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Can you cover a non-singular algebraic variety by pairwise disjoint singular closed subvarieties? Varieties are over an algebraically closed field of characteristic other than $2$ and $3$.

In characteristic $2$ and $3$ this is possible via quasi-elliptic surfaces as pointed out by abx.

If we allow locally closed subvarieties then $\mathbb{A}^2$ can be covered by cuspidal cubics $(y-t)^2=x^3$ (whenever two cubics intersect remove the finitely many intersection points from one of them).

Can you cover a non-singular algebraic variety by pairwise disjoint singular closed subvarieties? Varieties are over an algebraically closed field of characteristic other than $2$ and $3$.

In characteristic $2$ and $3$ this is possible via quasi-elliptic surfaces as pointed out by abx.

If we allow locally closed subvarieties then $\mathbb{A}^2$ can be covered by cuspidal cubics $(y-t)^2=x^3$ (whenever two cubics intersect remove the finitely many intersection points from one of them).

Can you cover a non-singular algebraic variety by pairwise disjoint singular closed subvarieties? Varieties are over an algebraically closed field of characteristic other than $2$ and $3$.

In characteristic $2$ and $3$ this is possible via quasi-elliptic surfaces as pointed out by abx.

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Can you cover a non-singular algebraic variety over an algebraically closed field by pairwise disjoint singular closed subvarieties? Varieties are over an algebraically closed field of characteristic other than $2$ and $3$.

Oops after a little thoughtIn characteristic $2$ and $3$ this is possible via quasi-elliptic surfaces as pointed out by abx.

If we allow locally closed subvarieties then $\mathbb{A}^2$ can be covered by cuspidal cubics. There is at most two values $t$ such that a point $(x, y)$ lies on $(y-t)^2=x^3$ so by axiom of choice we can make this into a partition.

However we can ask for a map(whenever two cubics intersect remove the finitely many intersection points from a non-singular variety to a possibly singular variety such that all fibers are singularone of them).

Can you cover a non-singular algebraic variety over an algebraically closed field by pairwise disjoint singular subvarieties?

Oops after a little thought $\mathbb{A}^2$ can be covered by cuspidal cubics. There is at most two values $t$ such that a point $(x, y)$ lies on $(y-t)^2=x^3$ so by axiom of choice we can make this into a partition.

However we can ask for a map from a non-singular variety to a possibly singular variety such that all fibers are singular.

Can you cover a non-singular algebraic variety by pairwise disjoint singular closed subvarieties? Varieties are over an algebraically closed field of characteristic other than $2$ and $3$.

In characteristic $2$ and $3$ this is possible via quasi-elliptic surfaces as pointed out by abx.

If we allow locally closed subvarieties then $\mathbb{A}^2$ can be covered by cuspidal cubics $(y-t)^2=x^3$ (whenever two cubics intersect remove the finitely many intersection points from one of them).

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Can you cover a non-singular algebraic variety over an algebraically closed field by pairwise disjoint singular subvarieties?

Oops after a little thought $\mathbb{A}^2$ can be covered by cuspidal cubics. There is at most two values $t$ such that a point $(x, y)$ lies on $(y-t)^2=x^3$ so by axiom of choice we can make this into a partition.

However we can ask for a map from a non-singular variety to a possibly singular variety such that all fibers are singular.

Can you cover a non-singular algebraic variety over an algebraically closed field by pairwise disjoint singular subvarieties?

Can you cover a non-singular algebraic variety over an algebraically closed field by pairwise disjoint singular subvarieties?

Oops after a little thought $\mathbb{A}^2$ can be covered by cuspidal cubics. There is at most two values $t$ such that a point $(x, y)$ lies on $(y-t)^2=x^3$ so by axiom of choice we can make this into a partition.

However we can ask for a map from a non-singular variety to a possibly singular variety such that all fibers are singular.

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