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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?

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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.

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