This is a continuation of a question asked by me previously with some added hypothesis. Let $X, Y$ be irreducible projective schemes over $\mathbb{C}$, $W \subset X \times Y$ a closed irreducible subscheme. Suppose that the natural projection map $pr_2:W \to Y$ is surjective on the underlying topological spaces, and every fiber is reduced. Note that for all $x \in W$, there is a natural map of tangent spaces $\phi_x:T_{W,x} \to T_{Y,pr_2(x)}$. If there is a point $y \in Y$ such that the $\mathbb{C}$-vector space generated by the union $\cup_{x \in pr_2^{-1}(y)} \phi_x(T_{W,x})$ does not coincide with $T_{Y,y}$, does this mean that $Y$ is nonreduced at the point $y$?


1 Answer 1


This is not true. Let $X$ be $\mathbb{A}^1$ with coordinate $s$. Let $\mathbb{A}^2$ have coordinates $t$ and $u$. Let $Y\subset \mathbb{A}^2$ be the irreducible, reduced closed subset with defining equation $u^2 - t^2(t+1) = 0$, i.e., $Y$ is a nodal cubic curve. Let $W\subset X\times Y$ be the closed subset with defining equations, $$s(u+t) - t = 0, \ \ s^2(1+t)-(s-1)^2 = 0.$$ On the open subset where $t$ is nonzero, then $W$ has defining equation $s=(1+(u/t))^{-1}$, so that $W$ is smooth and irreducible. On the open set where $t+1$ is nonzero, then the second defining equation gives that $W$ is smooth over the $t$-line.

So, altogether, $W$ is smooth. Except over the point $y$ with $(t,u)=(0,0)$, the projection from $W$ to $Y$ is an isomorphism. The fiber of $W$ over $y$ is one reduced point $x$, namely $(s,t,u) = (1/2,0,0)$. The Zariski tangent space $T_yY$ is two-dimensional generated by $\partial/\partial t$ and $\partial/\partial u$. However, $T_xW$ is one-dimensional, since $W$ is a smooth curve near $x$. Thus, the image of $T_xW$ does not generate $T_yY$. However, $Y$ is reduced.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.