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Let $(X,\mathcal{O}_X)$ be a quasi-compact ringed space which is locally isomorphic to an affine variety.

Is it true then that $(X,\mathcal{O}_X)$ is always locally isomorphic to an irreducible affine variety?

Intuitively I'd like to say yes, if we take $x\in X$ then we have some open $U\subseteq X$ with $x\in U$ and $(U,\mathcal{O}_X)\cong(V,\mathcal{O}_V)$ for some affine $V$ via an isomorphism $\varphi$. We can then find some closed irreducible $W\subseteq V$ with $\varphi(x)\in W$, but then $\varphi^{-1}(W)$ is not necessarily open in $X$, and I can't seem to find a useful open set related to it.

However on the other hand I'm struggling to come up with a counterexample, any help would be much appreciated.

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  • $\begingroup$ I guess you could further pass to a distinguished open affine of $W$ (open subsets of irreducible sets are again irreducible)? $\endgroup$ – Jane Doé May 4 at 5:37
  • $\begingroup$ @JaneDoé Would we not need our open set of $A\subseteq W$ to be open in $V$ then also? If it were, and if $(A,\mathcal(O)_W)$ were isomorphic to an irreducible affine variety then we could take our open set in $X$ to be $\varphi^{-1}(A)$, but I’m not sure we can do this if $A$ isn’t open in $V$? $\endgroup$ – Dave May 4 at 6:28
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    $\begingroup$ Ah having given it some more thought I think the statement is false. If we take the affine variety $V(XY)$, this will be the union of two lines meeting only at $0$. Then if we take $x=0$, any open set containing $x$ must just be $V(XY)$ minus a finite set of points. Say this open set is isomorphic to an affine variety $W$ by $\psi$, then $W=\psi(V(X))\cup\psi(V(Y))$. These sets will both be closed in $W$ and so it can’t be irreducible. Apologies for not spotting this earlier. $\endgroup$ – Dave May 4 at 7:03
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As in my comment, the answer is no.

If we take our space to be $V(XY)$ over some field $k$, this will be two lines meeting only at $(0,0)$. Taking $x=(0,0)$, any open set containing $x$ will just be $V(XY)$ minus a finite set of points. If this open set is isomorphic via a map $\psi$ to some affine $W$, then we would have $W=\psi(V(X))\cup\psi(V(Y))$. These sets are both closed, non-empty, and neither can be equal to the whole of $W$, and so $W$ cannot be irreducible.

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