 Solution

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.