Hartshorne EX. III.10.3 
A morphism $f: X\to Y$ of schemes of finite type over $k$ is etale if it is smooth of
  relative dimension $0$. It is unramified if for every $x\in X$, letting $y = f(x)$, we have
  $m_y\cdot\mathcal{O}_x = m_x$, and $k(x)$ is a separable algebraic extension of $k(y)$. Show that the
  following conditions are equivalent:
(i) $f$ is etale;
(ii) $f$ is flat, and $\Omega_{X/Y}= 0$;
(iii) $f$ is flat and unramified.

To show (i) and (ii) are equivalent: recall $f$ is smooth of relative dimension $0$ (i.e. etale) if 
(1) $f$ is flat; 
(2) if $X'\subset X$ and $Y'\subset Y$ are irreducible components such that $f(X')\subset Y'$, then $\dim X' = \dim Y'$;
(3) For any $x\in X$, $\dim_{k(x)} \Omega_{X/Y}\otimes k(x)=0$. 
I wonder why $f$ is etale implies $\Omega_{X/Y}=0$: I know if $X$ is integal then $\Omega_{X/Y}$ is locally free, but here we don't have $X$ integal, then how to guarantee $ \Omega_{X/Y}$ doesn't have torsion? More precisely, I think $\dim \Omega_{X/Y}\otimes k(x)=0$ can only guarantee $ \Omega_{X/Y}$ has rank $0$, how to see it doesn't have torsion. 
For the converse part, I know $f(X')$ is irreducible in $Y$ but I think I need some facts to deduce $\dim X' = \dim Y'$, can you give me some ideas?
For (2) and (3) are equivalent, since it refers to the separable algebraic extension, I think we need to use II.8.6.A:

Let $K$ be a finitely generated extension field of afield $k$. Then
  $\dim_K\Omega_{K/k}\ge\text{tr.d.} K/k$, and equality holds if and only if $K$ is separably
  generated over $k$. 

But I think in order to get $K$, we need $X$ to be integral, and $\Omega_{K/k}$ is corresponding to the stalk of $\Omega_{X/k}$ at the generic point. However, I can't go further. 
 A: Your strategy is inappropriate to this question. This first part is a straightforwards application of Nakayama's lemma: $\Omega_{X/Y}$ is coherent, so $(\Omega_{X/Y})_x$ is a finitely generated module over the local ring $\mathcal{O}_{X,x}$. Then the statement about the dimension is the same as $(\Omega_{X/Y})_x\otimes_{\mathcal{O}_{X,x}} k(x)= (\Omega_{X/Y})_x\otimes_{\mathcal{O}_{X,x}} \mathcal{O}_{X,x}/\mathfrak{m}_x = (\Omega_{X/Y})_x/\mathfrak{m}_x(\Omega_{X/Y})_x=0$, or $\mathfrak{m}_x(\Omega_{X/Y})_x=(\Omega_{X/Y})_x$, so by Nakayama, this means that $(\Omega_{X/Y})_x=0$. This means that the stalk of $\Omega_{X/Y}$ is zero at every point, or that $\Omega_{X/Y}$ is the zero sheaf. So we've shown (i) implies (ii).
The rest of the argument will make frequent use of some basic facts about  $\Omega_{X/Y}$:


*

*Hartshorne II.8.2A: For a ring map $A\to B$ and a base extension $A\to A'$ which gives a ring map $A' \to B':= B\otimes_A A'$, we have $\Omega_{B'/A'}=\Omega_{B/A}\otimes_B B'$.


*Hartshorne II.8.2A: For $S\subset B$ multiplicatively closed, $S^{-1}\Omega_{B/A}=\Omega_{S^{-1}B/A}$.


*If $R\subset A$ is multiplicatively closed and $R$ maps to invertible elements of $B$, then $\Omega_{B/A}=\Omega_{B/R^{-1}A}$. (Proof: apply the Leibniz rule to $1=f(s)f(s)^{-1}$.)


*For any surjective morphism of rings $A\to B$, we have $\Omega_{B/A}=0$.


*Hartshorne II.8.3A: If $A\to B \to C$ are maps of rings, we have a natural exact sequence of $C$-modules $$\Omega_{B/A}\otimes_B C \to \Omega_{C/A} \to \Omega_{C/B}\to 0$$


*Hartshorne II.8.7A: Let $A$ be a local ring with maximal ideal $\mathfrak{m}$ containing a field $k$ isomorphic to it's residue field. Then there is an isomorphism $\mathfrak{m}/\mathfrak{m}^2\to \Omega_{B/k}\otimes_B k$.


*If $A\to B$ is a ring map with $\Omega_{B/A}=0$, then the induced map on reductions $A_{red}\to B_{red}$ also has $\Omega_{B_{red}/A_{red}}=0$.

The final statement comes from an application of II.8.3A to the sequences of ring maps $A\to A_{red}\to B_{red}$ and $A\to B\to B_{red}$: the first gives you that $\Omega_{B_{red}/A}\cong \Omega_{B_{red}/A_{red}}$ since $\Omega_{A_{red}/A}=0$ as the ring map $A\to A_{red}$ is surjective, and the second gives you that $0=\Omega_{B/A}\otimes_B B_{red}$ surjects on to $\Omega_{B_{red}/A}\cong\Omega_{B_{red}/A_{red}}$ because $\Omega_{B_{red}/B}=0$.
To show (ii) implies (i), the fact that our map is flat and of finite type implies it is open because of exercise III.9.1. Let $Y'$ be an irreducible component of $Y$ and let $X'$ be an irreducible component of $X$ which maps in to $Y'$. Now take an affine open $Y''$ of $Y'$ which is open in $Y$ and some affine open $X''$ of $X'$ open in $X$ mapping in to it. As our map is open, the image of $X''$ in $Y''$ is open and thus dense, so the generic point of $X''$ maps to the generic point of $Y''$ (further, this map of generic points is exactly the same as the map of generic points one gets with $X'\to Y'$).
By the final quoted statement above and the fact that taking reductions does not change dimensions nor the fact that the generic point of $X''$ maps to the generic point of $Y''$, we may replace $X''$ and $Y''$ by their reductions while maintaining their dimensions and $\Omega_{X''/Y''}=0$. Now look at the map on generic points: by the localization property above, we get that the module of differentials associated to the induced map of fraction fields vanishes, so by II.8.6a, we have that these fraction fields are of the same transcendence degree and thus $X''$ and $Y''$ (and therefore $X'$ and $Y'$) have the same dimension. (Explicitly, if $A\to B$ is the map of domains induced by $X''\to Y''$, then let $S=B\setminus \{0\}$ and let $R=A\setminus\{0\}$, so $0=S^{-1}\Omega_{B/A}=\Omega_{S^{-1}B/A}=\Omega_{S^{-1}B/R^{-1}A}$, and this last module is the module of differentials associated to the induced map on fraction fields.)
To show that (ii) and (iii) are the same, we argue affine-locally, which lets us treat this as an algebra problem. To be clear: we want to show that a finite type map of $k$-algebras $f:A\to B$ has $\Omega_{B/A}=0$ iff it is unramified. By the localization properties above, it is not hard to see that this is equivalent to the stalk-local problem: if we have a finite type local map of local rings $A\to B$, then unramified is equivalent to $\Omega_{B/A}=0$.
Suppose $A,B$ are local rings with maximal ideals $\mathfrak{m},\mathfrak{n}$ and residue fields $E,F$ respectively and $f:A\to B$ is a local ring map of finite type between them. If we assume unramified, then by base change we get that $\Omega_{B/A}\otimes_A E = \Omega_{(B/\mathfrak{m}B)/E}$, and as $\mathfrak{m}B=\mathfrak{n}$ by assumption, we get that this last module is just the module of differentials associated to the map of residue fields $E\to F$. By assumption, this is separable algebraic, so by II.8.6a we have that it vanishes, and then by Nakayama we get that $\Omega_{B/A}=0$ as requested.
For the reverse direction, we'll show that if either the conditions $\mathfrak{m}B=\mathfrak{n}$ or "$F$ is a separable extension of $E$" is violated, then $\Omega_{B/A}\neq 0$. Start with the extension of fields: write $E\to B/\mathfrak{m}B \to F$, and applying II.8.3A, we get the following exact sequence.
$$ \Omega_{(B/\mathfrak{m}B)/E} \otimes_{B/\mathfrak{m}B} F \to \Omega_{F/E}\to \Omega_{(B/\mathfrak{m}B)/F}\to 0$$
The final term vanishes because the ring map $B/\mathfrak{m}B\to B/\mathfrak{n}=F$ is surjective, so the first map is a surjection. In particular, if the extension of fields $E\subset F$ is not algebraic separable then by II.8.6a we get the middle term is nonzero, then the left term must be nonzero, and as $\Omega_{(B/\mathfrak{m}B)/E}\cong \Omega_{B/A}\otimes_A E$, we get $\Omega_{B/A}\neq 0$.
There are two ways we can fail to have $\mathfrak{m}B=\mathfrak{n}$ - either $\sqrt{\mathfrak{m}B}$ is equal to $\mathfrak{n}$ or not. If they're not equal, II.8.3a applied to $E\to B/\mathfrak{m}B\to B/\sqrt{\mathfrak{m}B}$ combined with the observation that $B/\sqrt{\mathfrak{m}B}$ has a fraction field which is of positive transcendence degree over $E$ gives that $\Omega_{B/A}\neq 0$ similarly to the above conclusion.
In the other case, $\sqrt{\mathfrak{m}B}=\mathfrak{n}$ but $\mathfrak{m}B\neq \mathfrak{n}$, we have that $B/\mathfrak{m}B$ is Artinian and thus finite-dimensional as an $E$-vector space. Let $K$ be an algebraic closure of $A/\mathfrak{m}$. Then $B/\mathfrak{m}B \otimes_{E} K$ is again Artinian and a finite product of Artinian local rings with $K$ as their residue field. As $B/\mathfrak{m}B$ has a nilpotent element, $(B/\mathfrak{m}B)\otimes_{E} K$ has a nilpotent element and there exists some local ring factor $B'\subset K\otimes_{E} B/\mathfrak{m}B$ with nonzero nilpotent maximal ideal $\mathfrak{q}$. By II.8.7, we have that $\mathfrak{q}/\mathfrak{q}^2\cong \Omega_{B'/K}\otimes_{B'} K$ and thus the RHS is nonzero, and so $\Omega_{B'/K}$ is nonzero by Nakayama. As $\Omega_{B'/K}$ is a localization of $\Omega_{((B/\mathfrak{m}B)\otimes_{E} K)/K}\cong \Omega_{(B/\mathfrak{m}B)/E}\otimes_{E} K$ at a maximal ideal, $\Omega_{(B/\mathfrak{m}B)/E}$ is nonzero and we are done.

The equivalence of (ii) and (iii) seems like kind of a pain to me if you're trying to do it with just the tools Hartshorne has available at this point. I prefer the StacksProject definitions which define all of these things through standard smooth/etale/unramified ring maps, which makes everything a bit more straightforwards.
