ag.algebraic geometry - A closed subscheme of an open subscheme that is not an open subscheme of a closed subscheme?

Does this work?

Let $k$ be a field, and let $R$ be the subring of $prod_{i=1}^{infty} k[t]$ consisting of those sequences that stabilize. Let $I$ be the ideal comprised of those sequences that stabilize at $0$. Set $X=mathrm{Spec}(R)$, and let $U$ be the complement of $V(I)$ in $X$. (The inclusion $Uto X$ is quasiseparated, but not quasicompact; here $j_*mathcal{O}_U$ is not quasicoherent.)

If $e_i$ denotes the sequence that is $0$ except in the $i$th spot, where it is $1$, then the $e_i$'s generate $I$, and since $e_ie_j=delta_{ij}$, we deduce that $U=coprod_{i=1}^{infty}X_{e_i}=coprod_{i=1}^{infty}mathbf{A}_k^1$.

Now consider the ideal $(t^i)subset k[t]$ as an ideal sheaf on each $X_{e_i}$; these glue to give a quasicoherent ideal sheaf $I$ on $U$, which in turn defines a closed subscheme $Z$ of $U$, which is open in $X$.

I claim that there is no closed subscheme $V$ of $X$ such that $Z$ is open in $V$. If there were, it would be $V=V(J)$ for an ideal $Jsubset R$ such that $Jcdot R_{e_i}=(t^i)$. But this is impossible, since any element $f=(f_i)in J$ would have to have $f_iin (t^i)$ but would nevertheless have to stabilize. So it would have to stabilize at zero; hence $Jsubset I$, so $V(I)subset V(J)$, whence we ... EDIT: do not have a contradiction.

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