Return to Question

Let $f:X \to Y$ be a finite surjective morphism of quasi-projective schemes over $\mathbb{C}$, $X$ is reduced and $Y$ is integral. Suppose that there exists an integer $n$ such that for every closed point $y \in Y$, the fiber $f^{-1}(y)$ is reduced and consists of $n$ distinct closed points. Is $f$ flat?

Let $f:X \to Y$ be a finite surjective morphism of quasi-projective schemes over $\mathbb{C}$, $X$ is reduced and $Y$ is integral. Suppose that there exists an integer $n$ such that for every closed point $y \in Y$, the fiber $f^{-1}(y)$ is reduced and consists of $n$ distinct closed points. Is it true that $f$ is flat?

Let $f:X \to Y$ be a finite surjective morphism of quasi-projective scheme, $X$ is reduced and $Y$ is integral. Suppose that there exists an integer $n$ such that for every closed point $y \in Y$, the fiber $f^{-1}(y)$ is reduced and consists of $n$ distinct closed points. Is $f$ flat?

Let $f:X \to Y$ be a finite surjective morphism of quasi-projective schemes over $\mathbb{C}$, $X$ is reduced and $Y$ is integral. Suppose that there exists an integer $n$ such that for every closed point $y \in Y$, the fiber $f^{-1}(y)$ is reduced and consists of $n$ distinct closed points. Is $f$ flat?

Let $f:X \to Y$ be a finite surjective morphism of quasi-projective scheme, $X$ is reduced and $Y$ is integral. Suppose that there exists an integer $n$ such that for every closed point $y \in Y$, the fiber $f^{-1}(y)$ is reduced and consists of $n$ distinct closed points. Is $f$ flat?

Let $f:X \to Y$ be a finite surjective morphism of quasi-projective scheme, $X$ is reduced and $Y$ is integral. Suppose that there exists an integer $n$ such that for every closed point $y \in Y$, the fiber $f^{-1}(y)$ is reduced and consists of $n$ distinct closed points. Is $f$ flat?