Why is an étale $k$-algebra a finite product of separable field extensions? Let $k$ be a field, and let $k\to A$ be an étale ring map, then I wish to prove that $A$ is a finite product of separable field extensions of $k$. I can prove the converse, by using the definition of étale-ness in which we write $A=k[x_1,...,x_n]/I$ and consider $k\to A$ to be étale precisely when the differential map $$I/I^2\to\bigoplus_{i=1}^n A\mathrm{d}x_i$$ is an isomorphism, but this doesn't seem to be a useful definition for this direction. 
 A: Consider the map $\operatorname{Spec} A \to \operatorname{Spec} k$. We'll use  "of relative dimension 0" first - this means that the fiber of this morphism is zero-dimensional. Since the base is also zero dimensional, $\operatorname{Spec} A$ is also zero dimensional, which means it is a finite union of points. This implies that $A$ is a finite product of algebraic field extensions of $k$.
Smoothness will give the separability condition. Since every algebraic field extension of $k$ can be expressed as $k(\alpha_1,\cdots,\alpha_n)=k[x_1,\cdots,x_n]/(m_{\alpha_1}(x_1),m_{\alpha_2}(x_2),\cdots,m_{\alpha_n}(x_n))$ where $m_{\alpha_i}(x_i)$ is the minimal polynomial of $\alpha_i$ in the variable $x_i$, we may apply the Jacobian criteria for smoothness. If there exists an element $\beta$ which is not separable, by taking $\alpha_{n+1}=\beta$, we see that the final row of the Jacobian will be zero, so the Jacobian is not full rank and the map is not smooth. Since the map is smooth, there can be no inseparable elements, so the field extension is separable.
