I apologize for the naivity of this question but I am a beginner of algebraic geometry. Moreover, I realized that the initial question (quoted at the end) was not formulated very clearly. Hopefully, I can do better now involving the things learned from Zhen Lin's comments.

One sometimes sees morphism of affine or projective schemes over a base field defined by some kind of coordinate notation, e.g. \begin{equation} f:\mathbb{A^2}\to \mathbb{A^3}, (a,b)\mapsto (a^2+b,b^2-2,4)\hspace{20pt} (*) \end{equation} but formally a morphism of schemes is a priori a pair consisting of a map $f:\mathbb{A^2}\to \mathbb{A^3}$ of topological spaces and a morphism $f^\sharp:\mathcal{O_{\mathbb{A^3}}}\to f_*\mathcal{O_{\mathbb{A^2}}}$ of structure sheaves.

As Zhen Lin explains in the comments, the map of topological spaces $f$ is determined by what is does on the maximal ideals of $\mathbb{A}^2$. If the base field is algebraically closed, these maximal ideals are of the form $(x_1-a,x_2-b)$ by the Hilbert nullstellensatz. Hence, in this case it suffices to specify an image for the pairs $(a,b)$ to get the topological morphism.

I have (still) two problems with this notion.

Firstly, why exactly does such an assignment $(a,b)\mapsto (f_1(a,b),f_2(a,b),f_3(a,b))$ lead to a morphism of schemes and not only of topological spaces? I guess, the thing is that the $f_i$ are given by polynomials and not by arbitrary functions $k^2\to k$. Let me please look at a concrete example to see how this works.
Consider $f:Spec~k[x]\to Spec~k[y]$ and the assignment $a\mapsto a^2-1$. To construct a map of the structure sheaves, I have to build a $k$-algebra map $\tilde f:k[y]\to k[x]$ such that the pre-image of the ideal $(x-a)$ is the ideal $(y-(a^2-1))$. Is it correct to set $y\mapsto a^2-a-1$ and is it obvious that one finds such a $k$-algebra morphism for the ''higher dimensional'' analogues like $\bar f:k[y_1,y_2,y_3]\to k[x_1,x_2]$?

Secondly, when the base field is not algebraically closed, why does it suffice to specify an image only for the pairs $(a,b)$, too? Maybe, this is because for the special case of the ring $k[x_1,x_2]$, the maximal ideals are still precisely $(x_1-a,x_2-b)$, is this true?

This was the initial question to which Zhen Lin's comments refer to.

What is the very formal legitimation of defining a morphism $f:X\to Y$ of affine or projective schemes by a kind of coordinate notion (explained below) instead of applying the spectrum functor to a ring morphism (in the other direction) or instead of working with the prime ideals?

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    As you know, a scheme consists of two pieces of data: a topological space and a sheaf of local rings (locally isomorphic to spectra of commutative rings, of course). A morphism of schemes is just a morphism of locally ringed spaces, which again consists of two parts: a continuous map of topological spaces and a map of the structure sheaves. For the case of affine and projective varieties we can get away with only considering the closed points because there are enough of them to uniquely determine a morphism of the corresponding schemes. – Zhen Lin Jan 21 '12 at 17:09
  • Thank you for the comment. Can you be a little more precise with what you mean by your last sentence? Does this work also for non-algebraically closed fields? – Daniel Dreiberg Jan 21 '12 at 17:17
  • I believe a necessary condition in the affine scheme case is that the ring must be a Jacobson ring: in this case every prime ideal is uniquely determined as the intersection of maximal ideals containing it. Every $k$-algebra of finite type is a Jacobson ring. – Zhen Lin Jan 21 '12 at 17:24
  • Thanks again. Ok, this means that the morphism $f:\mathbb{A}^2\to \mathbb{A}^2$ of the question is determined if we are over an algebraically closed field. If the field isn't closed, there may be more maximal ideals and the morphism may not be determined by this assignment, right? – Daniel Dreiberg Jan 21 '12 at 17:49
  • No, I didn't say anything about algebraically closed fields there: $k$ is an arbitrary field. The trouble becomes a matter of understanding what the maximal ideals are. For example, the affine scheme $\{ x^2 + y^2 + 1 = 0 \}$ over $\mathbb{R}$ has points, even though $\{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 + 1 = 0 \}$ is empty! – Zhen Lin Jan 21 '12 at 17:59
up vote 3 down vote accepted

The category of affine schemes is equivalent to the opposite of the category of commutative rings, so to specify a morphism between two affine schemes it suffices to specify a morphism in the other direction between their rings of functions. For affine space over an arbitrary base ring $S$ a morphism $$f : \mathbb{A}^n \ni (x_1, ... x_n) \mapsto (f_1, ... f_m) \in \mathbb{A}^m$$

where $f_1, ... f_n$ are polynomials with coefficients in $S$ is merely the morphism corresponding to the morphism $S[y_1, ... y_m] \mapsto S[x_1, ... x_n]$ sending $y_i$ to $f_i$. Note that $S$ does not even need to be a field, much less an algebraically closed field.

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    Thank you for answering. Unfortunately, I think that your answer somehow implicitly assumes what I asked for: You identify $\mathbb{A}^n$ with tuples $(x_1,\ldots,x_n)$ in $S^n$ but I think of $\mathbb{A}^n$ as the spectrum of $S[x_1,\ldots,x_n]$ as a scheme or by the correspondence you mentioned as an element of $S[x_1,\ldots,x_n]$ itself. – Daniel Dreiberg Jan 22 '12 at 9:47
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    @Daniel: I do not. I identify $\mathbb{A}^n$ with the spectrum of $S[x_1, ... x_n]$, and I am giving a definition of what the polynomial notation means for a morphism. (You can verify that it specializes to the naive definition for points over any extension of any residue field of $S$.) – Qiaochu Yuan Jan 22 '12 at 20:02
  • I see and I apologize for my stupidity. – Daniel Dreiberg Jan 22 '12 at 20:19

Allow me to focus on (irreducible) affine varieties over an algebraically closed field $k$, for simplicity. Let $X$ and $Y$ be two such: that is to say, $X$ and $Y$ are integral affine schemes of finite type over $k$ (or $\operatorname{Spec} k$, if we're being pedantic), with affine coordinate rings $A(X)$ and $A(Y)$ respectively. I write $X(k)$ and $Y(k)$ for the subspace of $k$-valued points of $X$ and $Y$; since $k$ is algebraically closed, every closed point of $X$ and $Y$ are $k$-valued points. Note that $X(k)$ and $Y(k)$ are irreducible affine varieties in the classical sense.

I claim the following are true:

  1. If $\phi : X(k) \to Y(k)$ is a morphism of varieties in the classical sense, then the map $$f \mapsto f \circ \phi$$ is a $k$-algebra homomorpism $\phi^* : A(Y) \to A(X)$, when elements of $A(Y)$ are regarded as genuine functions $Y(k) \to k$. On the other hand, if $\phi^* : A(Y) \to A(X)$ is a $k$-algebra homomorphism, there is a map $\phi : X(k) \to Y(k)$ so that $\phi^*$ is of the above form. (Simply choose generators of $A(Y)$ and $A(X)$ and use them to embed $Y$ and $X$ in some affine space $k^n$.) In other words, there is a natural bijection $$\textrm{Hom}(X, Y) \cong \textrm{Hom}(A(Y), A(X))$$ where the LHS is the set of morphisms $X \to Y$ and the RHS is the set of $k$-algebra homomorphisms $A(Y) \to A(X)$.

    This is explained in detail in Hartshorne's Algebraic Geometry [Ch. I, §3]. This is the only part where the hypothesis that $k$ is algebraically closed is truly indispensable.

  2. If $A$ and $B$ are any two (commutative) rings, then there is a natural bijection $$\textrm{Hom}(A, B) \cong \textrm{Hom}(\operatorname{Spec} B, \operatorname{Spec} A)$$ where the LHS is the set of ring homomorphisms $A \to B$ and the RHS is the set of scheme morphisms $\operatorname{Spec} B \to \operatorname{Spec} A$.

    This is proven in Hartshorne [Ch. II, §2]. For our purposes, this implies that any $k$-algebra homomorphism $\phi^* : A(Y) \to A(X)$ gives rise to a unique morphism of $k$-schemes $\phi : X \to Y$ such that $\phi^\sharp_Y : \mathscr{O}_Y(Y) \to \phi_* \mathscr{O}_X(Y)$ is exactly the map $\phi^*$, once we have identified $\mathscr{O}_Y(Y) = A(Y)$ and $\phi_* \mathscr{O}_X(Y) = A(X)$. Here we may drop the hypothesis that $k$ is algebraically closed.

  3. If $\phi : X \to Y$ is a morphism of $k$-schemes, then $\phi |_{X(k)}$ has image contained in $Y(k)$. Moreover, if $\psi : X \to Y$ is another, then $\phi = \psi$ if and only if $\phi |_{X(k)} = \psi |_{Y(k)}$.

    The first claim is a straightforward consequence of the fact that a $k$-valued point of $X$ is exactly the same thing as a morphism of $k$-schemes $\operatorname{Spec} k \to X$. The second claim just comes from putting together the natural bijections of (1) and (2) and recognising that this is simply the correspondence between $\phi$ and $\phi |_{X(k)}$.

Here is a modified form of (3). Let $A$ and $B$ be $k$-algebras of finite type, where $k$ is no longer assumed to be algebraically closed. It is a fact of commutative algebra that a $k$-algebra of finite type is a Jacobson ring. Let $X = \operatorname{Spec} A$ and $Y = \operatorname{Spec} B$, and let $X_m = \operatorname{MaxSpec} A$ and $Y_m = \operatorname{MaxSpec} B$. I claim that if $\phi, \psi : X \to Y$ are two morphisms of $k$-schemes, then $\phi = \psi$ (as maps of topological spaces) if and only if $\phi |_{X_m} = \psi |_{Y_m}$ (as maps of topological spaces).

Indeed, by the definition of Jacobson ring, if $\mathfrak{p} \in X$, then $$\mathfrak{p} = \bigcap_{\mathfrak{m} \in X_m, \mathfrak{p} \subseteq \mathfrak{m}} \mathfrak{m}$$ and by the natural bijection in (2) above, $$\phi(\mathfrak{p}) = {\phi^*}^{-1} \mathfrak{p}$$ where $\phi^* : A \to B$ is the $k$-algebra homomorphism corresponding to $\phi : X \to Y$. But it is a fact of set theory that $${\phi^*}^{-1} \bigcap_{\mathfrak{m}} \mathfrak{m} = \bigcap_{\mathfrak{m}} {\phi^*}^{-1} \mathfrak{m}$$ and our hypothesis is that $\phi (\mathfrak{m}) = \psi (\mathfrak{m})$, so indeed $\phi = \psi$, at least as maps of topological spaces.

Unfortunately, it need not be true that $\phi = \psi$ as morphisms of $k$-schemes. For a simple example, consider $A = k[t]/(t^2)$ and $B = k[x, y]/(x^2, x y, y^2)$. It is clear that $\operatorname{Spec} A$ and $\operatorname{Spec} B$ are both single points, but there are plenty of $k$-morphisms $\operatorname{Spec} A \to \operatorname{Spec} B$: in fact, at least a whole $k^2$ worth, given by $$t \mapsto a x + b y$$ where $(a, b) \in k^2$. One may object that $A$ and $B$ are not integral domains, but it turns out that this hypothesis is not sufficient when $k$ is not algebraically closed. For example, let $K$ be any finite field extension of $k$ with a non-trivial $k$-automorphism $\sigma : K \to K$; as usual $\operatorname{Spec} K$ is just a point but $\sigma$ gives rise to a non-trivial $k$-morphism $\operatorname{Spec} K \to \operatorname{Spec} K$.

Fortunately, there is still a way to rescue the situation: in addition to assuming that $A$ and $B$ are integral domains, we also assume $\phi^\sharp$ and $\psi^\sharp$ agree on the stalks of $\mathscr{O}_Y$ over the closed points. This lets us embed $A$ and $B$ into their respective fields of fractions $K(A)$ and $K(B)$, and it follows that $\phi^\sharp$ and $\psi^\sharp$ must agree on the stalks of $\mathscr{O}_Y$ over the generic points in addition to the closed points: after all, if $\mathfrak{p} \subseteq \mathfrak{m}$ then $B_\mathfrak{p} \subseteq B_\mathfrak{m}$ (considered as subrings of $K(B)$). This is enough to conclude that $\phi^\sharp = \psi^\sharp$ as maps of sheaves, and so $\phi = \psi$ as morphisms of schemes.

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