# Non-reducedness of schemes and projective morphisms(revisited)

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

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.