= Solution
Let $k\to A$ be a map of <commutative rings>. The <Kähler differentials> $\Omega_{A/k}$ are generated as an $A$-<module> by symbols $da$, with relations
$$
d(a+b)=da+db,\qquad d(ab)=a\,db+b\,da,\qquad dc=0\quad(c\in k).
$$
These relations make $d:A\to\Omega_{A/k}$ the universal $k$-<derivation>. Every $k$-<derivation> $D:A\to M$ into an $A$-module factors uniquely through the map $da\mapsto D(a)$. Thus the <universal property of Kähler differentials> is
$$
\operatorname{Hom}_A(\Omega_{A/k},M)\cong\operatorname{Der}_k(A,M).
$$
This constructs the module and proves its universal property, rather than merely listing a formal derivative rule.
For the <polynomial ring> $P=k[X_i:i\in T]$, the <Kähler differentials of a polynomial algebra> form the free module $\bigoplus_{i\in T}P\,dX_i$. The usual formal partial derivatives prove that assigning arbitrary images to $dX_i$ defines a <derivation>, and every polynomial involves only finitely many variables. For $A=P/I$, the <Conormal exact sequence for Kähler differentials> gives
$$
I/I^2\longrightarrow\Omega_{P/k}\otimes_PA\longrightarrow\Omega_{A/k}\longrightarrow0,
\qquad [f]\longmapsto df.
$$
The map is well-defined because $d(I^2)$ vanishes after reduction modulo $I$. Its cokernel has exactly the universal property of <derivations> on $P$ that kill $I$, so is $\Omega_{A/k}$. In a finite polynomial presentation this gives
$$
\Omega_{A/k}\cong
\frac{\bigoplus_iA\,dX_i}{\left\langle\sum_i\frac{\partial f}{\partial X_i}dX_i:f\in I\right\rangle}.
$$
There need not be injectivity at $I/I^2$: in characteristic $p$, the relation $X^p$ has derivative zero.
<Localization of Kähler differentials> commutes with <localization of a ring>:
$$
\Omega_{S^{-1}A/k}\cong S^{-1}\Omega_{A/k},\qquad
d(a/s)=\frac{s\,da-a\,ds}{s^2}.
$$
The quotient rule follows by differentiating $s\cdot s^{-1}=1$. It extends every derivation uniquely and proves the isomorphism by the universal property. Likewise <base change for Kähler differentials> gives $\Omega_{(A\otimes_k k')/k'}\cong\Omega_{A/k}\otimes_k k'$: a $k'$-linear derivation is determined by its values on $A\otimes1$, and the product rule extends those values to the tensor product.
For a tower $k\to F\to L$, the <Transitivity exact sequence for Kähler differentials> is
$$
L\otimes_F\Omega_{F/k}\longrightarrow\Omega_{L/k}\longrightarrow\Omega_{L/F}\longrightarrow0.
$$
Quotienting $\Omega_{L/k}$ by the submodule generated by differentials of elements of $F$ represents precisely the $F$-derivations, which proves exactness. The first map need not be injective in general. To relate this to a <transcendence basis>, we need the stronger property supplied by a <separable field extension>.
If $L/F$ is finite separable and $u$ has minimal polynomial $g(T)=T^r+c_{r-1}T^{r-1}+\cdots+c_0$, differentiating its equation forces
$$
D(u)=-\frac{\sum_{i=0}^{r-1}u^iD(c_i)}{g'(u)}.
$$
The denominator is nonzero by separability. Conversely this formula extends an arbitrary $k$-<derivation> $F\to M$, for an $L$-<module> $M$, to $F[u]$; it kills the relation $g$ and hence descends to $L$. A tower of simple separable extensions proves unique extension for all finite separable $L/F$. Therefore <Kähler differentials under a separable field extension> satisfy
$$
\boxed{\Omega_{L/F}=0,\qquad \Omega_{L/k}\cong L\otimes_F\Omega_{F/k}.}
$$
This proves injectivity in this case, which would not follow from right exactness alone.
A <transcendence basis> $t_1,\ldots,t_d$ for a finitely generated field extension $L/k$ has <algebraic independence> over $k$ and makes $L/k(t_1,\ldots,t_d)$ algebraic, hence finite. A <separating transcendence basis> additionally makes that finite extension separable. Apply the polynomial computation and the quotient rule to the <rational function field> $F=k(t_1,\ldots,t_d)$, and then the separable-extension isomorphism. We obtain the central link:
$$
\boxed{\Omega_{L/k}=\bigoplus_{i=1}^dL\,dt_i,
\qquad\dim_L\Omega_{L/k}=\operatorname{trdeg}_kL}
$$
when the basis is separating. The dual statement says that any prescribed values of the $D(t_i)$ extend uniquely to a $k$-<derivation> of $L$ with values in $L$. In characteristic zero every transcendence basis of a finitely generated field extension is separating, so <Kähler differentials> measure <transcendence degree> exactly.
There is also a characteristic-zero test for <algebraic independence>. If $t_1,\ldots,t_r$ are algebraically dependent, choose a nonzero polynomial relation of minimum total degree. Some formal partial derivative is nonzero in characteristic zero, and it cannot also vanish on the tuple, since it has smaller degree. Differentiation therefore gives a nontrivial linear relation among the $dt_i$. Conversely, an algebraically independent tuple extends to a transcendence basis, whose differentials form a basis as just proved. Thus <differentials detect algebraic independence in characteristic zero>: a finite tuple is algebraically independent exactly when its differentials are linearly independent. Its differentials form a basis of $\Omega_{L/k}$ exactly when the tuple is a transcendence basis.
The separability qualification is essential in positive characteristic. For $k=\mathbb F_p(s)$ and $L=k(u)$ with $u^p=s$, the extension is purely inseparable of degree $p$, with <transcendence degree> zero. The polynomial presentation yields $\Omega_{L/k}=L\,du\ne0$, because the defining relation has derivative zero. Also, in $\mathbb F_p(t)/\mathbb F_p$, the tuple $\{t^p\}$ is a transcendence basis but $d(t^p)=0$, while $dt$ is a basis of the one-dimensional differential module. A chosen arbitrary transcendence basis therefore need not give differential coordinates in characteristic $p$.
Finally, the presentation makes the relation useful geometrically. For $A=k[x,y]/(y^2-x^3)$ in characteristic zero,
$$
\Omega_{A/k}=(A\,dx\oplus A\,dy)/(2y\,dy-3x^2\,dx).
$$
After passage to the <fraction field>, it has dimension one, the <transcendence degree> of the curve. At the origin, tensoring with its residue field leaves both $dx,dy$ independent, so the differential fibre has dimension two. For a $k$-rational point with maximal ideal $\mathfrak m$, that fibre is the <cotangent space of a local ring> $\mathfrak m/\mathfrak m^2$: write elements as their constant value plus an element of $\mathfrak m$, and note that derivations into $k$ kill $\mathfrak m^2$. This explains how <Kähler differentials> record both generic transcendental parameters and the extra tangent direction at a singular point.
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