How does the failure of integral closure of a coordinate ring relate to the cusp/singularity we see in the corresponding variety?

Consider the affine variety with coordinate ring $R=\mathbb{C}[x,y]/(y^2-x^3)$.

It is clear that the coordinate ring $R$ is not integrally closed since $t=y/x$, an element of the corresponding quotient field, is a root of an integral equation.

What is the geometric intuition relating the failure of $y/x$ to belong to $R$ and the cusp we see on the graph of $y^2-x^2=0$?

Solution 1:

The rational function $y/x$ is regular (i.e., defined) at every point on your curve $C$ other than the origin $P = (0,0)$, but at $P$ it is undefined. In order to get a function defined at the origin, we can blow up the point $P$, which will resolve the singularity at $P$.

The idea behind blowing up is the following. At a singular point $P$, there are multiple tangent lines. So we can replace that single point by several points, one for each tangent line, by adding another coordinate (a copy of $\mathbb{P}^1$) that keeps track of the slope of the tangent line. When $P = (0,0)$ is the origin, the slope is given by the function $y/x$, so our new coordinate $z$ will satisfy $z = y/x$, or $y = xz$ if we clear denominators. So really the blowup is (the closure of) the graph of the function $y/x$ on $C \setminus P$; see section 7.2 of Fulton's Algebraic Curves for details.

Your example is a bit confusing since the cuspidal cubic $y^2 = x^3$ has just one tangent line ($y = 0$) with multiplicity $2$ at the origin, so let's first consider the nodal cubic $y^2 = x^2(x + 1)$. Since its homogeneous term of lowest order is $y^2 - x^2 = (y - x)(y + x)$, then the nodal cubic has tangent lines $y = x$ and $y = -x$ at the origin. The function $y/x$ allows us to distinguish between the two branches of the curve. For points along the branch approximated by $y = x$ we have $y/x \approx 1$, and for those on the other branch, we will have $y/x \approx -1$.

So we will "pull apart" these branches by making a curve in $\mathbb{A}^2 \times \mathbb{P}^1$ whose $z$-coordinate is $y/x$. To find the equations for the blowup $\newcommand{\wt}{\widetilde} \wt{C}$, we simply substitute $y = xz$ in to the equations for $C$: \begin{align*} x^2(x+1) &= y^2 = x^2 z^2 \implies 0 = x^2(z^2 - (x+1)) = 0 \end{align*} The factor of $x^2$ corresponds to the extra factor of $\mathbb{P}^1$ we introduced, while the other factor gives us the relation $x = z^2 - 1$. Since $y = xz$, we find the parametric curve $$ \newcommand{\C}{\mathbb{C}} \wt{C} = \{(z^2 - 1, z(z^2 - 1), z) : z \in \C\} \, . $$ Below is the plot (made in SageMath) of the original curve, the blowup, and the points where it intersects the added copy of $\mathbb{P}^1$.

$\hspace{3cm}$

We can do the same thing in the case of the cuspidal cubic, obtaining the parametric curve $$ \wt{C} = \{(z^2 , z^3, z) : z \in \C\} $$ as shown in Zach Teitler's answer.

$\hspace{3cm}$

It's kind of hard to see what's going on from these still images, so it might be more illuminating to play with the interactive plots (made with SageMathCell) yourself: nodal cubic, cuspidal cubic.

The code for these plots is below.

# nodal cubic
var('t')
p1 = parametric_plot((t^2-1,t*(t^2-1),0), (t,-1.5,1.5), thickness = 2)
p2 = parametric_plot((t^2-1,t*(t^2-1),t), (t,-1.5,1.5), color = "green", thickness = 2)
p3 = parametric_plot((0,0,t), (t,-1.5,1.5), color = "black")
pts = point([(0,0,-1), (0,0,1)], color = "red", size = 5)
show(p1+p2+p3+pts)

# cuspidal cubic
var('t')
p1 = parametric_plot((t^2,t^3,0), (t,-1.5,1.5), thickness = 2)
p2 = parametric_plot((t^2,t^3,t), (t,-1.5,1.5), color = "green", thickness = 2)
p3 = parametric_plot((0,0,t), (t,-1.5,1.5), color = "black")
pts = point([(0,0,0)], color = "red", size = 5)
show(p1+p2+p3+pts)

This paper and this webpage have some great images, too.

Solution 2:

Let $R' = R[y/x]$. We can think of $R \subset R' \subset \operatorname{Frac}(R)$, the field of fractions of $R$.

I claim that $R' \cong \mathbb{C}[t]$. Write $t = y/x \in R'$. In $R'$, $xt=y$, so $x^2 t^2 = y^2 = x^3$. Since $R'$ is an integral domain, $t^2 = x$. And then $t^3 = y$. So (with some hand-waving) $R' = \mathbb{C}[x,y,t]/(y^2-x^3) \cong \mathbb{C}[x,y,t]/(t^2-x,t^3-y) \cong \mathbb{C}[t]$.

If you are familiar with prime spectra of rings, the inclusion $R \subset R'$ gives a map $\operatorname{Spec} R' \to \operatorname{Spec} R$.

If you would rather work with affine varieties, $R = \mathbb{C}[x,y]/(y^2-x^3)$ means $\operatorname{Spec}(R) = V(y^2-x^3)$. What about $R'$? Here $\operatorname{Spec}(R') = V(t^2-x,t^3-y) \subset \mathbb{C}^3$. This is a smooth (nonsingular) curve, parametrized by $t \mapsto (t^2,t^3,t)=(x,y,t)$.

The projection $(x,y,t) \mapsto (x,y)$ gives a map of curves $V(t^2-x,t^3-y) \to V(y^2-x^3)$.

The projection $(x,y,t) \mapsto (x,y)$ has a critical point when the tangent line of the curve is "vertical", i.e., in the kernel of the projection to the $xy$-plane. The derivative of the parametrization is $(2t,3t^2,1)$, which is equal to $(0,0,1)$ when $t=0$. So $t=0$ (or $(x,y,t) = (0,0,0)$) is a critical point of the projection from this curve to the $xy$-plane, and $(x,y)=(0,0)$ is a critical value.