Without the noetherian hypothesis, (3.3) and (3.4) are false. Let \(A=k[x_0, x_1, x_2, \ldots]\) with the relations \(x_0^n x_n=0\) for \(n=1,2, \ldots\). Let \(I\) be an injective \(A\)-module containing \(A\). Show that \(I \to I_{x_0}\) is not surjective.
Solution
Write \[ A=k[x_0,x_1,x_2,\ldots]/(x_0^n x_n:n\ge1), \] and use the same symbols for the residue classes of the variables. Identify \(A\) with its given submodule of \(I\), so \(1\in I\) means the image of \(1_A\).
1. For every \(n\ge0\), \(x_0^n x_{n+1}\) is nonzero in \(A\), while \(x_0^{n+1}x_{n+1}=0\).In particular, \(A\) is not noetherian.
For fixed \(n\), define a homomorphism \[ A\longrightarrow k[t,u]/(t^{n+1}u) \] by \(x_0\mapsto t\), \(x_{n+1}\mapsto u\), and \(x_j\mapsto0\) for every other \(j\ge1\). Every defining relation maps to zero: the one with index \(n+1\) maps to \(t^{n+1}u\), and all the others have a zero factor. The image of \(x_0^n x_{n+1}\) is \(t^nu\), which is not in the monomial ideal \((t^{n+1}u)\) of \(k[t,u]\). Thus the element is nonzero in \(A\). Its next multiple by \(x_0\) is zero by the defining relation.
Consequently the ideals \[ \Ann_A(x_0^0)\subsetneq\Ann_A(x_0)\subsetneq\Ann_A(x_0^2)\subsetneq\cdots \] form a strictly increasing chain: \(x_{n+1}\) belongs to the \((n+1)\)st annihilator and not to the \(n\)th. This violates the ascending chain condition for a noetherian ring.
2. The class \(1/x_0\in I_{x_0}\) is not in the image of \(I\to I_{x_0}\).
1.The inclusion \(A\hookrightarrow I\) makes that image nonzero.This is a contradiction, so the displayed class cannot lift.The argument works for every module containing \(A\), and in particular for the injective module in the statement.
3. The associated sheaf \(\widetilde I\) on \(X=\Spec A\) is not flasque.
For any ring and module, the associated-sheaf construction gives \[ \Gamma(X,\widetilde I)=I,\qquad \Gamma(D(x_0),\widetilde I)=I_{x_0}, \] with restriction the localization map [@Har10a, Proposition II.5.1]. Step 2 shows that this restriction is not surjective. Thus \(\widetilde I\) is not flasque even though \(I\) is an injective module, proving the failure of the noetherian conclusions invoked in the statement.
4. Q.E.D.
Step 2 proves the requested nonsurjectivity, with its explicit obstruction established in step 1; step 3 gives the corresponding sheaf-theoretic counterexample.