# A $k$ variety $X$ is smooth if it admits an unramified morphism to a smooth $k$-variety $Y$

I would like to show that given a $k$-variety $X$ of dimension $n$ along with an unramified morphism $f:X\rightarrow Y$ where $Y$ is a smooth $k$-variety of dimension $n$, then $X$ is smooth. (This is from Ravi Vakil's notes, namely 25.2.Ea.)

To this end I would like to use the cotangent exact sequence to show that $\Omega_{X/k}$ is locally free of rank $n$. Because $f$ is unramified, $\Omega{X/Y} = 0$, and so the contangent exact sequence yields:

$$f^* \Omega_{Y/k} \rightarrow \Omega_{X/k} \rightarrow 0.$$

Because $Y$ is smooth we now have a surjection from a locally free sheaf of rank $n$.

I'm not sure where to go from here. If this is left-exact, then by 6.2.10 in Qing Liu $f$ would be étale, which seems too strong. Vakil suggests using the conormal exact sequence, but I'm not sure how this applies as I don't have a closed immersion.

• It seems to me like this map needs to be an isomorphism, since it will be a surjection of locally free sheaves of the same rank. Using this reasoning I think you can show that the kernel is torsion, and that's probably a contradiction. – Hoot Aug 14 '16 at 21:57
• @Hoot How do we know that $\Omega_{X/k}$ is a locally free sheaf? – Takumi Murayama Aug 14 '16 at 22:32
• @Takumi I just meant that in the end I want to conclude that it is locally free, and I'm noting (correctly, I hope) that this implies the strong statement of the which the OP was suspicious. – Hoot Aug 14 '16 at 22:42
• @Hoot Ah, alright, I agree with you, then. My apologies if I sounded too harsh! – Takumi Murayama Aug 14 '16 at 22:57
• @Takumi Definitely not! – Hoot Sep 2 '16 at 1:28

To this end I would like to use the cotangent exact sequence to show that $\Omega_{X/k}$ is locally free of rank $n$. Because $f$ is unramified, $\Omega{X/Y} = 0$, and so the contangent exact sequence yields:
$$f^* \Omega_{Y/k} \rightarrow \Omega_{X/k} \rightarrow 0.$$
Because $Y$ is smooth we now have a surjection from a locally free sheaf of rank $n$.
This implies that $\Omega_{X/k} \otimes k(x)$ has dimension (over $k(x)$) at most $n$ for all $x$. On the other hand $\dim \Omega_{X/k} \otimes k(x)$ has at least the dimension of $X$ at $x$ (extend the ground field to its algebraic closure and prove the inequality over an algebraically closed field). So if we suppose $X$ is pure of dimension $n$, then$$\dim \Omega_{X/k} \otimes k(x) = n.$$This implies that $\Omega_{X/k}$ is locally free of rank $n$. We can also say that the map$$f^* \omega_{Y/k} \otimes k(x) \to \Omega_{X/k} \otimes k(x)$$is an isomorphism, so$$f^*\Omega_{Y/k} \to \Omega_{X/k}$$is an isomorphism by Nakayama.
Why does having all fibers of dimension $n$ imply that $\Omega_{X/k}$ is locally free?
This is a general fact on reduced Noetherian schemes $X$: if a coherent sheaf $\mathcal{F}$ on $X$ is such that $\dim_{k(x)} \mathcal{F} \otimes k(x)$ is constant for all $x$ in $X$, then $\mathcal{F}$ is locally free. You can find this statement as an exercise in Hartshorne.