Question about meaning behind a map of schemes

Question

I saw the following map of schemes in an example,

$${A}^1_\mathbb{C} \rightarrow \operatorname{Spec}(\mathbb{C}[x,y]/(y^2-x^3)),$$ $$t\mapsto (t^2, t^3).$$

Equivalently, this can be written as,

$$ \operatorname{Spec}(\mathbb{C}[T]) \rightarrow \operatorname{Spec}(\mathbb{C}[T^2,T^{3}]),$$ $$T\mapsto(T^2,T^3).$$

This is a map of topological spaces, and the points of $Spec(\mathbb{C}[T])$ are ideals of the form $(T - a)$ along with the generic point $(0).$ It is unclear to me what the map means. The only thing that made sense to me, is that the map is given by, $$(T-a)\mapsto((T-a)^2, (T-a)^3)$$ on prime ideals. Is this correct?

Answer 1

The notation $t\mapsto (t^2,t^3)$ is describing the map by its action on closed points, viewing the curve $\text{Spec}(\mathbb{C}[x,y]/(y^2-x^3))$ as being embedded in the plane $\mathbb{A}^2_{\mathbb{C}} = \text{Spec}(\mathbb{C}[x,y])$ as the curve cut out by the polynomial $y^2 - x^3 = 0$.

The point $a$ in $\mathbb{A}^1_{\mathbb{C}}$ corresponds to the maximal ideal $(T-a)$ in $\mathbb{C}[T]$, while its image $(a^2,a^3)$ in $\mathbb{A}^1_{\mathbb{C}}$ corresponds to the maximal ideal $(x-a^2,y-a^3)$ in $\mathbb{C}[x,y]$, which also descends to a maximal idea in $\mathbb{C}[x,y]/(y^2-x^3)$, since the ideal $(x-a^2,y-a^3)$ contains the ideal $(y^2-x^3)$.

So there's your action on maximal ideals. Formally, a map of affine varieties is given by a map of rings in the opposite direction, and the action on prime ideas is given by taking preimages. The map $\mathbb{C}[x,y]/(y^2-x^3)\to \mathbb{C}[T]$ is given by $x\mapsto T^2$ and $y\mapsto T^3$. Note that this sends $y^2-x^3$ to $0$, so it gives a ring homomorphism $f$. And indeed, the preimage of the ideal $(T-a)$ contains both $(x-a^2)$, since $f(x-a^2) = T^2-a^2\in (T-a)$, and $f(y-a^3) = T^3-a^3\in (T-a)$. Since the preimage of $(T-a)$ is proper and contains the maximal ideal $(x-a^2,y-a^3)$, it's equal to this ideal.

An alternative perspective, also presented in your question, is to view the curve as $\text{Spec}(\mathbb{C}[T^2,T^3])$. Indeed, $\mathbb{C}[x,y]/(y^2-x^3)$ is isomorphic to $\mathbb{C}[T^2,T^3]$ by $x\mapsto T^2$, $y\mapsto T^3$. But it seems misleading to me to write the map of schemes you're interested in as $T\mapsto (T^2,T^3)$. I don't know of a good information notation for this map, since $\text{Spec}(\mathbb{C}[T^2,T^3])$ is not presented to us as a curve embedded in any affine or projective space. On the other hand, in this perspective, the map of rings in the opposite direction is obvious: it's just the natural inclusion $\mathbb{C}[T^2,T^3]\hookrightarrow \mathbb{C}[T]$.

Answer 2

This map is really informal notation for the map $$ \mathbb{A}^1_{\mathbb{C}} = \operatorname{Spec} \mathbb{C}[t] \to \operatorname{Spec} (\mathbb{C}[x,y] / (y^2 - x^3)) $$ corresponding to the ring homomorphism $$ \mathbb{C}[x,y] / (y^2 - x^3) \to \mathbb{C}[t], \\ x \mapsto t^2, y \mapsto t^3. $$ Now, tracing through definitions, if you consider the topological part of this scheme morphism, then it sends the prime ideal $(t-a)$ of $\mathbb{C}[t]$ to the inverse image under this ring homomorphism - which turns out to be the ideal generated by $(x-a^2, y-a^3)$. This, in turn, is the closed point corresponding to $(a^2, a^3) \in \{ (x, y) \in \mathbb{C}^2 \mid y^2 = x^3 \}$ which explains why the original notation is sometimes used.

On the other hand, the zero ideal gets sent to the zero ideal of $\mathbb{C}[x,y] / (y^2 - x^3)$ which is the generic point of this irreducible one-dimensional scheme.