# Standard etale maps are etale

We define a finitely presented $$R$$-algebra $$A$$ to be etale if for every $$R$$-algebra $$S$$, every ideal $$I\subset S$$ such that $$I^2=0$$ and every homomorphism $$\phi: A\to S/I$$ there exists a unique homomorphism $$\psi: A\to S$$ such that $$\pi\circ\psi=\phi$$, where $$\pi: S\to S/I$$ is the standard projection map.

We call an $$R$$-algebra standard etale if it is isomorphic to $$R[X]_g/(f)$$ for some $$f,g$$ such that $$f'$$ is invertible in $$R[X]_g/(f)$$.

It is not obvious to me that a standard etale algebra is etale. How would one show this, or where would one start?

Let $$S$$ be any $$R$$-algebra, and $$A=R[X]_g/(f)$$ be any standard étale algebra. The $$R$$-morphisms $$A \rightarrow S$$ are given by the $$x \in S$$ such that $$f(x)=0$$ and $$g(x) \in S^{\times}$$.
Now let $$S$$ be any $$R$$-algebra, and $$I \subset S$$ an ideal with square zero. We need to show that for each $$x \in S$$ such that $$f(x) \in I$$ and $$g(x)$$ is invertible mod $$I$$ ($$x\,\mathrm{mod}\,I$$ represents one morphism $$A \rightarrow S/I$$), there is a unique $$x' \in S$$ congruent to $$x$$ mod $$I$$ such that $$g(x') \in S^{\times}$$ and $$f(x')=0$$.
Note that since $$I^2=0$$, $$y \in S$$ is invertible iff it is invertible mod $$I$$. So we need to show that there is a unique $$x' \in x +I$$ such that $$f(x')=0$$. But since $$I^2=0$$, it is easy to see that for $$y \in I$$, $$f(x+y)=f(x)+f'(x)y$$.
To conclude, one only needs to note that $$f'(x) \in S^{\times}$$, which is easy since $$f' \in A^{\times}$$ as $$A$$ is standard étale.