I was trying to solve exercise III.10.6 of Hartshorne and found this post about it. My problem is that I am struggling to show that the morphism $f : X \longrightarrow Y$ (following notation of the post, $X$ is the étale covering and $Y$ the nodal curve) is flat.
In the post it is said that we can use Proposition III.9.7 of the book. The Proposition assumes that $Y$ is regular, but he nodal cubic curve $Y$ is not regular at (0,0), right?
Then, how could we prove that this morphism is flat?
Thanks in advance.
Best Answer
Good catch! The nodal curve is not regular at the origin, and hence proposition III.9.7 does not apply. In order to use the equations at that post, we'll need to assume that we're not in characteristic two, which we'll do here. We can break up the check in to two parts:
Restricting the map $X\to Y$ to $f^{-1}(Y^{sm})\to Y^{sm}$, we see that this map becomes a disjoint union of isomorphisms from each component on to $Y^{sm}$, which is etale. It remains to check what happens over the singular point.
Over the singular point, we verify the criteria of exercise III.10.4, which says $f:X\to Y$ is etale iff for each $x\mapsto y$, the residue field $k(x)$ is a separable algebraic extension of $k(y)$ and the map $\widehat{\mathcal{O}_{Y,y}}\otimes_{k(y)} k(x) \to \widehat{\mathcal{O}_{X,x}}$ is an isomorphism. Both points mapping to the singular point of $Y$ are $k$-rational, so both have residue field $k$ and it remains to demonstrate that the map $X\to Y$ induces isomorphism on the completed local rings. We'll look at $(1,0)\mapsto(0,0)$, with the other map being exactly the same up to swapping $1$ and $-1$. The map on completed local rings is $k[[x,y]]/(y^2-x^3-x^2)\to k[[(a-1),b]]/(b^2-(a^2-1)^2)$ by $x\mapsto a^2-1$ and $y\mapsto ab$, which has inverse $(a-1)\mapsto -1+\sqrt{x+1}$, $b\mapsto\frac{y}{\sqrt{x+1}}$ where by $\sqrt{x+1}$ we mean the formal power series for $\sqrt{x+1}$, $\sum_{n=0}^\infty \binom{1/2}{n}x^n$ which is defined in our field because $2\neq 0$.