What is the tangent space of $\mathbb{Z}$ at $\mathfrak{p}=(5)$?

Question

What is the tangent space of the scheme $\text{Spec}\,\mathbb{Z}$ at the point $\mathfrak{p}= (5)$ ?

• the local ring of $\mathbb{Z}$ at $(5)$ is $\mathcal{O}_{\mathbb{Z},5}=\mathbb{Z}_5$ the 5-adic numbers
• and the ring of regular functions is the set of functions $f \in \mathbb{Z}$ vanishing at $\mathfrak{p}=5$, so we have $m_{\text{Spec}\, \mathbb{Z}, (5)} = 5 \,\mathbb{Z}_5$.
• therefore the co-tangent space of $\mathbb{Z}$ at $\mathfrak{p}=(5)$ is $m_x/m_x^2 = 5 \mathbb{Z}_5 / 5^2 \mathbb{Z}_5 \simeq \mathbb{Z}/5\mathbb{Z}$.

Originally I had thought it would be $\mathbb{Q}_5^\times / (\mathbb{Q}_5^\times)^2$ but oddly $\mathbb{Q}_p$ doesn't factor anywhere into this discussion.

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The localization of the ring $\mathbb{Z}$ at the place $(5)$ is $\mathfrak{m}_5=\mathbb{Z}[\frac{1}{5}]$, so perhaps the tangent space should be the quotient ring $\mathfrak{m}_5/\mathfrak{m}_5^2=\mathbb{Z}[\frac{1}{5}]/\mathbb{Z}[\frac{1}{5}]^2$.

Answer 1

The local ring $R$ is $\mathbb Z$ localized away from the prime ideal $5\mathbb Z$, i.e., the set $\{q\in \mathbb Q:q=\frac{a}{b}$ for some $a,b\in \mathbb Z, (5,b)=1\}$. Its maximal ideal is $5R$, and its Zariski tangent space is the dual of $5R/25R$, a vector space over $\mathbb Z/5\mathbb Z$ of dimension $1$.