Visualizing Nilpotents (Brainstorming)

Warning: this is a free-writing scratch so might be  incoherent. 

Geometric Interpretation of non-radical I

For a ring $A$, radical ideals of $A$ are in bijective correspondence with closed subsets of $\text{Spec} A$. How should we think about a general ideal $I$ geometrically? 

Every ideal $I$ and its power $I^n$ of $A$ cut out the same closed subsets $\mathbb{V}(I)$ of $\text{Spec} A$ as $\sqrt{I}$. Thus we distinguish them instead by the closed scheme structures each puts on $\mathbb{V}(I)$  (i.e. by the structure sheaves). In particular, we think of $I$ instead as the closed subschemes $\text{Spec} (A/I)$. 

Non-reduced rings and Differential Informations.

Let $A$ be a finitely generated $k$-algebra. If $I$ as above is not a radical ideal, then $A/I$ is not a reduced ring. Classically, every reduced finitely generated $k$-algebra is the coordinate ring of some (classical) affine variety. Its elements corresponds to regular functions of the variety. How should we think about the elements of a non-reduced f.g. $k$-algebra geometrically? 

Taylor Expansion.

Classically if $A = A(V)$ is the coordinate ring of a variety $V\subset \mathbb{A}^n(k)$, then its elements are polynomials of $k[x_1, \ldots, x_n]$ up to the relation $f \sim g$ iff $f|_V = g|_V$ as functions.  If $A$ is not reduced then its elements will correspond to polynomials of $k[x_1, \ldots, x_n]$ up to a stricter relation. If $V$ is the variety associated to $A/\sqrt{0_A}$ the reduction of $A$, then for two polynomials $f$ and $g$ correspond to the same element of $A$, they not only need to be equal on $V$, but also need to satisfy some extra differential condition. 

Example.  $\text{Spec}\mathbb{C}[x]/(x)$ and $\text{Spec}\mathbb{C}[x]/(x^2)$ both cut out the origin in $\mathbb{A}^1$. For every function $f$ in $\mathbb{C}[x]$, its image in $\mathbb{C}[x]/(x)$ gives us $f(0)$, whereas its image in $\mathbb{C}[x]/(x^2)$ gives us the linear term of $f$

In other words, $\mathbb{C}[x]/((x-a)^2)$ gives us the linear term of the Taylor explansion of $f$ at $a$, and hence first order differential information about $f$ at $a$. 

 Thus

$$\mathbb{C}[x]/(x^2) = \{ f \in \mathbb{C}[x] \} / \sim$ where

$$f \sim g \iff  f(0) = g(0) \text{ and (the Taylor expansions at $0$) of $f$ and $g$  have the same linear term}  .$$

Example.  Let $f \in \mathbb{C}[x,y]$. When we reduce $f$ mod $(x, y)$, we lose all the term of (the Taylor expansion at $0$) of $f$ except $f(0)$.

In the above list, on the left we list the rings and on the right we list the information preserved by the image of $f$ in those rings, i.e. the monomials of (the Taylor expansion at $(0,0)$)  of $f$ whose coefficients will be preserved.  

• $\mathbb{C}[x,y]/(x,y) \leftrightarrow  x^0, y^0$
• $\mathbb{C}[x,y]/(x^2,y)= \{(0,0)\} \leftrightarrow  x, x^0, y^0$
• $\mathbb{C}[x,y]/(x,y)^2= \{(0,0)\} \leftrightarrow x, y, x^0, y^0$
• $\mathbb{C}[x,y]/(x^2,y^2)= \{(0,0)\} \leftrightarrow xy, x, y, x^0, y^0$
• $\mathbb{C}[x,y]/(y^2)= \{(0,0)\} \leftrightarrow x^2, xy, x, y, x^0, y^0$

Recall the the coefficient of $x^i y^j$ in the Taylor explansion of $f$ at $(0,0)$ is $$\frac{1}{i! j!} \frac{\partial^{i+j}}{\partial_x^i \partial_y^j} f.$$

Thus image of $f$ in the above rings gives differential information about $f$.

Remark: we see that the first derivatives of $f$ at $0$ are just coefficient of $x, y$ in $f$. Thus we can view them as ``functions'' on $x$, $y$, or more sensibly as linear functions, on the $k$-vector space with basis $x, y$. In other words, $\frac{\partial f}{\partial x}(0)$ and $\frac{\partial f}{\partial y}(0)$ are just elements of the $k$-vector space

$$\left(\frac{(x, y)}{(x, y)^2}\right)^*.$$

Extra-infinitesimal Vectors

We can rephrase the above examples by adding infinitesimal points. We think of the points of $\text{Spec}\mathbb{C}[x]/(x^2)$ as $0$ together with an infinitesimal point $\epsilon_x$, such that $f (\epsilon_x)$ is just the coefficient of $x$ in (the Taylor expansion at $(0,0)$) of $f$. In that case the ring $\mathbb{C}[x]/(x^2)$ is now indeed just the polynomials of $\mathbb{C}[x]$ up to the equivalence relation

$$f \sim g \iff f(0) = g(0) \text{ and } f(\epsilon_x) = g(\epsilon_x).$$

Notice that these new ``points'' are not points in the usual sense since they do not preserve multiplication, i.e. $$f(\epsilon_x) g(\epsilon_x) \neq (fg)(\epsilon_x).$$

Points in the usual sense corresponds to evaluation map $\mathbb{C}[x]/(x^2) \to \mathbb{C}$. The vector $\epsilon_x$ is instead, a derivation $\partial_x: \mathbb{C}[x]/(x^2) \to \mathbb{C}$ as it satisfies the Leibnitz rule (coefficient of $x$ is $fg$ is ....- use this, polynomial multiplication and coefficient, to explain Leibnitz rule to calculus students later).

How many additional points?

Be careful though. Using the above the logic, we could think of the points of $\text{Spec}\left(\mathbb{C}[x,y]/(x^2,y^2)\right)$ as the origin together with an infinitesimal vectors $\epsilon_{xy}$, $\epsilon_{x}$, $\epsilon_{y}$. Intuitively, if our ring is over $\mathbb{R}$, when we know the derivative $$\frac{\partial f}{\partial_{x}}(0) \text{ and } \frac{\partial f}{\partial_{y}}(0) ,$$

we know all the ``directional derivatives'' of $f$ at $0$ since the above two vectors generated the tangent space of $\mathbb{A}^2$ at $0$. In general, the span of the two first order partial derivatives of $f$ at $0$ is canonically isomorphic to the space of all derivations on $\mathbb{C}[x,y]$.

So instead of adding just say $\epsilon_{x}$ and $\epsilon_{y}$, we should add their ``span''. In other words, we should think of  $\epsilon_{x}$ and $\epsilon_y$  as the tangent vectors $(1,0)$ and $(0, 1)$ in the tangent space of $\mathbb{A}^2$ at $(0,0)$ which we identify with $\mathbb{C}^2$. So when we add the points $(1,0)$ and $(0,1)$ to our space, we should also add their span in $\mathbb{C}^2$, i.e. the whole of $\mathbb{C}^2$.

How about  $\epsilon_{xy}$? What are the points we should add if we add $\epsilon_{xy}$? Remember $f(\epsilon_{xy})$ is just the coefficient of $xy$ in (the Taylor expansion at $(0,0)$) of $f$, and thus is (a multiple of ) $\frac{\partial f}{\partial x \partial y} (0, 0)$.  Using the same idea as before, we want to think of all second derivatives in two variables as a vector space, and then take the span of  $\frac{\partial}{\partial x \partial y}$ in that space. Namely we take the $3$-dimensional vector space with basis  $\frac{\partial}{\partial x \partial y}, \frac{\partial}{\partial x^2}, \frac{\partial}{\partial y^2}$. This is exactly the space

$$\frac{(x,y)^2}{(x, y)^3}.$$

Spec of Nilpotent rings has additional Infinitesimal vector

Example.

Let $\mathfrak{m}$ denote the ideal $(x,y)$ in $\mathbb{C}[x,y]$

Thus we can describe the sets of ``points''  of the following affine schemes as follows.

• $\text{Spec}\mathbb{C}[x,y]/(x,y) = \{(0,0)\}$
• $\text{Spec}\mathbb{C}[x,y]/(x^2,y)= \{(0,0)\}$ and span of $\{\frac{\partial }{\partial x}|_{(0,0)} \}$ in the tangent space of $\mathbb{C}[x,y]$ at $0$. The latter is also the space of derivations on $\mathbb{C}[x,y]$, as well as $(\mathfrak{m}/\mathfrak{m}^2)^*$. 
• $\text{Spec}\mathbb{C}[x,y]/(x,y)^2= \{(0,0)\}$ together with all "directional derivatives'' at $(0,0)$, viewed as first order derivations, or points of $(\mathfrak{m}/\mathfrak{m}^2)^*$
• $\text{Spec}\mathbb{C}[x,y]/(x^2,y^2)= \{(0,0)\}$, all points of $(\mathfrak{m}/\mathfrak{m}^2)^*$ and $\text{span} (xy)$ in $(\mathfrak{m}^2/\mathfrak{m}^3)^*$
• $\text{Spec}\mathbb{C}[x,y]/(y^2)= \{(0,0)\}$, (span of ) $\frac{\partial}{\partial_y}$, and for each $i$, (span of) $\frac{\partial^i}{\partial x^i}.$