I am reading the beginning of Sernesi's book "Deformations of algebraic schemes", and I am stuck on an example of a versal extension.
Background: $A\to R$ is a fixed ring homomorphism. By an $A$-extension of $R$ (by $I$) we mean an exact sequence $0\to I\to R'\to R\to 0$, where $R'$ is an $A$-algebra and $I^2=0$. By a morphism of extensions we mean a commutative diagram
where $h$ is an $A$-algebra homomorphism. An $A$-extension is said to be versal if it has a morphism to any other $A$-extension.
In the book, it is claimed the following: if $P$ is a polynomial $A$-algebra and $R=P/I$, then $$0\to I/I^2\to P/I^2\to R\to 0$$ is a versal extension.
Questions: I would like to see why the above exact sequence admits a morphism to any other extension. Furthermore, it is also claimed that such a $P$ always exists (for a fixed $A$-algebra $R$). Why is that true? It seems that any $A$-albegra is the quotient of a polynomial $A$-algebra.
Thank you for any help.
First, I don't think it is necessary to assume that $R$ is finitely generated over $A$. Just let $P = A[\{x_i\}_{i\in \Lambda}]$ be a polynomial ring over any index set $\Lambda$ such that $R$ can be generated by elements $\{r_i\}_{i\in \Lambda}$, and map $x_i\mapsto r_i$. Denote the kernel by $I$. Let $\pi\colon P \to R$ and $\overline\pi\colon P/I^2 \to R$ be the quotient maps.
Let $0\to J\to R''\to R\to 0$ be any extension as in your diagram. Since $\psi$ is surjective, for any $i\in \Lambda$ there is some $r_i''\in R''$ such that $\psi(r_i'') = r_i = \pi(x_i) = \overline\pi(\overline x_i)$. Make some choice of $r_i''$'s, and define $\tilde h\colon P\to R''$ by $x_i\mapsto r_i''$. Since $P$ is polynomial, this determines a homomorphism of $A$-algebras. To get our $h$, we want to show that $\tilde h(I^2) = (0)$.
Note that $\pi = \psi\circ \tilde h$ by construction, since this holds for all $x_i$. Let $a\in I$. Then we have $0 = \pi(a) = \psi\circ \tilde h(a)$, which implies $\tilde h(a)\in \operatorname{Ker}(\psi) = J$. Hence, if $a\in I^2$, then $\tilde h(a)\in J^2 = (0)$, which shows that $\tilde h(I^2) = (0)$. Hence $\tilde h$ factors through a map $h\colon P/I^2\to R''$ satisfying $h(I)\subseteq J$.
Note also that this homomorphism $h$ is possibly non-unique.