Relationship between étale maps and polynomials Let $k$ be a field, not necessarily algebraically closed, and suppose you have a map of $k$-schemes $\varphi: \mathbb{A}^n \to \mathbb{A}^n$, induced by some $\psi: k[y_1, \dots y_n] \to k[x_1, \dots x_n]$. Let $p_i(x_1, \dots x_n) := \psi(y_i)$, and define $J(p) := \left(\frac{\partial p_i}{\partial x_j}(p)\right)_{i, j \leq n} \in M_{n \times n}(\kappa(p))$.
Suppose you know that $\varphi$ is étale at some point $p \in \mathbb{A}^n$, i.e. that $\varphi$ is smooth of relative dimension zero at $p$. Is it true that $J(p)$ has maximum rank as a matrix in $M_{n\times n}(\kappa(p))$? The converse is, to the best of my knowledge, true, and this should also be "morally" true (since the $p_i$s are essentially "defining" $\varphi$), but I don't know how to prove it.
As a bonus: can we formulate a similar statement for $\varphi: X \to Y$, where this time $X = \operatorname{Spec} k[x_1, \dots x_m]/I \subseteq \mathbb{A}^m$ and $Y = \operatorname{Spec} k[y_1, \dots y_n]/J \subseteq \mathbb{A}^n$ are of finite type (but not necessarily reduced)?
 A: Question: "Suppose you know that $\phi$ is étale at some point $p\in\mathbb{A}^n$, i.e. that $\phi$ is smooth of relative dimension zero at $p$. Is it true that $J(p)$ has maximum rank as a matrix in $Mat(n\times n,\kappa(p))$?"
Answer: let $A:=k[x_1,..,x_n]$ with $k$ a field and let $\phi: A \rightarrow A$ be a map of $k$-algebras with $\phi(x_i):=f_i(x_1,..,x_n)$. You may construct the Jacobian matrix $J:=(\partial f_i/\partial_{x_j})$, which is a matrix in $Mat(n\times n, A)$ - $n\times n$-matrices with coefficients in $A$. If $f: \mathbb{A}^n_k \rightarrow \mathbb{A}^n_k$ is the corresponding map of "varieties" with $f(\mathfrak{p}):=\mathfrak{q}$, you get an induced map
$$f^{\#}_{\mathfrak{q}}: A_{\mathfrak{q}} \rightarrow A_{\mathfrak{p}}.$$
There is a canonical map $v: A \rightarrow \kappa(\mathfrak{p}):=A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$ and you get canonically a matrix
$$v(J):=(v(\partial f_i/\partial_{x_j})) \in Mat(n\times n, \kappa(\mathfrak{p})).$$
You may alternatively define the map $f$ to be "etale at $\mathfrak{p}$" iff $\det(v(J)) \neq 0$.
You can replace $k$ by any commutative ring $R$ and use this as a definition.
Note: You find this definition explained (with proofs) in Mumfords book
"The red book..", page 345. If you need this result you should verify that this definition is equivalent to the one in Hartshorne.
Let $A:=R[x_1,..,x_n]/I$ where $I:=(f_1,..,f_n)$. Let
$J:=(\frac{\partial f_i}{\partial_{x_j}})$. It follows
$$ \Omega^1_{A/R} \cong A\{dx_1,..,dx_n\}/((d(f_1),..,d(f_n))$$
where
$$d(f_i):=\partial f_i/\partial_{x_1}dx_1+\cdots + \partial f_i/\partial_{x_n}dx_n.$$
On page 346 (chapter 3, section 5 on etale morphisms) it is proved that $\Omega^1_{S/R}=0$ iff $\det(J) \in A^*$ is a unit.
There is another result that explains what an etale morphism is:
Let $k$ be an algebraically closed field and let $X$ be a non-singular variety over $k$ with an open subset $U \subseteq X$. Let $f_1,..,f_n \in H^0(U, \mathcal{O}_X)$. with induced map
$$\phi: U \rightarrow \mathbb{A}^n_k$$
induced by $\phi^*: k[x_1,..,x_n]\rightarrow H^0(U, \mathcal{O}_X)$ with $\phi^*(x_i):=f_i$.
(1) The map $\phi$ is etale.
(2) For all closed points $x\in U$ let $t_i:=f_i-f_i(x)$ Then $t_i$ generate $\mathfrak{m}_x/\mathfrak{m}_x^2$.
(3) For all closed points $x\in U$ it follows the canonical map
$$\phi: k[[x_1,..,x_n]]\rightarrow \tilde{\mathcal{O}_{X,x}}$$
defined by $\phi(x_i):=t_i$ is an isomorphism
(4) $\Omega^1_{U/k} \cong \oplus_i \mathcal{O}_Udf_i $
Again you find a proof in the "red book".
