The main virtue of the Albanese variety is its universal property: given any compact torus $A$, any morphism $X\to A$ factors uniquely through $T=Alb(X)$.
The easiest application of Albanese varieties I can think of is that if $H^0(X,\Omega^1_X)=0$, then every morphism $X\to A$ from $X$ into a compact torus $A$ is constant: indeed it must factor through $T=H^0(X,\Omega_{X}^1)^*/H_1(M,\mathbb{Z})$, which is just a point if $H^0(X,\Omega^1_X)=0$.
This applies in particular to $\mathbb P^n_\mathbb C$, whose holomorphic maps into compact tori are thus all constant.
A more sophisticated use of Albanese varieties is in the proof that any non ruled projective surface has a unique minimal model: see Beauville's book, theorem V.19
Let me answer the literal questions the OP asked, and then give a solution to the question he was really getting at (elucidated in the chat linked in the above comments).
The literal question
Let's suppose that $X\to Y$ is a finite flat map (let's assume that $Y$ is Noetherian to avoid headaches). Then, to check that $X\to Y$ has vanishing differentials is equivalent to checking that $X\to Y$ is etale. To do this without actually computing the differentials is, in general, not easy. There are a few ways one can do it in practice:
- Show that its unramified in the sense that for every $x\in X$ and $y=f(x)\in Y$ we have that the map $\mathcal{O}_{Y,y}\to\mathcal{O}_{X,x}$ has the property that $\mathfrak{m}_y\mathcal{O}_{X,x}=\mathfrak{m}_x$ and $k(x)/k(y)$ is finite separable.
- Show that the map $X\to Y$ satisfies Grothendieck's infinitesimal lifting criterion (a priori you actually have to show unicity of the lifts to get etaleness, but since our map is finite showing smoothness is fine since smooth plus relative dimension $0$ implies etale).
- You can show that the discriminant of the trace form $(x,y)\mapsto \mathrm{tr}(xy)$ (defined affine locally on $Y$) is a perfect pairing.
Of course, these might all be harder than just computing the differentials given what you know about $X\to Y$.
Another useful technique (which I implement below) is that to check that $X\to Y$ is etale it suffices to check this over some fpqc cover of $Y$. More concretely, since you like commutative algebra, if we have a ring map $A\to B$ and we have some faithfully flat map $A\to R$ (e.g. like $A\to\widehat{A}$ if $A$ is local) then $A\to B$ is etale if and only if $R\to B\otimes_A R$ is. This falls into the general notion of 'fpqc descent'. But, one concrete way of seeing it is that the non-degeneratness of the trace pairing (mentioned in the third bullet above) can be checked after base changing to a faithfully flat extension.
As for the question in the comments (paraphrasing):
"Is it enough that $Y=\mathrm{Spec}(A)$, $A$ a Dedekind domain, and $X=\mathrm{Spec}(B)$ with $B$ a product of normal rings?"
The answer is no as $\mathrm{Spec}(\mathbb{Z}[i])\to\mathrm{Spec}(\mathbb{Z})$ shows. If you want them to be (even excellent!) DVRs you can consider $\mathrm{Spec}(\mathbb{Z}[i]_{(1+i)})\to\mathrm{Spec}(\mathbb{Z}_{(2)})$.
The question the OP was actually interested in
I will write here the actual question the OP was interested. If the OP decides to transfer it from our chat to a question on the main forum I will transfer this answer there.
Exercise: Let $Y=\mathrm{Spec}(A)$ with $A$ a DVR with uniformizer $\pi$. Let $G$ be a finite flat group scheme over $Y$. Show that $G$ is etale if and only if $G=\mathrm{Spec}(B)$ with $\displaystyle B\cong \prod_{i=1}^n B_i$ with each $B_i$ a normal domain.
My proof might be overly complicated. I'm sure there is some simpler proof just messing around with augmentation ideals. I also tried to write the below as commutative algebra-y as possible at the request of the OP--but I don't like commutative algebra, so it didn't end up optimally algebra-y.
Solution: Let us suppose first that $G$ is finite etale over $Y$. We can decompose $G$ into connected components $\displaystyle G=\bigsqcup_i \mathrm{Spec}(B_i)$. We claim that each $B_i$ is a domain. The point is that since $B_i$ is Noetherian we know that it suffices to show that all the local rings of $B_i$ at its maximal ideals are domains. Note though that since $A\to B_i$ is unramified we have for any maximal $\mathfrak{q}$ of $B_i$ the map $A\to B_\mathfrak{q}$ has the property that $\pi B_\mathfrak{q}$ is equal to $\mathfrak{q}$. This means that $B_\mathfrak{q}$ is a regular local ring (since $\dim B_\mathfrak{q}=1$ since it's finite over $A$) and thus a domain.
Remark: The above is a half-way proof to the general fact that the connected finite etale covers of an integral normal scheme are integral normal.
Suppose, conversely, that $\displaystyle G=\bigsqcup_i \mathrm{Spec}(B_i)$ with $B_i$ normal domains. Note that since $\dim B_i=1$ we see that normality implies that for each closed point $\mathfrak{q}$ of $B_i$ we have that $(B_i)_\mathfrak{q}$ is a DVR and so is its completion. Note then that we have that
$$H:=G_{\widehat{A}}\cong \bigsqcup_i \mathrm{Spec}(B_i\otimes_A \widehat{A})=\bigsqcup_i \bigsqcup_{\mathfrak{q}\in\mathrm{MaxSpec}(B_i)}\mathrm{Spec}(\widehat{(B_i)_{\mathfrak{q}}})$$
We deduce that $H$ has connected components which are normal domains.
But, by the usual connected-etale sequence we have that $H$ is etale over $\widehat{A}$ iff its connected component of the identity $H^\circ$ is trivial (e.g. see section 8.1 of this). Note though that by our above decomposition, all the connected components of $H$ are normal and integral. In particular, $H^\circ=\mathrm{Spec}(B_0)$ for some normal domain $B_0$. This then implies that its generic fiber $H_\eta^\circ:=\mathrm{Spec}(B_0\otimes_{\widehat{A}}K)$
(with $K:=\mathrm{Frac}(\widehat{A})$) is normal and integral.
But, since $H^\circ_\eta$ is finite over $\mathrm{Spec}(K)$ and integral we know that $H^\circ_\eta=\mathrm{Spec}(C)$ where $C$ is a field. Note though that we have an identity section $\mathrm{Spec}(K)\to H^\circ_\eta$ since $H^\circ_\eta$ is a group over $\mathrm{Spec}(K)$. This implies that we have a map $C\to K$ of $K$-algebras. Since $C$ is a field this must be an isomorphism. Thus, $C=K$. So we see that $B_0\otimes_{\widehat{A}} K\cong K$. Since $B_0$ is finite free over $\widehat{A}$ this implies that $B_0$ is rank $1$. The conclusion follows. $\blacksquare$
Best Answer
The general fact you're looking for is that if $E/F$ is a field extension and $t_1,\cdots,t_n$ is a transcendence basis for $E$ over $F$ (not necessarily finite), then $dt_i$ are a basis for $\Omega_{E/F}$. All the facts necessary for this should be in your favorite commutative algebra textbook, and I'll sketch the proof below:
$\Omega_{k[x_1,\cdots,x_n]/k}$ is free on $dx_i$. This follows from Yoneda and the product rule: we can arbitrarily assign any value to all the $dx_i$ and that determines a derivation. Note that there is no reason to restrict to finite $n$ here.
$\Omega$ commutes with localization. This lets us pass from $k[x_1,\cdots,x_n]/k$ to $k(x_1,\cdots,x_n)/k$, and see that $\Omega$ is still free on the $dx_i$.
$\Omega$ plays well with algebraic field extensions, ie if $E/F/k$ is a tower of field extensions with $E/F$ algebraic, then $E\otimes_k \Omega_{F/k} \to \Omega_{E/k}$ is an isomorphism. This can be proven from the first exact sequence: one treats the case of a finite extension using the theorem of the primitive element, and then uses the fact that every algebraic field extension is the colimit of it's finite subextensions to get the result.
So if we set up a tower $\Bbb R/\Bbb Q^{tr}/\Bbb Q$ where $\Bbb Q^{tr}$ is a maximal purely transcendental subextension of $\Bbb R/\Bbb Q$ chosen so that $\pi$ is in it's transcendence basis over $\Bbb Q$, we see the result we were after.