# Exercise II 2.8 on Hartshorne's Algebraic Geometry

There are many people asking about exercises on this book. I've tried to check existing similar questions as many as I can, but I can't promise to have read everything. Sorry if this is duplicated.

Let $X$ be a scheme. For any point $x \in X$, we define the Zariski tangent space $T_x$ to $X$ at $x$ to be the dual of the $k(x)$-vector space $\mathfrak m_x/\mathfrak m_x^2$. Now assume that $X$ is a scheme over a field $k$, and let $k[\varepsilon]/\varepsilon^2$ be the ring of dual numbers over $k$. Show that to give a $k$-morphism of $\text{Spec} k[\varepsilon]/\varepsilon^2$ to $X$ is equivalent to giving a point $x \in X$, rational over $k$ (i.e., such that $k(x) =k$), and an element of $T_x$.

My Attempts: Suppose that $x \in X$ and $\alpha \in T_x$ is given. I think of determining the map of underlying topological spaces $f: \text{Spec} k[\varepsilon]/\varepsilon^2 \rightarrow X$. $k[\varepsilon]$ is a PID, so is $k[\varepsilon]/\varepsilon^2$. The nonzero ideals of $k[\varepsilon]/\varepsilon^2$ are of the form $(\varepsilon+a)$ for $a \in k$. (Right?) Some $(\varepsilon+a) = f^{-1}(x)$. Then $$f^{\sharp}: \mathcal O_x \rightarrow (k[\varepsilon]/\varepsilon^2)_{(\varepsilon+a)}.$$ Also $$\mathfrak m_x \rightarrow (\varepsilon+a)\text{ (in } k[\varepsilon]/\varepsilon^2 \text{)}$$ $$\mathfrak m_x/\mathfrak m_x^2 \rightarrow (\varepsilon+a)/(\varepsilon+a)^2 \text{ (in } k[\varepsilon]/\varepsilon^2 \text{)}.$$ Even if this map is given by $\alpha \in T_x$, how can we get the whole $f$? In the other direction, when the $k$-morphism $f: \text{Spec}k[\varepsilon]/\varepsilon^2 \rightarrow X$ is given, how can it determine $x \in X$ and some $\alpha \in T_x$?

• $(\epsilon + a)$ cannot be an ideal in the quotient because it doesn't even contain $(\epsilon^2)$ (unless $a = 0$). The ring of dual numbers defined above is local and hence only one prime ideal.
– user38268
Oct 27, 2013 at 15:04

Denote $$A=k[\epsilon]/\epsilon^2$$ the regular local ring. We can see that $$A$$ has only one prime ideal, namely $$m=\epsilon A$$ the maximal ideal. Each element of $$m$$ is in the form of $$\lambda\epsilon,\lambda\in k$$. Use the isomorphism $$m/m^2\to k,\lambda\epsilon\mapsto \lambda$$ to identify the two. Let $$\eta\in \mathrm{Spec} A$$ be the closed point.
1.Given $$f:\mathrm{Spec} A\to X$$ a morphism, we set $$x=f(\eta)$$, then local homomorphism $$f_{\eta}^{\#}:O_x\to A_m=A$$ induces a $$k-$$morphism of residues fields $$k(x)\to k$$, hence an isomorphism. So, $$x$$ is a rational point. And $$f_{\eta}^{\#}$$ induces $$m_x/m_x^2\to m/m^2=k$$, which is an element of $$T_x$$.
2.Conversely, let $$j:k\to O_x$$ and $$\phi:O_x\to O_x/m_x$$ be the natural maps. As $$x\in X$$ is rational, $$\phi j$$ is an isomorphism with the inverse $$\Phi:O_x/m_x\to k$$. Now given $$\psi:m_x/m_x^2\to k$$ an element of $$T_x$$, we define a ring morphism $$O_x\to A,h\mapsto \Phi\phi(h)+\psi(h-j\Phi\phi(a))\epsilon$$, and this gives us $$\mathrm{Spec} A\to \mathrm{Spec} O_x$$. There's a natural morphism $$\mathrm{Spec} O_x\to X$$, thus by composition we get $$\mathrm{Spec} A\to X$$.