Interpretation of $\mathbb{Q}[\varepsilon]/(\varepsilon^3)$-points of a scheme $X=\text{spec}\frac{\mathbb{Q}[x_1,...,x_n]}{(f)}$

Question

I'm beginning to study the functor of points of a scheme and ran into the classical example that, given an affine scheme $X=\text{Spec}\dfrac{\mathbb{Q}[x_1,...,x_n]}{(f)}$, which is (in the clear fashion) a $\mathbb{Q}$-scheme, choosing a $\mathbb{Q}[\varepsilon]/(\varepsilon^2)$-point of $X$ is equivalent to choosing a point $(a_1,...,a_n)\in\mathbb{Q}^n$ such that $f(a_1,...,a_n)=0$ and a vector $(b_1,...,b_n)$ tangent to the surface $f=0$. As hinted in the thread Correspondence between $k[\epsilon]/(\epsilon^2)$-points of a scheme, and points with tangent vectors, the method of expanding $f$ as a taylor series works as well for $\mathbb{Q}[\varepsilon]/(\varepsilon^n)$-points of $X$, but I'm having problems generalizing beyond the case $n=2$. For example: a $\mathbb{Q}[\varepsilon]/(\varepsilon^3)$-point of $X$ is "the same" as a morphism of $\mathbb{Q}$-algebras $\mathbb{Q}[x_1,...,x_n]/(f)\longrightarrow\mathbb{Q}[\varepsilon]/(\varepsilon^2)$. Of course, the only freedom in choosing such morphisms is in $x_i\longmapsto a_i+b_i\varepsilon+c_i\varepsilon^2$. But knowing that $f(x_1,...,x_n)=0$ implies (using the taylor expansion of $f$ first around ) that \begin{align} 0&= f(a_1+b_1\varepsilon+c_1\varepsilon^2,...,a_n+b_n\varepsilon+c_n\varepsilon^2)\\ &= f(a_1+b_1\varepsilon,...,a_n+b_n\varepsilon)+\sum_i\frac{\partial f}{\partial x_i}(a_1+b_1\varepsilon,...,a_n+b_n\varepsilon)c_i\varepsilon^2\\ &= f(a_1,...,a_n)+\sum_i\frac{\partial f}{\partial x_i}(a_1,...,a_n)b_i\varepsilon+\sum_i\left(\frac{\partial f}{\partial x_i}(a_1,...,a_n)+(...)\varepsilon+(...)\varepsilon^2\right)c_i\varepsilon^2\\ &= f(a_1,...,a_n)+\sum_i\frac{\partial f}{\partial x_i}(a_1,...,a_n)b_i\varepsilon+\sum_i\frac{\partial f}{\partial x_i}(a_1,...,a_n)c_i\varepsilon^2, \end{align} which implies that $(a_1,...,a_n)$ is a point (in the usual sense) of $f=0$, but both $(b_1,...,b_n)$ and $(c_1,...,c_n)$ are tangent vectors to that surface! Shouldn't $(c_1,...,c_n)$ have something to do with the second derivatives (or perhaps the hessian) of $f$?

Answer 1

The solution to your extremely clever paradox is that the calculation of $f(a_1+b_1\varepsilon+c_1\varepsilon^2,...,a_n+b_n\varepsilon+c_n\varepsilon^2)$ does involve, despite appearances, the second derivatives of $f$ !
Take for example $f(x_1,\dots,x_n)=\lambda x_1^2-x_1$. Then your formula or a direct calculation yield $$ f(a_1+b_1\varepsilon+c_1\varepsilon^2,...,a_n+b_n\varepsilon+c_n\varepsilon^2)=\lambda(a_1^2+[b_1^2+2a_1b_1]\epsilon+2a_1c_1\epsilon^2 )-(a_1+b_1\epsilon +c_1\epsilon ^2)$$ from which you get the conditions $$\lambda a_1^2-a_1= \lambda [b_1^2+2a_1b_1] -b_1= 2\lambda a_1c_1-c_1=0 \quad \quad (\bigstar)$$ for a point $(a_1+b_1\varepsilon+c_1\varepsilon^2,...,a_n+b_n\varepsilon+c_n\varepsilon^2)\in (\mathbb Q[\varepsilon]/(\varepsilon^3))^n$ to correpond to a morphism $\dfrac{\mathbb{Q}[x_1,...,x_n]}{(f)}\to \mathbb Q[\varepsilon]/(\varepsilon^3)$
Solution of the paradox
The conditions $(\bigstar)$ do involve a second derivative of $f$, namely $$\large\lambda =\frac 1 2 \frac {\partial^2 f}{\partial x_1^2}(0)$$