Past exam of the introduction to schemes course of the University of Oxford / 2021 / 3

Choose an affine open subscheme $V=\mathrm{Spec}A\subseteq Y$ containing $y$. Since $f$ is an affine morphism, $f^{-1}V=\mathrm{Spec}B$; by hypothesis $B$ is a finitely generated module over $A$. If $\mathfrak{m}\subset A$ corresponds to the closed point $y$, then $\kappa (y)=A/\mathfrak{m}$. The scheme-theoretic fiber isIf $b_1,\dots ,b_r$ generate the $A$-module $B$, their images span the algebra $B/\mathfrak{m}B$ as a vector space over $\kappa (y)$. It is therefore finite-dimensional. By the allowed fact about the spectrum of a finite-dimensional commutative algebra, it has only finitely many prime ideals. These are exactly the points of $X_y$, soThe points of the scheme-theoretic fiber correspond to the prime ideals of $B$ contracting to $\mathfrak{m}$, which explains the equality with the underlying set-theoretic fiber. An empty fiber is allowed. In fact, replacing $A/\mathfrak{m}$ by $\kappa (y)$ gives the same finite-dimensional argument for every point, without requiring it to be a closed point.

Failure in general. Let $f:{\mathbb{A}}_k^1=\mathrm{Spec}k[t]\to \mathrm{Spec}k$ be the structure morphism and take the trivial line bundle $\mathcal{L}={\mathcal{O}}_{{\mathbb{A}}_k^1}$. The polynomial ring $k[t]$ is a Noetherian ring, so this is a coherent sheaf. Its direct image on the one-point target has sectionsA coherent sheaf on $\mathrm{Spec}k$ corresponds to a finite-dimensional vector space over $k$, whereas $1,t,t^2,\dots$ are linearly independent. Thus

Quasi-coherence for the stated morphism. Now let $f$ satisfy (a), and let $\mathcal{F}$ be a quasi-coherent sheaf on $X$. Over any affine open subscheme $V=\mathrm{Spec}A$ of $Y$, write $f^{-1}V=\mathrm{Spec}B$ and $\mathcal{F}{|}_{f^{-1}V}=\tilde{M}$ for a $B$-module $M$. Regard $M$ as an $A$-module through the ring homomorphism $A\to B$. For $a\in A$,The last two localizations are identical: both invert the action of the same element on $M$. These identifications commute with restrictions, so on the basis of principal open subsets they identify $(f_{\ast }\mathcal{F}){|}_V$ with the affine module sheaf $\widetilde{{}_{A}M}$. Hence this direct image is a quasi-coherent sheaf on each $V$, and therefore on $Y$. This part uses only that $f$ is an affine morphism.

Coherence under the Noetherian hypotheses. If $X,Y$ are Noetherian schemes and $\mathcal{F}$ is a coherent sheaf, then $M$ is a finitely generated module over $B$. Choose generators $m_1,\dots ,m_s$ of $M$ over $B$, and generators $b_1,\dots ,b_r$ of $B$ over $A$, supplied by (a). Then the finite familygenerates $M$ as an $A$-module: expand each coefficient of an expression $m={\sum }_{\ell }{c}_{\ell }{m}_{\ell }$ in the $b_j$. Since $A$ is a Noetherian ring, the affine module sheaf $\widetilde{{}_{A}M}$ is a coherent sheaf. This holds on an affine open cover of $Y$, proving

The exact sequence and associated modules. Put $A=k[x]$ and $\mathfrak{m}=(x)$. Evaluation at zero gives the short exact sequence of $A$-modulesThe affine module sheaf associated to $(x)$ is the ideal sheaf $\mathcal{J}$ of the closed point $p$, and the one associated to $A$ is the structure sheaf ${\mathcal{O}}_{{\mathbb{A}}_k^1}$. To identify the third term, on a principal open subset $D(g)$,which equals $\Gamma (i^{-1}D(g),{\mathcal{O}}_p)$. The restrictions also agree, so $\tilde{k}\cong i_{\ast }{\mathcal{O}}_p$. Exactness of localization proves that applying the affine module sheaf construction to the module sequence remains exact: on every stalk the sequence is its localization. ThusEach term is an affine module sheaf, hence a quasi-coherent sheaf. In this example each is also a coherent sheaf, since its defining module is finitely generated over the Noetherian ring $k[x]$.

The failure of left exactness. For an affine module sheaf $\tilde{M}$, the pullback of a sheaf of modules along $i$ corresponds to the tensor product of modules $k{\otimes }_{A}M$. In particular,The first vector space is nonzero, with basis the class of $x$. But $i^{\ast }j$ sends this class to the class of $x$ in $A/(x)$, which is zero. Thus $i^{\ast }j$ is the zero map $k\to k$, and is not injective. More explicitly, the pulled-back sequence iswhere exactness fails at its first copy of $k$. Thus $i^{\ast }$ is not a left-exact functor:Although $(x)$ is a free module over $A$ with one generator, its inclusion into $A$ becomes multiplication by $x$ after choosing its generator; the tensor product of modules with $A/(x)$ kills that map. This makes explicit why local freeness of $\mathcal{J}$ does not preserve this injection under the given pullback.