Polynomial differentials are free

Sections of a square-zero thickening. Let $E(M)$ be the commutative ring whose underlying abelian group is $P\oplus M$ with product $(p,m)(p^{\mathrm{\prime }},m^{\mathrm{\prime }})=(p{p}^{\mathrm{\prime }},p{m}^{\mathrm{\prime }}+p^{\mathrm{\prime }}m)$, made into an $A$-algebra by $a\mapsto (\phi (a),0)$. The first projection $\pi \text{\hspace{0.17em}⁣}:E(M)\to P$ is an $A$-algebra homomorphism with kernel the square-zero ideal $M$. If $s\text{\hspace{0.17em}⁣}:P\to E(M)$ is an $A$-algebra homomorphism with $\pi \circ s={\mathrm{i}\mathrm{d}}_P$, write $s(f)=(f,D_s(f))$; additivity of $s$ gives $D_s(f+g)=D_s(f)+D_s(g)$, the identity $\pi \circ s=\mathrm{i}\mathrm{d}$ and $A$-linearity give $D_s(\phi (a))=0$, and multiplicativity $s(fg)=s(f)s(g)$, expanded with $m{m}^{\mathrm{\prime }}=0$ in $E(M)$, gives $D_s(fg)=f{D}_s(g)+g{D}_s(f)$; conversely these three laws make the formula $s(f)=(f,D_s(f))$ multiplicative and unital. So sections of $\pi$ over the identity correspond bijectively to the elements of ${\mathrm{Der}}_A(P,M)$ by [F2].

Every tuple of values is realised. Let $m_1,\dots ,m_n\in M$. By [F3] applied to $\phi \text{\hspace{0.17em}⁣}:A\to E(M)$ and the family $(x_i,m_i)\in E(M)$, $i=1,\dots ,n$, there is a unique $A$-algebra homomorphism $\theta \text{\hspace{0.17em}⁣}:P\to E(M)$ with $\theta (x_i)=(x_i,m_i)$ and $\theta (\phi (a))=(\phi (a),0)$. The composite $\pi \circ \theta \text{\hspace{0.17em}⁣}:P\to P$ is an $A$-algebra endomorphism of $P$ with $x_i\mapsto x_i$, so by the uniqueness clause of [F3] it is the identity; hence $\theta (f)=(f,D(f))$ for the map $D\text{\hspace{0.17em}⁣}:P\to M$ given by the second coordinate, and $D(x_i)=m_i$. By step 1.1 the map $D$ is an $A$-derivation of $P$ into $M$.

Uniqueness of the values on generators. If $D^{\mathrm{\prime }}\in {\mathrm{Der}}_A(P,M)$ satisfies $D^{\mathrm{\prime }}(x_i)=m_i$ for all $i$, then $f\mapsto (f,D^{\mathrm{\prime }}(f))$ is an $A$-algebra homomorphism $P\to E(M)$ by step 1.1, it agrees with $\theta$ on $A$ and on each $x_i$, and hence equals $\theta$ by the uniqueness clause of [F3]; therefore $D^{\mathrm{\prime }}=D$. So for every $P$-module $M$ the evaluation map ${\mathrm{Der}}_A(P,M)\to M^n$, $D\mapsto (D(x_1),\dots ,D(x_n))$, is a bijection; it is $P$-linear, and natural in $M$ because a $P$-linear $t\text{\hspace{0.17em}⁣}:M\to N$ sends $D$ to $t\circ D$ with values $t(D(x_i))$.

Freeness. Composing the natural bijections of [F1] and of step 3.1 gives natural bijections ${\mathrm{Hom}}_P({\mathrm{\Omega }}_{P\mathrm{/}A},M)\cong M^n\cong {\mathrm{Hom}}_P(P^n,M)$ for every $P$-module $M$. The image of the identity of $P^n$ is a $P$-linear map $\phi \text{\hspace{0.17em}⁣}:{\mathrm{\Omega }}_{P\mathrm{/}A}\to P^n$, and its inverse image is a $P$-linear map $\psi \text{\hspace{0.17em}⁣}:P^n\to {\mathrm{\Omega }}_{P\mathrm{/}A}$ with $\phi \circ \psi =\mathrm{i}\mathrm{d}$ and $\psi \circ \phi =\mathrm{i}\mathrm{d}$: both identities are checked on generating sets, the standard basis of $P^n$ and, by the explicit construction of the bijection in step 3.1, the elements $\mathrm{d}{x}_i$. Hence $\psi$ is an isomorphism sending $e_i$ to $\mathrm{d}{x}_i$, so ${\mathrm{\Omega }}_{P\mathrm{/}A}$ is free with basis $\mathrm{d}{x}_1,\dots ,\mathrm{d}{x}_n$. For $n=0$ we have $P=A$ and $M^0=0$, so step 3.1 says that every $A$-derivation of $A$ into any $A$-module is zero, and [F1] gives ${\mathrm{\Omega }}_{A\mathrm{/}A}=0$.