2
$\begingroup$

This question already has an answer here:

I want to prove that the Zariski tangent space at $x\in X$ ($X$ is an affine scheme) is isomorphic to $Hom_K(X,K[\epsilon]/(\epsilon^2))$ (K is the residue field at $x$). I want to say that $$Hom_K(\mathfrak{m}_{X,x}/\mathfrak{m}_{X,x}^2,K)\simeq Hom_K(\mathcal{O}_{X,x}/\mathfrak{m}_{X,x}^2,K[\epsilon]/(\epsilon^2))$$ so I need splitting $\mathcal{O}_{X,x}$ into $\mathfrak{m}_{X,x}/\mathfrak{m}_{X,x}^2\oplus K$, and I build this splitting using the exact sequence $$\mathfrak{m}_{X,x}/\mathfrak{m}_{X,x}^2\rightarrow \mathcal{O}_{X,x}/\mathfrak{m}_{X,x}^2\rightarrow K$$ so I need a splitting map $\sigma:K\rightarrow \mathcal{O}_{X,x}/\mathfrak{m}_{X,x}^2$. How can I do to find $\sigma$?

$\endgroup$

marked as duplicate by Zhen Lin, Dietrich Burde, Thomas Andrews, Vedran Šego, user61527 Oct 8 '13 at 20:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4
$\begingroup$

Let $K$ be a field and $X$ a $K$-scheme i.e. a scheme provided with a morphism $f:X\to \text {Spec} (K)$.
In particular all local rings $\mathcal O_{X,x}$ are $K$-algebras and by abuse of language we write $K\subset \mathcal O_{X,x}$.
That said, a rational point is a point $x\in X$ such that $\mathcal O_{X,x}=K\oplus \mathfrak m_x$.
[Of course $K$ and $\mathfrak m_x$ always intersect in zero, but for a non-rational point we have $\mathcal O_{X,x}\supsetneq K\oplus \mathfrak m_x$. Think of $\text {Spec} (\mathbb C)$ over $\mathbb R$: its only point is non rational because $\mathbb C \supsetneq \mathbb R\oplus 0$]

Given such a rational point $x$, the Zariski tangent space $T_x(X)$ is the set of $K$-morphisms $\text{Spec}(K[\epsilon ] )\to X$ whose image is the point $x$.
(For that reason Mumford calls the $K$-scheme $\text{Spec}(K[\epsilon ] )$ a desincarnated tangent vector over $K$)

Equivalently $T_x(X)$ is the set of local $K$-morphisms of local rings $$\mathcal O_{X,x}=K\oplus \mathfrak m_x\to K[\epsilon]:q+m\mapsto q+l(m)\cdot\epsilon$$ These morphisms in turn correpond to $K$-linear maps $l:\mathfrak m_x\to K: m\mapsto l(m)$ sending $\mathfrak m_x^2$ to $0$ or finally to $K$-linear maps $\overline l:\mathfrak m_x/\mathfrak m_x^2 \to K: \overline m\mapsto l(m)$.
And we have rediscovered the Zariski definition $T_x(X)=(\mathfrak m_x/\mathfrak m_x^2)^*$ !

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.