Dimension of Zariski Tangent over $\mathbb{Z}[2i]$

Question

I'm trying to compute the dimension of the Zariski Tangent space of the affine scheme $\operatorname{Spec} A$, where $A = \mathbb{Z}[2i] \cong \mathbb{Z}[x]/(x^2 + 4)$, at the point $(2, 2i)$. Using the isomorphism in the previous sentence, we associate this point to the maximal ideal $(2, x)$ so that our residue field is simply $\kappa(p) = \mathbb{Z}_2$. I know that the Zariski Tangent space in this case may be identified with the set of morphisms $$ \operatorname{Spec} \mathbb{Z}_2[\epsilon]/(\epsilon^2) \to \operatorname{Spec} A $$ mapping the unique point $[(\epsilon)]$ to $p$. Any such map of affine schemes should correspond to some $\mathbb{Z}_2$-linear map of global sections $$ \mathbb{Z}[x] / (x^2 + 4) \to \mathbb{Z}_2[\epsilon]/(\epsilon^2) $$ but now I think I'm starting to get a bit lost in the details of $\mathbb{Z}_2$-linearity, where our generators should go and what the dimension as a $\mathbb{Z}_2$ vector space is. Is there any clean way to get the dimension of $T_{X,p}$ from this? Thanks for any help in advance

Answer 1

The procedure you're describing here is both incomplete and doesn't quite match what you say you're trying to do. As best I can tell, you're importing the calculation of the relative tangent space to a morphism from the case of varieties over a field, but some details have been lost along the way.

Here's what the method you're looking at should look like in the general case. Let $X\to S$ be a morphism of schemes and $x\in X$ a point mapping to $s\in S$. Consider the following diagram:

$$\require{AMScd} \begin{CD} \operatorname{Spec} \kappa(x) @>>> X\\ @VVV @VVV \\ \operatorname{Spec} \kappa(s) @>>> S \end{CD}$$

A relative tangent vector (an element of $T_{X/S,x}$) is a way of factoring the top arrow $\operatorname{Spec} \kappa(x) \to X$ as $\operatorname{Spec} \kappa(x) \to \operatorname{Spec} \kappa(x)[\varepsilon]/(\varepsilon^2) \to X$ where the first arrow is the spectrum of the natural projection to $\kappa(x)$. When you're working with a variety over a field, i.e. $X\to \operatorname{Spec} k$ with some conditions, the relative tangent space $T_{X/\operatorname{Spec} k,x}$ is (canonically?) isomorphic to the absolute tangent space since $\operatorname{Spec} k$ has zero tangent space. But this isn't the situation you find yourself in here - you haven't specified a morphism (you'd probably want $\operatorname{Spec} \Bbb Z[x]/(x^2+4)\to\operatorname{Spec} \Bbb Z$, I think) and depending on the morphism the relative tangent space generally won't be equal to the absolute tangent space.

To see this, taking $X=\operatorname{Spec} A$ and $S=\operatorname{Spec} \Bbb Z$, filling in the above diagram appropriately ends up reducing to finding morphisms of rings $\Bbb Z[x]/(x^2+4)\to \Bbb F_2[\varepsilon]/(\varepsilon^2)$ which commute with the natural projections to $\Bbb F_2$ given by sending $1\mapsto 1$. There aren't too many potential choices: the map is determined by where $x$ goes, and wherever $x$ goes must square to zero, so you get a 1-dimensional relative tangent space given by the two morphisms $x\mapsto 0$ and $x\mapsto \varepsilon$.

To compute the absolute tangent space, first compute the cotangent space then take the $\kappa(x)\cong\Bbb F_2$ dual. The Zariski cotangent space is $m/m^2$, which in our case is $(2,x)/(2,x)^2=(2,x)/(4,2x,x^2)=(2,x)/(4,2x)$ as $x^2=-4$ in $\Bbb Z[x]/(x^2+4)$. This gives you a 2-dimensional $\Bbb F_2$ vector space, with elements $\{0,2,x,x+2\}$. So the absolute tangent space is two dimensional and does not match our calculation of the relative tangent space to $\operatorname{Spec} A\to\operatorname{Spec} \Bbb Z$ from above.