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I'm trying to work out a basic example where formal smoothness should fail.

I'm considering $\mathbb{R} \to \mathbb{R}[x,y]/(x^2-y^2)$.
The idea is that not every $\mathbb{R}$-homomorphism $\mathbb{R}[x,y]/(x^2-y^2) \to \mathbb{R}$ should lift to a homomorphism $\mathbb{R}[x,y]/(x^2-y^2) \to \mathbb{R}[\varepsilon]/(\varepsilon^2)$. But I can't see that ever being possible: after all, I can just take the exact same homomorphism, with image $\mathbb{R} \subset \mathbb{R}[\varepsilon]/(\varepsilon^2)$; this gives a valid lift.

Do I need to consider something else than $R = \mathbb{R}[\varepsilon]/(\varepsilon^2)$ with $I = (\varepsilon)$ in order to witness the failure of formal smoothness? I would think that was enough, given that it should fully explain lifting of points to tangent vectors.

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Indeed, you need to consider a different artinian ring to witness the failure of formal smoothness.

Look instead at $\mathbb{R}[\epsilon]/\epsilon^3 \to \mathbb{R}[\epsilon]/\epsilon^2$ and map $\mathbb{R}[x,y]/(x^2-y^2)$ to $\mathbb{R}[\epsilon]/\epsilon^2$ by $x \mapsto \epsilon$, $y \mapsto 0$. If you wanted to lift this to a map to $\mathbb{R}[\epsilon]/\epsilon^3$, then you would have $x \mapsto \epsilon + a \epsilon^2$ and $y \to b \epsilon^2$, but then $x^2 = \epsilon^2$ and $y^2=0$.

The "geometric" intuition here is that the Zariski tangent space to $x^2=y^2$ at $0$ is two dimensional but those tangent vectors which have slope other than $\pm 1$ can not be extended to higher order jets.

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    That's perfect, thanks.2010-10-24