To put this question in precise language, let $X$ be an affine scheme, and $Y$ be an arbitrary scheme, and $f : X \rightarrow Y$ a morphism from $X$ to $Y$. Does it follow that $f$ is an affine morphism of schemes? While all cases are interesting, a counterexample that has both $X$ and $Y$ noetherian would be nice.
[Math] Are morphisms from affine schemes to arbitrary schemes affine morphisms
ac.commutative-algebraag.algebraic-geometry
Related Solutions
Begin with $Y$ equal to the affine plane, $\text{Spec}\ R$, for $R=k[s,t]$. Let $\overline{f}:\overline{X}\to Y$ be the blowing up of $Y$ at the ideal $\mathfrak{m} = \langle s,t \rangle$. Define $I\subset \mathfrak{m}$ to be the ideal $\langle s,t^2 \rangle$. Define $\mathcal{G}$ to be $\widetilde{R/\mathfrak{m}}$, define $\mathcal{F}$ to be $\widetilde{R/I}$, and define $p:\mathcal{F}\to \mathcal{G}$ to be the natural surjection. The pullback $\overline{f}^*\mathcal{G}$ is the structure sheaf of the exceptional divisor $E$. The pullback $$\overline{f}^*p:\overline{f}^*\mathcal{F}\to \overline{f}^*\mathcal{G},$$ is an isomorphism except at a single point $q$ on $E$.
Define $X$ to be the open complement of $\{q\}$ in $\overline{X}$. Define $f:X\to Y$ to be the restriction of $\overline{f}$ to $X$. Since we remove a single closed point from the positive dimensional fiber $E$ of $\overline{f}$, the morphism $f$ is still surjective. On $X$, the natural homomorphism $$f^*p:f^*\mathcal{F}\to f^*\mathcal{G},$$ is an isomorphism. Therefore, the pullback map $H^0(Y,\mathcal{F}) \to H^0(X,f^*\mathcal{F})$ factors through the map $$H^0(p):H^0(Y,\mathcal{F})\to H^0(Y,\mathcal{G}).$$ This map has a one-dimensional kernel.
This already fails in the affine case. For example, if $Y = \mathbf A^1$ and $X = \mathbf A^1 \setminus \{0\}$, then $Y$ is the scheme-theoretic image of $X \hookrightarrow Y$. But if $Z$ is the line with two origins, then the two inclusions $Y \to Z$ agree on $X$ but are not identical.
There is a positive result when $f$ has closed image, for then $X \to \operatorname{im}(f)$ is surjective (see also the introduction to this question). For schemes of finite type over a Jacobson ring (e.g. $\mathbf Z$ or a field $k$), this is the only way $X \to \operatorname{im}(f)$ can be an epimorphism:
Lemma. Let $f \colon X \to Y$ be an epimorphism of schemes, and let $y \in Y$ be a closed point. Then there exists $x \in X$ with $f(x) = y$. In particular, if $Y$ is Jacobson and $f$ of finite type, then $f$ is surjective.
Proof. For the first statement, consider the open $U = Y\setminus\{y\}$ and let $Z$ be two copies of $Y$ glued along $U$. The two natural inclusions $Y \to Z$ differ, so their restrictions to $X$ differ, showing that $y$ is in the (set-theoretic) image of $f$. The second statement follows from Chevalley's theorem. $\square$
However, there exist epimorphisms between non-Jacobson schemes that are not surjective:
Example. Let $A$ be a local domain of dimension $\geq 2$; for example $A = k[x,y]_{(x,y)}$. Let $A \subseteq B$ be a local homomorphism to a discrete valuation ring $B$; for example by blowing up the closed point of $A$ and localising at a generic point of the exceptional fibre. Let $X = \operatorname{Spec} B$ and $Y = \operatorname{Spec} A$. Then $f \colon X \to Y$ is not surjective (it only hits the generic point $\eta$ and the closed point $s$), but it is an epimorphism. Indeed, if $g,h \colon Y \rightrightarrows Z$ agree on $X$, then $g(s) = h(s)$. If $U \subseteq Z$ is an affine open neighbourhood of $g(s)$, then $g$ and $h$ both land in $U$, since the only open subset of $Y$ containing $s$ is $Y$ itself. The result then follows since $A \subseteq B$ is a monomorphism of rings. $\square$
Best Answer
No, here is an example of a morphism $f:X\to Y$ which is not affine although $X$ is affine.
Take $X=\mathbb A^2_k$, the affine plane over the field $k$ and for $Y$ the notorious plane with origin doubled: $Y=Y_1\cup Y_2$ with $Y_i\simeq \mathbb A^2_k$ open in $Y$ and $Y\setminus Y_i= \lbrace O_i\rbrace$, a closed rational point of $Y$.
We take for $f:X\to Y$ the map sending $X$ isomorphically to $Y_1$ in the obvious way.
Then, although the scheme $X$ is affine, the morphism $f$ is not affine because the inverse image $f^{-1}(Y_2)$of the affine open subscheme $Y_2\subset Y$ is
$X \setminus \lbrace 0 \rbrace=\mathbb A^2_k \setminus \lbrace 0 \rbrace$, the affine plane with origin deleted, well known not to be affine.