Example 28.25.2. We will give two examples showing that the first displayed map of Lemma 28.25.1 is not an isomorphism.

Let $k$ be a field. Consider the ring

$A = k[x, y, z_1, z_2, \ldots ]/(x^ nz_ n).$

Set $I = (x)$ and let $M = A$. Then the element $y/x$ defines a section of the structure sheaf of $\mathop{\mathrm{Spec}}(A)$ over $D(x) = \mathop{\mathrm{Spec}}(A) \setminus V(I)$. We claim that $y/x$ is not in the image of the canonical map $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _ A(I^ n, A) \to A_ x = \mathcal{O}(D(x))$. Namely, if so it would come from a homomorphism $\varphi : I^ n \to A$ for some $n$. Set $a = \varphi (x^ n)$. Then we would have $x^ m(xa - x^ ny) = 0$ for some $m > 0$. This would mean that $x^{m + 1}a = x^{m + n}y$. This would mean that $\varphi (x^{n + m + 1}) = x^{m + n}y$. This leads to a contradiction because it would imply that

$0 = \varphi (0) = \varphi (z_{n + m + 1} x^{n + m + 1}) = x^{m + n}y z_{n + m + 1}$

which is not true in the ring $A$.

Let $k$ be a field. Consider the ring

$A = k[f, g, x, y, \{ a_ n, b_ n\} _{n \geq 1}]/ (fy - gx, \{ a_ nf^ n + b_ ng^ n\} _{n \geq 1}).$

Set $I = (f, g)$ and let $M = A$. Then $x/f \in A_ f$ and $y/g \in A_ g$ map to the same element of $A_{fg}$. Hence these define a section $s$ of the structure sheaf of $\mathop{\mathrm{Spec}}(A)$ over $D(f) \cup D(g) = \mathop{\mathrm{Spec}}(A) \setminus V(I)$. However, there is no $n \geq 0$ such that $s$ comes from an $A$-module map $\varphi : I^ n \to A$ as in the source of the first displayed arrow of Lemma 28.25.1. Namely, given such a module map set $x_ n = \varphi (f^ n)$ and $y_ n = \varphi (g^ n)$. Then $f^ mx_ n = f^{n + m - 1}x$ and $g^ my_ n = g^{n + m - 1}y$ for some $m \geq 0$ (see proof of the lemma). But then we would have $0 = \varphi (0) = \varphi (a_{n + m}f^{n + m} + b_{n + m}g^{n + m}) = a_{n + m}f^{n + m - 1}x + b_{n + m}g^{n + m - 1}y$ which is not the case in the ring $A$.

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