Tangent vector of a scheme

Question

I am reading Mumford's book abelian varieties. On page 97, he defined a tangent vector at $x \in X$ (where $X$ is a scheme of finite type over an algebraically closed field $k$ and $x$ is a point of $X$) to be a derivation $\mathcal{O}_{X,x} \to k$. I thought you can only define a derivative of $A$ into $M$ is $M$ is an $A$ module, which doesn't seem to be the case here?

Then in the proof of the proposition on page 97, he makes the observation that, if $A$ is a $k$-algebra and $B$ is an $A$-algebra, the set of $k$-derivations of $A$ into $B$ is in bijection with the set of morphisms $\phi: A \to B[\epsilon]/(\epsilon^2)$ mapping $a$ to $a \cdot 1 +$ a multiple of $\epsilon$, and the bijection is defined as follows: a derivation $D$ corresponds to $\phi(a) = a \cdot 1 + (Da) \cdot \epsilon$. Going back to a tangent vector at a point of a scheme, since $k$ is not an $\mathcal{O}_{X,x}$-module, this "$a \cdot 1$" seems to be not well defined.

He then claims that given the above observation, giving a tangent vector at $x \in X$ is equivalent to giving a morphism $\operatorname{Spec} k[\epsilon]/(\epsilon^2) \to X$ such that the image of $\operatorname{Spec} k \to \operatorname{Spec} k[\epsilon]/(\epsilon^2) \to X$ is the point $x$, which I also can't really understand.

Thank you in advance for any help!

Answer 1

$k$ is an $\mathcal{O}_{X,x}$ module: there's a ring homomorphism $\mathcal{O}_{X,x} \to \mathcal{O}_{X,x}/\mathfrak{m} = k$ which makes $k$ in to a module over $\mathcal{O}_{X,x}$. This should resolve the issues in the first two paragraphs.

For the third paragraph, this is just an application of the second paragraph. Remember what a morphism of schemes is: it's a map of topological spaces $X\to Y$ plus a map of sheaves of rings $\mathcal{O}_Y \to f_*\mathcal{O}_X$. As $k[\epsilon]/(\epsilon^2)$ has one prime ideal $(\epsilon)$, we see that $\operatorname{Spec} k[\epsilon]/(\epsilon^2)$ is a single point, so let's call it's image $x$. Then via taking stalks of the sheaf map $\mathcal{O}_Y \to f_*\mathcal{O}_{\operatorname{Spec} k[\epsilon]/(\epsilon^2)}$ at the point $x$, we get a map $\mathcal{O}_{Y,x} \to k[\epsilon]/(\epsilon^2)$, and we see that paragraph 2 applies.