$\newcommand{\Spec}{\mathrm{Spec}\ }$ Let $(P)$ be a property of rings. I call $(P)$ local when $(P)$ satisfy these two conditions:

  • if $A$ is a ring satisfying $(P)$, then the distinguished rings $A_f$ also satisfy $(P)$;
  • If $\Spec A$ is covered by distinguished open $\Spec A_i$ with the $A_i$ having $(P)$, then $A$ satisfy $(P)$.

Now if $(P)$ is local, then it is natural to extend the property $(P)$ to schemes by saying that a scheme $X$ has property $(P)$ iff for all open affines $\Spec A$ of $X$, $A$ has property $(P)$.

By definition of locality, then

  • a ring $A$ satisfy $(P)$ iff $\Spec A$ satisfy $(P)$;
  • a scheme $X$ satisfy $(P)$ iff $X$ can be covered by affines open $\Spec A_i$ with $A_i$ satisfying $(P)$.

Likewise in the relative setting, local properties of morphisms of rings allow to define a corresponding notion for morphisms of schemes.

[I have to point out that sometimes extending a local property $(P)$ of rings to schemes this way is called $(\mathrm{locally}\ P)$, and a scheme $X$ is said to have property $(P)$ when $X$ is locally P and satisfy some finiteness condition. For instance $X$ is noetherian when it is locally noetherian and quasi-compact; a morphism is of finite presentation when it is locally of finite presentation and quasi-compact + quasi-separated.]

Now while this is a standard construction explained in all textbooks, it is harder to find references for what happen to the global sections of non affine open subschemes.

Indeed, if $X$ has a local property $(P)$, then an open scheme $U$ has also property $(P)$, but $\Spec O_X(U)$ may not have $(P)$ when $U$ is not affine.

For instance:

However there are properties that hold for sections over any open subschemes:

  • If $X$ is reduced, then for every open subscheme $U$, $O_X(U)$ is reduced;
  • If $X$ is integral, then for every open subscheme $U$, $O_X(U)$ is integral.

I am interested to what happens with other properties $(P)$. I am also interested to what happens in the relative case: if a morphism $X \to Y$ has property $(P)$, then does $\Spec_Y(X) \to Y$ also has $(P)$?

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    $\begingroup$ Integrality is not local by your definition, I don't think. A disjoint union is an open cover, and a disjoint union of integral schemes need not be integral. $\endgroup$
    – Will Sawin
    Commented Aug 21, 2013 at 20:31
  • $\begingroup$ Yes you are right of course! I just wanted to add another example than reduced, that's why I gave the integrality example. One could correct this as follows: a noetherian ring whose stalks are integral is a product of domain. This is a local condition, and so if I am not mistaken a "locally integral" locally noetherian scheme has global sections a product of domains also. $\endgroup$ Commented Aug 21, 2013 at 20:46
  • $\begingroup$ I misunderstood the question and thus deleted my answer. $\endgroup$ Commented Aug 21, 2013 at 21:02
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    $\begingroup$ It might be better to just assume all your schemes are connected for integrality. Then you don't need any noetherian hypothesis. $\endgroup$ Commented Aug 21, 2013 at 21:07
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    $\begingroup$ Yes but as Will pointed out, connectivity is not a local property, so that's why I used locally noetherian instead. But this is a good point: if a local property fails for the global sections of a non affine scheme, is there any sort of additional global property that makes it work? $\endgroup$ Commented Aug 22, 2013 at 11:19

1 Answer 1



For an integral scheme, being normal (integrally closed in ones own fraction field) satisfies this property. Indeed, suppose that $a, b \in A = \Gamma(X, O_X)$ are such that $a/b$ satisfy some polynomial $p(x) \in A[x]$. Then $a|_U, b|_U$ satisfy the same polynomial after restriction to each (affine) set $U \subseteq X$.

Of course this shows up in many applications of things like Stein factorization.


A reduced ring $R$ is seminormal if for any finite extension $R \subseteq S$ satisfying the following two properties is an isomorphism.

  • The induced map on $\text{Spec}$'s is an isomorphism
  • The induced residue field extensions $k(r) \subseteq k(s)$ are isomorphisms for all $s \in \text{Spec } S$ mapping to $r \in \text{Spec } R$.

The typical example of a seminormal ring is a node, the cusp $k[x^2,x^3]$ is not seminormal

Equivalently, $R$ is seminormal if and only if for any $a/b$ in the total ring of fractions of $R$, one has that if $(a/b)^2, (a/b)^3 \in R$ then $(a/b) \in R$ (see a paper by Swan, he might be assuming finitely many minimal primes, I forget the details). It follows similarly that seminormality satisfies this property.

Weak normality

Weak normality is similar to semi-normality. A reduced ring is called weakly normal if for any finite birational extension $R \subseteq S$ satisfying the following properties is an isomorphism:

  • The induced map on $\text{Spec}$'s is an isomorphism
  • The induced residue field extensions $k(r) \subseteq k(s)$ are purely inseparable for all $s \in \text{Spec } S$ mapping to $r \in \text{Spec } R$.

This can also be phrased as requiring that every birational universal homeomorphism is an isomorphism.

I do NOT know if weakly normal rings satisfy the sort of property asked for. I do not think it is in the literature (but perhaps I am wrong). I remember I convinced myself that they did not several years ago, but never wrote down an example carefully.

  • $\begingroup$ Second condition about seminormality: purely... what? $\endgroup$ Commented Aug 22, 2013 at 7:50
  • $\begingroup$ Must be "purely inseparable". $\endgroup$ Commented Aug 22, 2013 at 11:03

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