# Calculating normalization of the projective cusp $V(x^2z-y^3)$

I've understood the normalization of the affine cusp $$V(x^2-y^3)\subset\mathbb{A}^2$$ is just $$\phi:\mathbb{A}^1\to \operatorname{Spec}(\mathbb{C}[x,y]/(x^2-y^3))$$ coming from the algebra map $$\phi^*:\mathbb{C}[x,y]/(x^2-y^3) \to \mathbb{C}[t] \\ x \mapsto t^3 \\ y\mapsto t^2$$

How does this extend to the projective curve $$V=V(x^2z-y^3)$$? Do we get a similar map $$\phi:\mathbb{P}^1\to\operatorname{Proj}(\mathbb{C}[x,y,z]/(x^2z-y^3))$$ coming from the map $$\phi^*:\mathbb{C}[x,y,z]/(x^2z-y^3) \to \mathbb{C}[t] \\ x \mapsto t^3 \\ y\mapsto t^2 \\ z \mapsto 1$$ by just extending the affine algebra map? What happens for the point $$(1:0:0)$$ with $$z=0$$?

• The construction of the normalization of an integral scheme (your scheme is integral) is constructed locally: Cover your scheme $\cup U_i:=\cup Spec(A_i)=X$ and construct the integral closure $\tilde{A} \subseteq K(A)$ of $A$ in its quotient field $K(A)$ and glue. You get a normal scheme $\pi:\tilde{X} \rightarrow X$ where the map $\pi$ is finite and satisfies a universal property (Hartshorne, Ex.II.3.8). Sep 13 at 15:27
• I believe the curve is nonsingular (and therefore normal) in two charts and singular in the $D(z)$-chart. Something nontrivial happens in the $D(z)$-chart as you have observed: The normalization of $x^2-y^3$ is the affine line. You should end up with a non-singular curve and via a glueing it should be possible to check if this curve is the projective line. Sep 13 at 15:34

Since $$\DeclareMathOperator{\Proj}{Proj} \DeclareMathOperator{\Spec}{Spec} \newcommand{\C}{\mathbb{C}} \mathbb{P}^1 = \Proj(\C[t_0, t_1])$$, then $$\C[t_0, t_1]$$ is the homogeneous coordinate ring of $$\mathbb{P}^1$$, so the desired homomorphism of graded rings is \begin{align*} \phi^*:\mathbb{C}[x,y,z]/(x^2z-y^3) &\to \mathbb{C}[t_0, t_1] \\ x &\mapsto t_0^3 \\ y &\mapsto t_0^2 t_1 \\ z &\mapsto t_1^3 \, . \end{align*} From this, we see that the point $$(t_0 : t_1) = (1:0)$$ maps to the point $$(x:y:z) = (1:0:0)$$.
As a note, not every map of graded rings induces a map of $$\Proj$$s. In general, given a map $$\varphi: S_\bullet \to R_\bullet$$ of graded rings, there is an induced morphism of schemes $$\varphi^*: \Proj(R_\bullet) \setminus \mathbb{V}(\varphi(S_+)) \to \Proj(S_\bullet)$$ where $$S_+$$ is the irrelevant ideal. (See $$\S6.4$$ of Vakil's The Rising Sea or Tag 01MX of the Stacks Project.) Basically the problem is that, if $$\varphi$$ has a nontrivial kernel, then elements in this kernel will get mapped to "$$(0: 0 : \cdots : 0)$$", which is not a point in projective space. Fortunately in the example above, \begin{align*} \mathbb{V}(\phi^*(S_+)) &= \mathbb{V}(\phi^*(x,y,z)) = \mathbb{V}(t_0^3, t_0^2 t_1, t_1^3) = \varnothing \end{align*} so we obtain a map defined on all of $$\P^1$$.
• Thanks for the detailed answer! That means $V(x^2z-y^3)$ is singular in $\mathbb{P}^2$ and gluing together $(\phi \circ\psi_1)^*$ and $(\phi\circ\psi_2)^*$ gives the normalization $\mathbb{P}^1\to V(x^2z-y^3)$? Sep 15 at 7:06
• @LegNaiB Yes, that's right. $\mathbb{V}(x^2z - y^3)$ is singular at $(0:0:1)$, which can be checked by showing are the partial derivatives vanish there. Yes, those two maps glue to give $\phi: \mathbb{P}^1 \to \mathbb{V}(x^2z - y^3)$. Sep 15 at 17:09