Lemma 110.45.1. There exists an affine scheme $X = \mathop{\mathrm{Spec}}(A)$ and a closed subscheme $T \subset X$ such that $T$ is Zariski locally on $X$ cut out by ideals generated by idempotents, but $T$ is not cut out by an ideal generated by idempotents.

## 110.45 Ideals generated by sets of idempotents and localization

Let $R$ be a ring. Consider the ring

It is easy to show that every prime $\mathfrak q \subset B(R)$ is either of the form

or of the form

Hence we see that

where the topology is not just the disjoint union topology. It has the following properties: Each of the copies indexed by $n \in \mathbf{Z}$ is an open subscheme, namely it is the standard open $D(x_ n)$. The "central" copy of $\mathop{\mathrm{Spec}}(R)$ is in the closure of the union of any infinitely many of the other copies of $\mathop{\mathrm{Spec}}(R)$. Note that this last copy of $\mathop{\mathrm{Spec}}(R)$ is cut out by the ideal $(x_ n, n \in \mathbf{Z})$ which is generated by the idempotents $x_ n$. Hence we see that if $\mathop{\mathrm{Spec}}(R)$ is connected, then the decomposition above is exactly the decomposition of $\mathop{\mathrm{Spec}}(B(R))$ into connected components.

Next, let $A = \mathbf{C}[x, y]/((y - x^2 + 1)(y + x^2 - 1))$. The spectrum of $A$ consists of two irreducible components $C_1 = \mathop{\mathrm{Spec}}(A_1)$, $C_2 = \mathop{\mathrm{Spec}}(A_2)$ with $A_1 = \mathbf{C}[x, y]/(y - x^2 + 1)$ and $A_2 = \mathbf{C}[x, y]/(y + x^2 - 1)$. Note that these are parametrized by $(x, y) = (t, t^2 - 1)$ and $(x, y) = (t, -t^2 + 1)$ which meet in $P = (-1, 0)$ and $Q = (1, 0)$. We can make a twisted version of $B(A)$ where we glue $B(A_1)$ to $B(A_2)$ in the following way: Above $P$ we let $x_ n \in B(A_1) \otimes \kappa (P)$ correspond to $x_ n \in B(A_2) \otimes \kappa (P)$, but above $Q$ we let $x_ n \in B(A_1) \otimes \kappa (Q)$ correspond to $x_{n + 1} \in B(A_2) \otimes \kappa (Q)$. Let $B^{twist}(A)$ denote the resulting $A$-algebra. Details omitted. By construction $B^{twist}(A)$ is Zariski locally over $A$ isomorphic to the untwisted version. Namely, this happens over both the principal open $\mathop{\mathrm{Spec}}(A) \setminus \{ P\} $ and the principal open $\mathop{\mathrm{Spec}}(A) \setminus \{ Q\} $. However, our choice of glueing produces enough "monodromy" such that $\mathop{\mathrm{Spec}}(B^{twist}(A))$ is connected (details omitted). Finally, there is a central copy of $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(B^{twist}(A))$ which gives a closed subscheme whose ideal is Zariski locally on $B^{twist}(A)$ cut out by ideals generated by idempotents, but not globally (as $B^{twist}(A)$ has no nontrivial idempotents).

**Proof.**
See above.
$\square$

## 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.

## Comments (0)