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.

  • $\begingroup$ 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
  • 3
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.