Let $X$ be a nonsingular variety of dim $n $ over an algebraically closed field $k$. Let $Y$ be an irreducible closed subscheme defined by a sheaf of ideals $\mathscr I$.
Then I want to prove that
$ Y $ is a nonsingular variety over $k $ iff
(1) $ \Omega_{Y/k} $ is locally free, and
(2) the sequence $ 0 \rightarrow \mathscr I/\mathscr I^2 \rightarrow \Omega_{X/k} \otimes \mathcal O_Y \rightarrow \Omega_{Y/k} \rightarrow 0$ is exact.
This is a theorem from Hartshorne's Algebraic Geometry (Chapter II, theorem 8.17).Assuming (1) and (2) and proving Y is nonsingular is okay. But the other way is not clear.
So I will assume $Y$ is nonsingular. The subdivision (1) follows directly as $Y $ is nonsingular (In fact $ \Omega_{Y/k} $ will be locally free of rank $q = \operatorname{dim}Y$) . But I am not able to do the 2nd subdivision.
We have the following exact sequence $ \mathscr I/\mathscr I^2 \rightarrow \Omega_{X/k} \otimes \mathcal O_Y \rightarrow \Omega_{Y/k} \rightarrow 0 $ (which is called the second exact sequence in literature). Let us name the map $ \mathscr I/\mathscr I^2 \rightarrow \Omega_{X/k} \otimes \mathcal O_Y $ as $ \delta $ and the map $\Omega_{X/k} \otimes \mathcal O_Y \rightarrow \Omega_{Y/k} $ as $ \phi $.
My aim is to prove that $\delta$ is injective. I will mimic the proof of Hartshorne and try to tell what I have not understood there-
Let $y\in Y $ be a closed point. Then ker $\phi$ is locally free of rank $r= (n-q)$ at $y$, so it is possible to choose sections $ x_1, x_2,...x_r \in \mathscr I $ in a suitable neighbourhood of $y$, such that $dx_1,dx_2,...,dx_r$ generate ker $\phi$ (This is because $X $ is nonsingular, so $\Omega_{X/k} \otimes \mathcal O_Y $ is locally free $ \mathcal O_Y $- module of rank $n$, and $Y$ is nonsingular, so $\Omega_{Y/k}$ is locally free $\mathcal O_Y$-module of rank $q$). Let $\mathscr I'$ be the ideal sheaf generated by $x_1, x_2,...x_r $, and $Y'$ be the corresponding closed subscheme.
So far everything is fine. Now he says- By construction, the $dx_1,dx_2,...,dx_r$ generate a free subsheaf of rank $r$ of $\Omega_{X/k} \otimes \mathcal O_{Y'} $ in a neighbourhood of $y$. I don't understand why $dx_1,dx_2,...,dx_r$ generate a free subsheaf of rank $r$ of $\Omega_{X/k} \otimes \mathcal O_{Y'} $ in a neighbourhood of $y$？
If we can show $(dx_1)_y,\cdots,(dx_r)_y$ generate a free submodule of rank $r$ of $\mathcal O_{X,y}$-module $(\Omega_{X/k})_y$, then $dx_1,dx_2,...,dx_r$ generate a free subsheaf of rank $r$ of $\Omega_{X/k} \otimes \mathcal O_{Y'} $ in a neighbourhood of $y$.
But how to show $(dx_1)_y,\cdots,(dx_r)_y$ generate a free submodule of rank $r$ of $\mathcal O_{X,y}$-module $(\Omega_{X/k})_y$?
$Y'$ is a "larger" scheme than $Y$ in the sense that its ideal sheaf is a priori smaller. This means that if the $dx_1, \dots, dx_r$ locally satisfy no nontrivial relations in $\Omega_{X/k} \otimes \mathcal{O}_{Y}$, then they also satisfy no nontrivial relations in $\Omega_{X/k} \otimes \mathcal{O}_{Y'}$, since $\Omega_{X/k} \otimes \mathcal{O}_{Y}$ can be viewed as a quotient of $\Omega_{X/k} \otimes \mathcal{O}_{Y'}$. Hence, they also locally generate a free subsheaf in this second sheaf.