Question about Kähler differentials and derivations

Question

Let $x\in X$ be a closed point of a scheme $X$ of finite type over $k$. Then it's said that the following $k$-vector spaces are canonically isomorphic:

1) the tangent space $T=\operatorname{Hom}_k(m_x/m^2_x,k)$ to $X$ at $x$, where $m_x$ is the maximal ideal of local ring $\mathcal{O}_x$ ;

2) the space of point derivations $D$: $\mathcal{O}_x\to k$, i.e., $k$-linear maps that $D(fg)=f(x)Dg+g(x)Df$;

3) $\operatorname{Hom}_{\mathcal{O}_x}((\Omega _{X/k})_x, \mathbb{k}(x))$

According to proof, one should notice every point derivation $D$ kills $m^2_x$, and therefore induces a linear functional $l: m_x/m^2_x \to k$. Conversely, given any such $l$, define $D$ by $D(f)=l[f-f(x)]$ gives a point derivation.

I don't know why derivation always kills $m^2_x$, I can't see it follows from definitions. And conversely, the derivation defined above seems not satisfy the Leibniz Rule. Maybe something wrong with my calculation. Also, about the Kalher differetial, it's said that the square of elements in $\Omega _{B/A}$ is zero. I don't know why this follows too. Hope someone can help. Thanks!

Answer 1

Let $A = \mathcal{O}_{X,x},$ and let $\mathfrak{m} = m_x.$ Then $\left(\Omega_{X/k}\right)_x\cong\Omega^1_{A/k}.$ So to simplify notation, I will use $(A,\mathfrak{m})$ instead of $(\mathcal{O}_{X,x},m_x).$ In the following, let $i : k\to A$ and $\pi : A\to A/\mathfrak{m}\cong k$ be the natural structure maps. Note in particular that $\pi\circ i = \mathrm{id}_k,$ and that if $f\in A = \mathcal{O}_{X,x},$ then $f(x) = \pi(f),$ and by $f - f(x)$ we mean $f - i\circ\pi(f).$

1. To see that a derivation kills $\mathfrak{m}^2,$ let $m,n\in\mathfrak{m}.$ Then $D(mn) = m\cdot Dn + n\cdot Dm,$ but $Dn,Dm\in k.$ Let $a\in A$ and $b\in k;$ then the natural $A$-module structure on $k$ is given by $a\cdot b := \pi(a)b.$ If $a\in\mathfrak{m},$ this tells us that $a\cdot b = 0,$ so that $$D(mn) = m\cdot Dn + n\cdot Dm = \pi(m)Dn + \pi(n) Dn = 0 + 0 = 0.$$
2. Let's check that the Leibniz rule holds. First, observe that if $a\in A,$ then $a - i\circ\pi(a)\in\mathfrak{m}.$ We simply need to check that $\pi(a - i\circ\pi(a)) = 0,$ and indeed $$ \pi(a - i\circ\pi(a)) = \pi(a) - \pi\circ i\circ\pi(a) = \pi(a) - \mathrm{id}_k\circ\pi(a) = \pi(a) - \pi(a) = 0. $$ Now, if $a,b\in A$ and $\phi : \mathfrak{m}/\mathfrak{m}^2\to k$ is a $k$-linear map then \begin{align*} D(ab) &= \phi(ab - i\circ\pi(ab) + \mathfrak{m}^2)\\ &= \phi(ab + i\circ\pi(a)b - i\circ\pi(a)b - i\circ\pi(ab)+ \mathfrak{m}^2)\\ &= \phi(ab - i\circ\pi(a)b+ \mathfrak{m}^2) + \phi(i\circ\pi(a)b - i\circ\pi(ab)+ \mathfrak{m}^2)\\ &= \phi(b(a - i\circ\pi(a))+ \mathfrak{m}^2) + \phi(i\circ\pi(a)b - (i\circ\pi(a))(i\circ\pi(b))+ \mathfrak{m}^2)\\ &= \phi(b(a - i\circ\pi(a))+ \mathfrak{m}^2) + \phi(i\circ\pi(a)(b - i\circ\pi(b))+ \mathfrak{m}^2)\\ &= \pi(b)\phi(a -i\circ\pi(a)+ \mathfrak{m}^2) + \pi(a)\phi(b - i\circ\pi(b)+ \mathfrak{m}^2)\\ &= b\cdot Da + a\cdot Db. \end{align*} To obtain the second to last equality, we observe that if $b\in A$ and $m\in\mathfrak{m},$ then $(b - i\circ\pi(b))m\in\mathfrak{m}^2.$ Therefore, \begin{align*} \phi(bm + \mathfrak{m}^2) &= \phi(bm + i\circ\pi(b)m - i\circ\pi(b)m + \mathfrak{m}^2)\\ &= \phi(i\circ\pi(b)m + (b - i\circ\pi(b))m + \mathfrak{m}^2)\\ &= \phi(i\circ\pi(b)m + \mathfrak{m}^2)\\ &= \pi(b)\phi(m + \mathfrak{m}^2). \end{align*}

Before I answer your last question, let me make a general construction. Let $M$ be a $B$-module. We define the ring $B[M]$ to be $B\oplus M$ as a $B$-module, with ring structure given by $$(b_1,m_1)(b_2,m_2) := (b_1b_2, b_1m_2 + b_2 m_1).$$ Note in particular that elements of $M$ in $B[M]$ have square zero: $$(0,m)(0,m) = (0,0\cdot m + 0\cdot m) = (0,0).$$ So, the ring $B[M]$ is a generalization of the dual numbers where elements of the module $M$ have square zero. (To recover the dual numbers, let $B = M = k.$ You can even think of elements of $B[M]$ as being elements of the form $b + m\epsilon,$ where $b\in B,$ $m\in M,$ and $\epsilon^2 = 0$.)

There is a natural surjection \begin{align*} \pi_M : B[M]&\to B\\ (b,m)&\mapsto b \end{align*} In fact, if $B$ is an $A$-algebra, then $A$-algebra maps $f : B\to B[M]$ such that $\pi_M\circ f = \mathrm{id}_B$ are in bijection with $A$-derivations $D : B\to M.$ I leave it as an exercise to you to check that (1) if $f(b) = (b, D(b))$ is such morphism of $A$-algebras, then $D : B\to M$ is an $A$-derivation, (2) that any $A$-derivation $D : B\to M$ induces an $A$-algebra morphism $f_D : B\to B[M]$ given by $f_D(b) = (b,D(b)),$ and (3) that this correspondence is indeed a bijection.

3. To answer your last question, note that $\Omega^1_{B/A}$ is only a $B$-module, not a $B$-algebra. When Mumford talks about elements of $\Omega^1_{B/A}$ having square zero (page 143 in my edition, Theorem 4 chapter III section 1), he is talking about these elements having square zero in the ring $B[\Omega^1_{B/A}].$ He writes:

   To get a map backwards, we define a ring $R = B\oplus\Omega_{B/A},$ where $B$ acts on $\Omega_{B/A}$ through the module action and the elements of $\Omega_{B/A}$ have square zero.

$\quad\,\,\,\,$This is no more nor less than the construction I just outlined of $B[\Omega^1_{B/A}]$!