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
\[ B(R) = R[x_ n; n \in \mathbf{Z}]/(x_ n(x_ n - 1), x_ nx_ m; n \not= m) \]
It is easy to show that every prime $\mathfrak q \subset B(R)$ is either of the form
\[ \mathfrak q = \mathfrak pB(R) + (x_ n; n \in \mathbf{Z}) \]
or of the form
\[ \mathfrak q = \mathfrak pB(R) + (x_ n - 1) + (x_ m; n \not= m, m \in \mathbf{Z}). \]
Hence we see that
\[ \mathop{\mathrm{Spec}}(B(R)) = \mathop{\mathrm{Spec}}(R) \amalg \coprod \nolimits _{n \in \mathbf{Z}} \mathop{\mathrm{Spec}}(R) \]
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. $\square$
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