# Zariski Tangent Space and $k[\varepsilon]/\varepsilon^2$

Let $X$ be a scheme over a field $k$, and let $x\in X$ be a rational point, that is, we have $k(x):=\mathcal{O}_x/\mathfrak{m}_x\cong k$. Let $\alpha:\mathfrak{m}_x/\mathfrak{m}_x^2\rightarrow k$ be an element of $T_x=(\mathfrak{m}_x/\mathfrak{m}_x^2)^*$, the Zariski tangent space to $X$ at $x$. Using $x$ and $\alpha$, I'd like to define a $k$-morphism of schemes Spec $k[\varepsilon]/\varepsilon^2\rightarrow X$.

In the course of the construction, I need to define a local map of local rings $\mathcal{O}_x\rightarrow k[\varepsilon]/\varepsilon^2$. Using $\alpha$, what is a natural way to define such a map?

It might be helpful to note that I have an injection $k\hookrightarrow\mathcal{O}_x$ arising from the $k$-scheme structure on $X$. I think this leads to a decomposition $\mathcal{O}_x/\mathfrak{m}_x^2=k\oplus(\mathfrak{m}_x/\mathfrak{m}_x^2)$ as $k$ vector spaces, from which the desired map would follow, but I can't seem to explicitly see what this decomposition is inside of $\mathcal{O}_x/\mathfrak{m}_x^2$.

This is part of exercise II.2.8 in Hartshorne.

## 1 Answer

As you note, you have an injection $k \hookrightarrow \mathcal{O}_x$ of rings. Composing with the projection, this produces an injection $k \hookrightarrow \mathcal{O}_x / \mathfrak{m}_x^2$, with the property that further projecting given an isomorphism $k \cong \mathcal{O}_x / \mathfrak{m}_x$ (here you use the assumption that $x$ is $k$-rational). It follows that $\mathcal{O}_x/\mathfrak{m}_x^2 \cong k \bigoplus \mathfrak{m}_x/\mathfrak{m}_x^2$, so that your $\alpha$ gives you the morphism of local rings you need: explicitly, $(x,y) \mapsto x+\alpha(y) \epsilon$.

• What is the ring structure of the decomposition? In order for the map you've given to be a ring map, we would need $(x,y)\cdot(x',y')=(xx',xy'+x'y)$. How can we view the direct sum as internal, that is, given an element $a\in\mathcal{O}_x/\mathfrak{m}_x^2$, what is the unique decomposition $a=x+y$ inside of $\mathcal{O}_x/\mathfrak{m}_x^2$? – Jared Mar 24 '13 at 0:08
• @Jared, Well, each function $a$ has a value $a(x)$ at a $k$-rational point, $a(x) \in k(x)=k$, which by definition is just its class modulo $\mathfrak{m}_x$. The function $a$ can therefore be written $a=a(x)+b$ for a function $b$ which is $0$ at $x$. This is the decomposition you asked for. – Stephen Mar 24 '13 at 17:48
• @Jared, secondly, the ring structure equation you need is immediate from the fact that if $b(x)=0$ then $b^2=0$ in $\mathcal{O}_x / \mathfrak{m}_x^2$. – Stephen Mar 24 '13 at 17:49
• @Jared, basically, you already knew everything, it's just a matter of getting used to working with the definitions. Keep thinking about it, and in another month (or two, at most) you will be comfortable with it. – Stephen Mar 24 '13 at 17:50
• Awesome. Thanks Steve. This is all very helpful. – Jared Mar 24 '13 at 22:21