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I am trying to understand the definition of Zariski tangent space, let's try an example that could be in a number theory course.

  • $A = \mathbb{Z}[i]$ is a ring. E.g. $\mathbb{Z}[i] \simeq \mathbb{Z}[x]/(x^2 + 1)$ (solving the polynomial equation $x^2 + 1 = 0$.

  • The prime ideals $\mathfrak{p} \in \text{Spec}(A)$ e.g. $\{ 2+i, 7, 13, 5 - 2i \} \in \text{Spec}(A)$ called "points".

  • Then we construct the local rings, $\mathbb{Z}[i]_\mathfrak{p}$ such as $\mathbb{Z}[i]_7$ or $\mathbb{Z}[i]_{2+i} \simeq \mathbb{Z}_5$ (they are isomorphic as rings, obviously they are different).

  • The maximal ideal, $\mathfrak{m} = \mathfrak{p} \mathbb{Z}[i]_\mathfrak{p}$ is used to define the Zariski tangent space $\mathfrak{m}/\mathfrak{m}^2$, a vector space over the field: $k=A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$.

For each of these Gaussian primes $\mathfrak{p}$, I'd like the (Zariski) tangent space $\mathfrak{m}/\mathfrak{m}^2$ and co-tangent space $(\mathfrak{m}/\mathfrak{m}^2)^*$ of $\text{Spec}(A)$ at the various points. Let's check that $\mathfrak{p} = 2 + 3i$ is "prime" since $2^2 + 3^2 = 4 + 9 = 13$ we are OK.

hm2020
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cactus314
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    A minor note - I think you have things backwards. $\mathfrak m / \mathfrak m^2$ is the cotangent space and its dual is the tangent space. The analogy is in the (local) ring of germs of smooth functions at some point of a manifold. Here $(\mathfrak m / \mathfrak m^2)^*$ is naturally identified with the usual tangent space, defined say as infinitessimal curves at the point in question. – paul blart math cop Jun 05 '21 at 21:57
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    OP, could you clarify what it is you need help with? You've provided a list of statements rather than any questions, and it's not clear what is just exposition vs. what is confusing to you. – Tabes Bridges Jun 05 '21 at 22:27
  • @TabesBridges it's difficult to say... What's so "tangent" about this construction? This is ring theory. Literally a tangent is the tangent line to a circle, or the tangent line to a variety. – cactus314 Jun 05 '21 at 22:52
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    The fundamental idea in modern algebraic geometry is that every ring is the ring of functions on some geometric space (with "functions" and "space" suitably generalized in the process). If you think about smooth functions on a manifold $M$, the evaluation map that sends a function $f$ to its value at a fixed point $p$ is a surjective ring map $ev_p: C^\infty(M) \to \mathbb R$ whose target is a field; thus $\ker(ev_p)$ is a maximal ideal, and in fact all maximal ideals arise in this way... – Tabes Bridges Jun 06 '21 at 00:28
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    ...So $\mathfrak m$ is functions vanishing at $p$; this means that $\mathfrak m^2$ is functions vanishing to order $\ge 2$ at $p$, so elements of the quotient are functions which vanish at $p$ and have the same first derivative at $p$. – Tabes Bridges Jun 06 '21 at 00:28
  • OP this comment is a bit late, but I think you may find the discussion in this question helpful. The accepted answer talks about how Grothendieck style scheme theory allows for a unified geometric language for complex manifolds, varieties over finite fields, rings of integers of number fields, and all rings simultaneously. It doesn't specifically answer your question, but I think the ideas it presents are important for understanding the why of schemes. – paul blart math cop Jun 07 '21 at 03:30

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Question: "I am trying to understand the definition of Zariski tangent space, let's try an example that could be in a number theory course."

You find an "elementary" discussion of the tangent and cotangent space at this thread:

$aX + bY$ is an element of $M^2$ if and only if the line $aX + bY = 0$ is tangent to $W$ at $(0, 0)$

Example: A regular curve: If $C \subseteq \mathbb{A}^2_k$ is a plane regular curve over a algebraically closed field $k$ of characteristic zero, and if $p \in C$ is a point with maximal ideal $\mathfrak{m}_p:=(x-a,y-b)$ it follows the line

$$l(x,y):=u(x-a)+v(y-b)$$

is tangent to $C$ at $p$ iff $l(x,y) \in \mathfrak{m}_p^2$. A generator for the tangent line to $C$ at $p$ is the line defined by the polynomial

$$l_p(x,y):=F_x(p)(x-a)+F_y(p)(y-b)$$

where $F_x(p)$ is the partial derivative of $F(x,y)$ evaluated at $p$ etc. The curve $C$ is defined by the polynomial $F(x,y)$. Hence the line $l(x,y)$ is in $\mathfrak{m}_p^2$ iff $l(x,y)$ is a scalar mutiple of $l_p(x,y)$.

Example: Algebraic number theory: In your case of the ring of integers $\mathcal{O}:=\mathbb{Z}[i]$ in a number field $K:=\mathbb{Q}(i)$ you must check that $\mathfrak{m}_x:=(a+bi) \subseteq \mathcal{O}$ generate a maximal ideal and calculate the two $\kappa(x)$-vector spaces

$$Hom_{\kappa(x)}(\mathfrak{m}_x/\mathfrak{m}_x^2, \kappa(x)), \mathfrak{m}_x/\mathfrak{m}_x^2.$$

Note that if $\mathfrak{m}_x$ is prime ideal, it is maximal and there is a canonical isomorphism

$$\mathfrak{p}_x/\mathfrak{p}_x^2 \cong \mathfrak{m}_x/\mathfrak{m}_x^2$$

where $\mathfrak{p}_x:= \mathfrak{m}_x(\mathcal{O}_{\mathfrak{m}_x})$.

Note: In algebraic number theory, the "cotangent module" (the module of Kahler differentials) $\Omega^1:=\Omega^1_{\mathcal{O}/\mathbb{Z}}$ gives rise to the ramified primes. The different $\Delta_{L/K}$ is the annihilator ideal of the cotangent module $\Omega^1_{\mathcal{O}_L/\mathcal{O}_K}$, and the discriminant $\delta_{L/K}:=N_{L/K}(\Delta_{K/L})$ has the property that a prime ideal $\mathfrak{p} \subseteq \mathcal{O}_K$ is ramified iff $\mathfrak{p}\in Z:=V(\delta_{L/K})$. Hence the complement $$U:=Spec(\mathcal{O}_K) - V(\delta_{L/K})$$

is the open subscheme of unramified primes.

The canonical map

$$\pi: C:=Spec(\mathcal{O}) \rightarrow S:=Spec(\mathbb{Z})$$

is finite, and the closed subscheme $Z\subseteq S$ with open complement $U:=S-Z$ has the property that the induced map

$$\pi_U:\pi^{-1}(U) \rightarrow U$$

is etale. This is one of the usages of the cotangent module in algebraic number theory. The cotangent module $\Omega^1$ contains information on which primes that ramify. The tangent module

$$T:=(\Omega^1)^*$$

is the dual of the cotangent module, and when $\Omega^1$ is locally free you do not lose information when passing to $T$ since $T^*\cong (\Omega^1)^{**} \cong \Omega^1$. When not locally free, you lose information.

hm2020
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