They both have the same zero set, but the second model is singular at every single point!
See Silverman's "The Arithmetic of Elliptic Curves", Example 3.7 in Chapter 1, for a nice example of a birational map between $\mathbb{P}^1$ (smooth!) and a singular cubic.