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3 typo

Here is some intuitive propaganda for Anton's answer...

We know that a qsep (quasi-separated) scheme (over $\mathbb{Z}$) is precisely one where the intersection U∩V of any two open affines, U=Spec(A) and V=Spec(B), is quasi-compact. Looking at compliments gives a different perspective: that their differences U\V and V\U are cut out by finitely many elements in A,B respectively, meaning that these differences are "easy to see". I'd say this justifies the following credo:

• A quasi-separated scheme is one where any two open affines are "easy to distinguish".
• A non-qsep scheme is one containing some "subtle distinction" between open affines.

The two copies of $\mathbb{A}^\infty$ in Anton's answer differ only by the origin, which is "hard to see" in that it cannot be cut out by finitely many ring elements, and I'd say using infinitely many variables to cut out one point is about the most natural way to achieve this. Thus, I like to characterize non-qsep schemes as containing "(infinitely) subtle distinctions" such as this one.

Further tinkering yields a similar way to think about a qsep morphism $f:X\to Y$. I'd say the corresponding credo is that:

• A quasi-separated morphism in is one which preserves the existence of "subtle distinctions".
• A non-qsep morphism is one which destroys some "subtle distinctions".

This helps intuitivize things theorems like:

(1) " Any map from a qsep scheme is qsep ", because it has no subtle distinction that can be destroyed.

(2) " If $Y$ is qsep, then $f:X\to Y$ is qsep iff $X$ is qsep ", since $f$ destroys subtle distinctions iff $X$ has them.

(3) " If $g\circ f$ is qsep, then $f$ is qsep ", since if $f$ destroyed some subtle distinction, then $g$ could not recover it.

Here is a coarse and a fine justification for this credo in each direction...

Coarse version: By 1971 EGA I 6.1.11, for any cover of Y by qsep opens $V_{i}$, $f$ is qsep iff each preimage $f^{-1}(V_i)$ is qsep. Thus, $f$ is non-qsep iff there is some qsep open $V\subseteq Y$ such that $f^{-1}(V)$ is non-qsep, meaning it contains some subtle distinction which is lost after application of by $f$.

Fine version: Suppose $f$ is qsep. By 1971 EGA I 6.1.9, fibre products and compositions of qsep morphisms are qsep, and any universal injection is qsep (for example any immersion). Now suppose

$S\hookrightarrow X$

$T\hookrightarrow Y$

are any universal injections such that $f|_S$ factors through $T$, for example if $T$ is the scheme-theoretic image of $S$. Then $T$ qsep $\Rightarrow$ $S$ qsep, hence $S$ non-qsep $\Rightarrow$ $T$ non-qsep, meaning $f$ preserves the existence of subtle distinctions in passing from any such $S$ to $T$.

2 typesetting

Here is some intuitive propaganda for Anton's answer...

We know that a qsep (quasi-separated) scheme (over $\mathbb{Z}$) is precisely one where the intersection U∩V of any two open affines, U=Spec(A) and V=Spec(B), is quasi-compact. Looking at compliments gives a different perspective: that their differences U\V and V\U are cut out by finitely many elements in A,B respectively, meaning that these differences are "easy to see". I'd say this justifies the following credo:

• A quasi-separated scheme is one where any two open affines are "easy to distinguish".
• A non-qsep scheme is one containing some "subtle distinction" between open affines.

The two copies of $\mathbb{A}^\infty$ in Anton's answer differ only by the origin, which is "hard to see" in that it cannot be cut out by finitely many ring elements, and I'd say using infinitely many variables to cut out one point is about the most natural way to achieve this. Thus, I like to characterize non-qsep schemes as containing "(infinitely) subtle distinctions" such as this one.

Further tinkering yields a similar way to think about a qsep morphism $f:X\to Y$. I'd say the corresponding credo is that:

• A quasi-separated morphism in one which preserves the existence of "subtle distinctions".
• A non-qsep morphism is one which destroys some "subtle distinctions".

This helps intuitivize things like:

1)

(1) " Any map from a qsep scheme is qsep"qsep ", because it has no subtle distinction that can be destroyed.

2)

(2) " If $Y$ is qsep, then $f:X\to Y$ is qsep iff $X$ is qsep"qsep ", since f $f$ destroys subtle distinctions iff $X$ has them.

3)

(3) " If $g\circ f$ is qsep, then $f$ is qsep"qsep ", since if $f$ destroyed some subtle distinction, then $g$ could not recover it.

Here is a coarse and a fine justification for this credo in each direction...

Coarse version: By 1971 EGA I 6.1.11, for any cover of Y by qsep opens $V_{i}$, $f$ is qsep iff each preimage $f^{-1}(V_i)$ is qsep. Thus, $f$ is non-qsep iff there is some qsep open $V\subseteq Y$ such that $f^{-1}(V)$ is non-qsep, meaning it contains some subtle distinction which is lost after application of by f.$f$.

Fine version: Suppose f $f$ is qsep. By 1971 EGA I 6.1.9, fibre products and compositions of qsep morphisms are qsep, and any universal injection is qsep (for example any immersion). Now suppose

$S\hookrightarrow X$

$T\hookrightarrow Y$

are any universal injections such that $f|_S$ factors through $T$, for example if $T$ is the schematic scheme-theoretic image of $S$. Then $T$ qsep $\Rightarrow$ $S$ qsep, hence $S$ non-qsep $\Rightarrow$ $T$ non-qsep, meaning f $f$ preserves the existence of subtle distinctions in passing from any such $S$ to $T$.

1

Here is some intuitive propaganda for Anton's answer...

We know that a qsep (quasi-separated) scheme (over $\mathbb{Z}$) is precisely one where the intersection U∩V of any two open affines, U=Spec(A) and V=Spec(B), is quasi-compact. Looking at compliments gives a different perspective: that their differences U\V and V\U are cut out by finitely many elements in A,B respectively, meaning that these differences are "easy to see". I'd say this justifies the following credo:

• A quasi-separated scheme is one where any two open affines are "easy to distinguish".
• A non-qsep scheme is one containing some "subtle distinction" between open affines.

The two copies of $\mathbb{A}^\infty$ in Anton's answer differ only by the origin, which is "hard to see" in that it cannot be cut out by finitely many ring elements, and I'd say using infinitely many variables to cut out one point is about the most natural way to achieve this. Thus, I like to characterize non-qsep schemes as containing "(infinitely) subtle distinctions" such as this one.

Further tinkering yields a similar way to think about a qsep morphism $f:X\to Y$. I'd say the corresponding credo is that:

• A quasi-separated morphism in one which preserves the existence of "subtle distinctions".
• A non-qsep morphism is one which destroys some "subtle distinctions".

This helps intuitivize things like:

1) "Any map from a qsep scheme is qsep", because it has no subtle distinction that can be destroyed.

2) "If $Y$ is qsep, then $f:X\to Y$ is qsep iff $X$ is qsep", since f destroys subtle distinctions iff $X$ has them.

3) "If $g\circ f$ is qsep, then $f$ is qsep", since if $f$ destroyed some subtle distinction, then $g$ could not recover it.

Here is a coarse and a fine justification for this credo in each direction...

Coarse version: By 1971 EGA I 6.1.11, for any cover of Y by qsep opens $V_{i}$, $f$ is qsep iff each $f^{-1}(V_i)$ is qsep. Thus, $f$ is non-qsep iff there is some qsep open $V\subseteq Y$ such that $f^{-1}(V)$ is non-qsep, meaning it contains some subtle distinction which is lost after application of by f.

Fine version: Suppose f is qsep. By 1971 EGA I 6.1.9, fibre products and compositions of qsep morphisms are qsep, and any universal injection is qsep (for example any immersion). Now suppose

$S\hookrightarrow X$

$T\hookrightarrow Y$

are any universal injections such that $f|_S$ factors through $T$, for example if $T$ is the schematic image of $S$. Then $T$ qsep $\Rightarrow$ $S$ qsep, hence $S$ non-qsep $\Rightarrow$ $T$ non-qsep, meaning f preserves the existence of subtle distinctions in passing from any such $S$ to $T$.