Call a scheme $Y$ proper positive-dimensional over $\mathrm{Spec}\,\mathbb{C}$ abnormal if there exists an irreducible scheme $X$ affine of finite type over $\mathrm{Spec}\,\mathbb{C}$ and a $\mathbb{C}$-morphism $X\rightarrow Y$ bijective on the underlying spaces.

Call a scheme $Y$ proper positive-dimensional over $\mathrm{Spec}\,\mathbb{C}$ strongly abnormal if there exists an irreducible scheme $X$ affine of finite type over $\mathrm{Spec}\,\mathbb{C}$ and a $\mathbb{C}$-morphism $X\rightarrow Y$ inducing a homeomorphism on the underlying spaces.

Strongly abnormal schemes are obviously abnormal, and abnormal schemes can not be normal (by Zariski's main). By point-set topology, abnormal schemes are irreducible. An example of a strongly abnormal scheme is the projective nodal cubic.

Are there examples of strongly abnormal schemes of each dimension that are not direct products of positive-dimensional $\mathbb{C}$-schemes? Are there examples of abnormal schemes that are not strongly abnormal of each dimension that are not direct products of positive-dimensional $\mathbb{C}$-schemes?

  • 7
    $\begingroup$ Probably related: meta.mathoverflow.net/questions/4200/flood-of-new-users $\endgroup$
    – YCor
    Jun 1, 2019 at 16:51
  • 6
    $\begingroup$ why the down-votes? $\endgroup$
    – user140765
    Jun 1, 2019 at 20:16
  • 1
    $\begingroup$ It is entirely possible the OP meant nodal cubic, see mathoverflow.net/a/330544/74900 $\endgroup$
    – user74900
    Jun 2, 2019 at 6:47
  • 1
    $\begingroup$ @SándorKovács What do you mean, makes no sense at all? Are there no examples in higher dimensions? I don't think this question was badly formulated or something to deserve downvotes. The post defined its terms, some people may dislike these definitions on aesthetical grounds. That is not a valid reason to downvote probably. $\endgroup$
    – user74900
    Jun 2, 2019 at 10:15
  • 2
    $\begingroup$ OK, here is a cleaner version of those comments: 1) every singular proper curve is "abnormal": take the normalization and for each singular point throw away all but one of the pre-images. The remaining curve is affine and the morphism is bijective. 2) If homeomorphism is meant in the Zariski topology, then any two curves are homeomorphic and any bijective morphism between curves is a homeomorphism. And then any singular proper curve is "strongly abnormal". $\endgroup$ Jun 2, 2019 at 16:39

1 Answer 1


In [M. Vitulli, Corrections to: "Seminormal rings and weakly normal varieties'']


Vitulli argues this can only happen for stongly abnormal in dimension 1 (at least for reduced schemes). She was concerned specifically with this sort of question as it caused an issue in a previous paper of hers with Leahy.

Theorem 3.6: Let $f: Y \to X$ be a morphίsm of varieties without one- dimensional components defined over an algebraically closed field of arbitrary characteristic. If $f$ is a homeomorphism (with respect to the Zariski topology on both $X$ and $Y$), then f is a finite morphίsm.

Note that most of that paper is about characteristic 0 (ie, most theorems only work in characteristic 0, and they add the arbitrary characteristic to point out the argument doesn't need characteristic 0).

In your setting, if you have a proper strongly abnormal reduced scheme, then any such cover is also proper. I'm not quite sure about the abnormal version, I feel like I've seen it somewhere, but I don't recall a reference. I would also check some papers on seminormalization, at least for the case of reduced schemes.

  • $\begingroup$ what happens if you are over a non-algebraically closed field? Will there be no "strongly abnormal" schemes in dimension>1? $\endgroup$
    – user141498
    Jun 10, 2019 at 7:32
  • $\begingroup$ I haven't thought about it, but I don't see any new difficulties. $\endgroup$ Jun 18, 2019 at 3:36

You must log in to answer this question.