Normality

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.

Semi-normality

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 $r \in \text{Spec } R$.

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.

Normality

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.

Semi-normality

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.

Normality

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.

Semi-normality

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 for all $s \in \text{Spec } S$ mapping to $r \in \text{Spec } R$.

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.

Normality

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.

Semi-normality

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 $r \in \text{Spec } R$.

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.