Finding the image of a morphism between affine schemes

Question

I am confused about two things, one of my problem sets for a course on Abelian varieties, I was given a morphism $\mathbb A^1_k\rightarrow \mathbb A^2_k$ "in coordinates" as: $$f(t)=(t^2-1,t^3-t)$$ Now, since I have never taken a course in classical algebraic geometry, the only way I know how to think about varieties is as schemes (which in this course has been fine since we almost entirely use the language of schemes in lectures). I take this to mean that we are giving a morphism on the set of maximal ideals in $\mathbb A^1_k$ of the form $\langle z-t\rangle$. Here $k$ is a field, with characteristic not equal to $2$, but it's not necessarily algebraically closed, so these are not all the maximal ideals. So my first question, is when someone says that we give a morphism in coordinates, is this actually what it means? Does somehow defining a morphism on maximal ideals of this form uniquely determine a morphism of schemes?

Since I want to think about things as schemes, and these are affine schemes, I wanted to find the ring morphism that induces the morphism of affine schemes which would agree with $f$ on maximal ideals of the form $\langle z-t\rangle$. I deduce that the following ring homomorphism works: \begin{align} \phi:k[x,y]&\longrightarrow k[z]\\ x&\longmapsto z^2-1\\ y&\longmapsto z^3-z \end{align} Now I want to find the image of this morphism. In the solutions, the professor just states that it is the vanishing locus $Y=\mathbb V(y^2-x^2(x+1))$. I can see that $f^{-1}(Y)$ is equal to $\mathbb A^1_k$, as: \begin{align} f^{-1}(Y)=&\{\mathfrak{p}\in \mathbb A^1_k: y^2-x^2(x+1)\in \phi^{-1}(\mathfrak{p})\}\\ =&\{\mathfrak{p}\in \mathbb A^1_k:0\in \mathfrak p\}\\ =&\mathbb A^1_k \end{align} but I don't see why we should have that $Y$ is actually the image. So my second question is how do I actually see that this is the image?

Answer 1

Part one of your question is a duplicate and answered here on MSE. You have some misconceptions in your post, though, so let's take a little time to address those: you're not just specifying a morphism on a set of maximal ideals; you're responsible for a map on all prime ideals plus a morphism of sheaves; the easiest way to package that all up for a morphism of affine schemes is usually as the underlying ring map, since $\operatorname{Spec}$ tells you how to recover all of that information. You got there in the end, though - your identification of the corresponding ring homomorphism is correct.

Calculating the image depends on what you mean by image. Since you haven't specified set-theoretic versus scheme-theoretic, let me try to give you an answer which works for both at the same time. By your work, the image of $f$ is contained in $Y$. Next, any $k$-morphism of affine varieties over $k$ with source $\Bbb A^1_k$ is closed: $\Bbb A^1_k\to X$ corresponds to a map $k[X]\to k[t]$, and if there is a non-constant element of $k[t]$ in the image, the morphism $\Bbb A^1_k\to X$ is finite, while if not, the morphism $\Bbb A^1_k\to X$ is constant, so either way the morphism is closed. Continuing, the image of any irreducible space is irreducible, so the image of $f$ is an irreducible closed subscheme of $Y$. But $Y$ is irreducible of dimension one, so the only choices are a point or $Y$; as $f(0)=(-1,0)$ and $f(1)=(0,0)$, the image must be all of $Y$.

(To find the corresponding $Y$ for a different map, you can calculate the kernel of the map $k[x,y]\to k[z]$ by elimination theory. The reason this works is that if $f:\operatorname{Spec} B\to \operatorname{Spec} A$ is a map of schemes corresponding to $\varphi:A\to B$ and $\varphi$ factors as $A\to A/I \to B$, then $f$ factors as $\operatorname{Spec} B\to \operatorname{Spec} A/I\to \operatorname{Spec} A$, so the image of $\operatorname{Spec} B$ is contained in the closed subscheme $\operatorname{Spec} A/I\subset\operatorname{Spec} A$. The smallest such subscheme happens when $I$ is as large as possible, or when $I=\ker\varphi$.)