Example 27.25.2. We will give two examples showing that the first displayed map of Lemma 27.25.1 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 27.25.1. 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 $A$.

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01PN. Beware of the difference between the letter 'O' and the digit '0'.