I want to understand embedded points. Although I have some basic examples in mind, and I read through, for instance, the relevant chapter in Vakil's FOAG, I would like to understand them more deeply. So I asked myself this question:
Suppose we have an affine line in $\mathbb A^3$, say $L=V(x,y)\subset\mathbb A^3$, and we fix the origin $p=[(x,y,z)]\in L$. What is the datum of an embedded point at $p$? What do I have to specify, exactly?
So we have this line, and I want to produce another subscheme of $\mathbb A^3$ which is the same line, but with the origin now being non-reduced. But I think one can do this in several ways, e.g. the embedded point could "point" in different directions. Which directions? The other two axes only?
If you have any ideas on how to sort out this problem, or similar ones, I would be glad to learn from you.
Thanks!
Let's first study the question without an ambient space: Given a commutative ring $A$ and a prime ${\mathfrak p}\lhd A$, you may glue an "infinitesimal component" to $\text{Spec}(A)$ along $v({\mathfrak p})$ by replacing $A$ with the pullback $\tilde{A}$ of $(A/{\mathfrak p})[X]/(X^2)\to A/{\mathfrak p}\leftarrow A$. Geometrically, this means taking the pushout of the closed embedding $\text{Spec}(A/{\mathfrak p})\hookrightarrow\text{Spec}(A)$ along the infinitesimal thickening $\text{Spec}(A/{\mathfrak p})\hookrightarrow\text{Spec}((A/{\mathfrak p})[X]/(X^2))$. This makes ${\mathfrak p}$ an associated prime in the following sense: first, the element $t := (X,0)$ in $\tilde{A}$ satisfies $t^2=0$ and the projection $\tilde{A}\to A$ induces an isomorphism $\tilde{A}/(t)\cong A$, so that in particular $\text{Spec}(A)\cong\text{Spec}(\tilde{A})$. Under this bijection, ${\mathfrak p}$ corresponding to $\tilde{{\mathfrak p}} := \{(\overline{a}X,x)\ |\ a\in A, x\in{\mathfrak p}\}$, which is precisely the annihilator of $t$ in $\tilde{A}$; hence $\tilde{{\mathfrak p}}\in\text{Ass}(\tilde{A})$ as expected.
If $A=B/I$ so $\text{Spec}(A)$ arises as a closed subscheme of some $\text{Spec}(B)$, with ${\mathfrak p}\lhd B$ a prime containing $I$, then extending the embedding $\text{Spec}(A)\to\text{Spec}(B)$ to a morphism $\text{Spec}(\tilde{A})\to\text{Spec}(B)$ amounts, by the universal property of the pushout/pullback, to extending the embedding $\text{Spec}(B/{\mathfrak p})\to\text{Spec}(B)$ to a morphism $\text{Spec}((B/{\mathfrak p})[X]/(X^2))\to\text{Spec}(B)$, that is, to chosing a derivation $D: B\to B/{\mathfrak p}$. At the moment, I'm lacking the connection of that with the tangent space of $\text{Spec}(B)$ at $\mathfrak p$ in general (maybe someone else can help?), but if $B$ is a $k$-algebra and ${\mathfrak p}={\mathfrak m}$ is a $k$-rational point, then such a derivation $B\to B/{\mathfrak m}=k$ satisfying the additional constraint of $k$-linearity is the same as an element of the tangent space of $\text{Spec}(B)$ at ${\mathfrak m}$. Moreover, the resulting morphism $\text{Spec}(\tilde{A})\to\text{Spec}(B)$ is a closed embedding if and only if $D$ does not vanish completely on $I$, meaning that $D$ does not point "in the direction of $\text{Spec}(A)$ within $\text{Spec}(B)$". Any such choice of $D$ will then lead to a closed subscheme of $\text{Spec}(B)$ in which the point ${\mathfrak m}$ has been thickened "in the direction of $D$", and in particular is an associated prime.
Hence, in your example, you can thicken $L$ in any direction different from $L$ itself. Explicitly, let $\textbf{x}:=(x_0,y_0,z_0)\in{\mathbb A}^3_{\mathbb k}({\mathbb k})$ be a ${\mathbb k}$-rational point outside $L({\mathbb k})$, i.e. $(x_0,y_0)\neq (0,0)$. Then $\textbf{x}$ yields a derivation $D_{\mathbb x}: {\mathbb k}[x,y,z]\to {\mathbb k}$ (the latter viewed as an ${\mathbb k}[x,y,z]$-module by evaluation at $0$) given by $D_{\mathbb x}(p) := \left.\frac{\text{d}}{\text{d}\lambda}p(\lambda\textbf{x})\right|_{\lambda=0}=x_0 p_x(\textbf{0})+y_0 p_y(\textbf{0})+u_0 p_z(\textbf{0})$ not vanishing on $I := (x,y)$, and the ideal $J := \{p\in I\ |\ D_{\mathbf x}(p)=0\}\subset I$ defines the infinitesimal thickening of $v(I)$ at $\textbf{0}$ in the direction of $\textbf{x}$. Explicitly, if wlog $x_0=1$, then $\overline{x}\neq 0$ in ${\mathbb k}[x,y,z]/J$ and its annihilator is the maximal ideal $(x,y,z)$ corresponding to the origin. Note that having ${\mathbb k}[x,y,z]/J$ as function spaces means that a polynomial vanishes if and only if it vanishes on the line $L$ and has vanishing differential in the direction of $\textbf{x}$; the tangent in direction of $\textbf{x}$ is therefore 'remembered' on restriction to the subscheme associated to $J$.