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!

Notice that Cassels' transformation gives a birational map between the two curves, the quartic and the cubic in Weierstrass form (i.e., there are rational maps in both directions). However, this does not guarantee that both curves need to be non-singular, as the maps are not morphisms (regular rational maps).

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.


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