I asked the following question in the past. The focus on my question is the title itself, but I was little curious about the concept of the "double points" (or "fat points") at the sight of Ravi Valki's lecture note. To begin with, I read up on the concept of dual ring, because , like the comment of the following question, I was not ready to understand the concept of double points back then. And instead of the Valki's note, I referred the Gathman's lecture note in page 96. Here, I transcribe the Gattman's explanation about the double points:

Geometrically, one can think of $\operatorname{Spec}R$ as “a point that extends infinitesimally in one direction”: As on the affine line $\mathbb{A}_{k}^{1}$, there are polynomial functions in one variable on $\operatorname{Spec}R$, but the space is such an infinitesimally small neighborhood of the origin that we can only see the linearization of the functions on it, and that it does not contain any actual points except $0$.

If $R={k[x]}/{(x^2)}$ , then $\operatorname{Spec}R$ is just one point,$(x)$, and clearly the point would not mean a geometrical point like a point $P$ on the line or segment, because indeed the point is a principal ideal. Also, every (principle) ideal contains $(0)$, Hence, in my conclusion after I read the Gathman's note :

  • it is plausible $(x) \in \operatorname{Spec}R$ is fatter than the geometrical point.
  • the double point is located in the middle of affine line $\mathbb{A}_{k}^{1}$ , i.e it bloats throughout near the origin point.

However, I have yet to understand the geometrical meaning described on the Gathman's lecture, espeically, "infinitesimally" in one direction. Obviously, it is impossible that the double point eat out all affine line. However, I do not surely catch why the double points occupies only very small area of affine line.

Here is my attempt (why the point extends infinitesimally) :

the double point $(x)$ is a set and the set is written as $(x)=\left\{\bar{r}x : \bar{r} \in R\right\}$. (where $\bar{r}$ is a representation of $R$). when considering this set, the element of $(x)$ is likely to be dotted sporadically on $\mathbb{A}_{k}^{1}$.-- this is my thought I believe. But, it would be impossible, because $R$ is not reduced ring , that means $R=k[x]/(x^2)$ has a nilpotent element. When considering my previous question, whenever I pick an (nonzero) element $\bar{r} \in R ~,~ \bar{r}^{2}=\bar{0}$. Thus, it is impossible that the point is dotted affine line here and there. For example, when considering $\bar{y}= \bar{r}^{31}x \in (x)$, but $\bar{r}^{31}x= (\bar{r}^2)^{15}\cdot \bar{r} x=\bar{0}\cdot \bar{r} x = \bar{0}$. ( I just expect the point $\bar{y}$ would be located far away from the origin point, $\bar{0}$, but my expectation clearly fails) Anyway, the double point cannot expand around the origin point, $\bar{0}$.


1 Answer 1


You are taking the ball a little too literally. The word infinitesimal should be interpreted in the sense of calculus. If I have a function $f = 3+x+x^2$ at a point $x=0$, then $f(0) = 3$ is the basic information of the value of the function at the point where as $f'(0) = 1$ and $f''(0) = 2$ are what we think of as infinitesimal information. The idea is that if you reduce a function of $k[x]$ modulo a reduced ideal $I$, you lose all the infinitesimal information but if you reduce it modulo a non-reduced ideal $I$, you get to retain some of the infinitesimal information. Gathmann and Vakil just indicate this using a little ball around the point. A much more detailed explanation below:

For a ring $R$ and an ideal $I$, we have the natural map $ R \to R/I$ taking an element $f \mapsto \overline f$, it's reduction in the quotient. The term reduction itself implies that we are forgetting or throwing away some information. The intuition for fat points now is that when we reduce $f$, it remembers more than just it's value at the underlying scheme

The following examples just expand on this idea:

  • If $R = \mathbb Z$ and $I = (5)$, then which object do you think retains more information? The integer $7428$ or it's reduction $\overline 3$? Obviously the former, since an element of $\mathbb Z$ contains way more information than an element of $\mathbb Z/ 5 \mathbb Z$. So under the reduction map to $\mathbb Z/ 5 \mathbb Z$, the number $7428$ loses all information except it's remainder modulo $5$.

  • If $R = \mathbb Z$ and $I = (25)$, then which object do you think retains more information? The integer $7428$ or it's reduction $\overline 8$? Obviously the former, since an element of $\mathbb Z$ contains way more information than an element of $\mathbb Z/ 25 \mathbb Z$. This is still a lot less information than the original integer $7428$ but is still more information than the reduction of $7428$ modulo $5$.

  • Similarly, if $R = k[x]$ and $I = (x)$, which object do you think retains more information? The polynomial $f = x^3+3x^2-x+4$ or it's reduction ( or equivalently it's value at $x=0$) of $4$? Obviously the former since the full polynomial remembers a lot more: for instance it knows its values at all other field values in $k$, you can take it's derivatives, etc. So under the reduction map to $k[x]/(x)$, the polynomial $f$ loses all information except it's value at $x=0$.

  • Now lets deal with $R = k[x]$ and $I = (x^2)$. Then the reductions of a polynomial $f = a_nx^n+a_{n-1}x^{n-1} + \cdots a_1x+a_0$ has representative $a_1x+a_0$. So under the reduction, the function $f$ is remembering BOTH it's value at $0$, i.e. $a_0$ as well as it's derivative at $0$, i.e. $a_1$. We are still losing a lot of information but remembering more than if $I = (x)$.

  • More generally if $I = (x^n)$, we remember the value of $f$ at $0$ as well as the additional coefficients $a_1, a_2,\ldots a_{n-1}$.

  • So the punchline is that being a fatpoint means that under the reduction map $R \to R/I$, (where $I$ is now not a reduced ideal) the functions remember a bit more than just their value.


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