First of all, I am considering a simple example where inflection point(s) are easy to find. I have constructed a cubic of one real variable, $$f(x) = x^3-6x^2+11x-6$$ which has roots $1,2,$ and $3$. Furthermore, taking two derivatives ($f^{\prime \prime}$ (x)) and solving for the roots, gives us that there is an inflection at $$x = 2$$

I am trying to see how the geometry of this inflection/undulation point for this example is related to the algebra of schemes.

Take the functor $Z_T: \mathbf{Alg_R} \rightarrow \mathbf{Sets}$ , which gives us the simultaneous solution set to all polynomials in $T\subset R[x_1,\ldots,x_s]$, in any R-Algebra $C$.

Functorialising our problem above means setting $R = \mathbb{R}$, $T = \langle x^3-6x^2+11x-6\rangle\subset \mathbb{R}[x]$, and $$A = \frac{\mathbb{R}[x]}{\langle (x-1)(x-2)(x-3) \rangle}$$ for the natural correspondence $\text{Spec}_{\mathbb{R}}(A)\cong Z_T$.

I have constructed the above because I am hoping that there is some precise way of viewing the inflection at $x=2$ in the algebra. I feel that it is natural that I should be looking at some property of the representing ring $A$, or Spec but I have no intuition for what.

At risk of being hand-wavy, I know that 'tangential intersections' corresponds to 'multiple roots', which corresponds to nilpotents, and I was hoping to find a similar property here.



1 Answer 1


I think rather than the three points that are the zeroes of $f$, you are really interested in the plane curve given by $C: y = x^3-6x^2+11x-6$. The coordinate ring (over $\mathbb{R}$) of this curve is $$ A = \mathbb{R}[C] = \frac{\mathbb{R}[x,y]}{\langle y - (x^3-6x^2+11x-6)\rangle} \, . $$ The point $P = (2,0)$ corresponds to the maximal ideal $\mathfrak{p} = \langle x-2,y \rangle$. The tangent line $\ell$ to $C$ at $P$ is $$ \ell = F_x(P)(x - 2) + F_y(P)(y - 0) = x-2 + y $$ where $F = y - (x^3-6x^2+11x-6)$.

The fact that $P$ is an inflection point can be detected by calculating the order of vanishing of the tangent line $\ell$ in the local ring $A_\mathfrak{p}$. Since $C$ is nonsingular at $P$, then $A_\mathfrak{p}$ is a discrete valuation ring (DVR). Let $\mathfrak{m} = \mathfrak{p} A_\mathfrak{p}$ denote its unique maximal ideal. Since the vertical line $x-2 = 0$ is not tangent to $C$ at $P$, then $x-2$ is a uniformizing parameter for $A_\mathfrak{p}$, i.e., $\mathfrak{m} = \langle x-2 \rangle$.

Computing the formal Taylor expansion of $F$ at $P$, we find $$ F = (x-2) + y - (x - 2)^3 $$ which shows that $$ \ell = (x-2) + y = (x-2)^3 \in \mathfrak{m}^3 $$ in the local ring $A_\mathfrak{p}$. (Recall that $F = 0$ in $A$ since we are quotienting by its principal ideal.) This shows that $\DeclareMathOperator{\ord}{ord} \ord_P(\ell) = 3$, so $P$ is an inflection point, aka an ordinary flex.

If you're interested, Fulton's curve book has some nice exercises on flexes, namely 3.12 and 5.23.


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