What does this notion of scheme morphism mean?

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?

$\begingroup$ 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. $\endgroup$

$\begingroup$ 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? $\endgroup$

$\begingroup$ 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. $\endgroup$

$\begingroup$ 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 $\mathbb$ has points, even though $\< (x, y) \in \mathbb^2 : x^2 + y^2 + 1 = 0 \>$ is empty! $\endgroup$

2 Answers 2

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 $$f : \mathbb^n \ni (x_1, . x_n) \mapsto (f_1, . f_m) \in \mathbb^m$$

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.

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 $\operatorname k$, 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 $$\textrm(X, Y) \cong \textrm(A(Y), A(X))$$ 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 $$\textrm(A, B) \cong \textrm(\operatorname B, \operatorname A)$$ where the LHS is the set of ring homomorphisms $A \to B$ and the RHS is the set of scheme morphisms $\operatorname B \to \operatorname A$. 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 $\phi^\sharp_Y : \mathscr_Y(Y) \to \phi_* \mathscr_X(Y)$ is exactly the map $\phi^*$, once we have identified $\mathscr_Y(Y) = A(Y)$ and $\phi_* \mathscr_X(Y) = A(X)$. Here we may drop the hypothesis that $k$ is algebraically closed.
3. If $\phi : X \to Y$ is a morphism of $k$-schemes, then $\phi |_$ has image contained in $Y(k)$. Moreover, if $\psi : X \to Y$ is another, then $\phi = \psi$ if and only if $\phi |_ = \psi |_$. 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 $\operatorname k \to X$. 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 $\phi |_$.

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 $X = \operatorname A$ and $Y = \operatorname B$, and let $X_m = \operatorname A$ and $Y_m = \operatorname B$. 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 $\phi |_ = \psi |_$ (as maps of topological spaces).

Indeed, by the definition of Jacobson ring, if $\mathfrak

\in X$, then $$\mathfrak

= \bigcap_ <\mathfrak\in X_m, \mathfrak

\subseteq \mathfrak> \mathfrak$$ and by the natural bijection in (2) above, $$\phi(\mathfrak

) = <\phi^*>^ \mathfrak

$$ where $\phi^* : A \to B$ is the $k$-algebra homomorphism corresponding to $\phi : X \to Y$. But it is a fact of set theory that $$<\phi^*>^ \bigcap_<\mathfrak> \mathfrak = \bigcap_<\mathfrak> <\phi^*>^ \mathfrak$$ and our hypothesis is that $\phi (\mathfrak) = \psi (\mathfrak)$, 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 $\operatorname A$ and $\operatorname B$ are both single points, but there are plenty of $k$-morphisms $\operatorname A \to \operatorname B$: 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 $\operatorname K$ is just a point but $\sigma$ gives rise to a non-trivial $k$-morphism $\operatorname K \to \operatorname K$.

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 $\mathscr_Y$ 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 $\mathscr_Y$ over the generic points in addition to the closed points: after all, if $\mathfrak

\subseteq \mathfrak$ then $B_\mathfrak

\subseteq B_\mathfrak$ (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.