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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 $z=0$?

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  • $\begingroup$ 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). $\endgroup$
    – hm2020
    Sep 13 at 15:27
  • $\begingroup$ 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. $\endgroup$
    – hm2020
    Sep 13 at 15:34
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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 $(x:y:z) = (1:0:0)$.

Now $\newcommand{\P}{\mathbb{P}} \newcommand{\A}{\mathbb{A}} \P^1$ is covered by two copies of $\A^1$, which map homeomorphically onto the subsets of $\P^1$ where $t_0 \neq 0$ and $t_1 \neq 0$, respectively. These two maps $\psi_0, \psi_1: \mathbb{A}^1 \hookrightarrow \P^1$ correspond to the ring homomorphisms \begin{align*} \psi_0^*: \mathbb{C}[t_0, t_1] &\to \mathbb{C}[t]\\ t_0 &\mapsto 1\\ t_1 &\mapsto t \end{align*} and \begin{align*} \psi_1^* : \mathbb{C}[t_0, t_1] &\to \mathbb{C}[t]\\ t_0 &\mapsto t\\ t_1 &\mapsto 1 \, . \end{align*} The map you wrote in the question is the composition $\psi_1^* \circ \phi^* = (\phi \circ \psi_1)^*$, corresponding to the map \begin{align*} \mathbb{A}^1 &\overset{\psi_1}{\hookrightarrow} \P^1 \overset{\phi}{\to} \mathbb{V}(x^2 z - y^3)\\ t &\mapsto [t : 1] \mapsto [t^3 : t^2 : 1] \, . \end{align*} You noted that the point $(1:0:0)$ does not lie in the image of this map, which is because the point $(1:0)$ which maps to it under $\phi$ is not contained in the image of $\psi_1$, but rather in the image of $\psi_0$.

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 $\P^1$.

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    $\begingroup$ 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 $\mathbb{P}^1\to V(x^2z-y^3)$? $\endgroup$
    – LegNaiB
    Sep 15 at 7:06
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    $\begingroup$ @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 $\phi: \mathbb{P}^1 \to \mathbb{V}(x^2z - y^3)$. $\endgroup$ Sep 15 at 17:09
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    $\begingroup$ Thanks again, that really helped for my understanding! $\endgroup$
    – LegNaiB
    Sep 15 at 17:25

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