On his book Commutative Algebra, Matsumura defines the notion of an extension of a ring $C'$ by a $C'$-module $N$ (the definition is here). In (25.B), he defines the trivial extension $C'*N$ and then he claims that an extension $(C,\varepsilon,i)$ of $C'$ by $N$ is trivial if and only if $\varepsilon$ has a section that is a ring homomorphism.
On the one hand, it is easy to verify that the section $C'\to C'*N,c'\mapsto(c',0)$ is a ring homomorphism. But I am having trouble proving the other implication. Here are my thoughts so far: If $s:C\to C'$ is a ring homomorphism such that $\varepsilon\circ s=\text{id}_C$, then the map \begin{align*}\varphi:C&\to C'*N\\ c&\mapsto(\varepsilon(c),c-s(\varepsilon(c'))) \end{align*} is an isomorphism of abelian groups that sends $1_{C'}$ to $1_{C'*N}$. Hence, it suffices to check multiplicativity of $\varphi$.
But this is what I am having trouble with. How does one proceed?
We have $\varphi(cd)=(\varepsilon(cd),cd-s\varepsilon(c)s\varepsilon(d))$ and \begin{align*} \varphi(c)\varphi(d) &=(\varepsilon(c),c-s\varepsilon(c))(\varepsilon(d),d-s\varepsilon(d))\\ &=(\varepsilon(cd),c(d-s\varepsilon(d))+d(c-s\varepsilon(c)))\\ &=(\varepsilon(cd),2cd-s\varepsilon(d)c-s\varepsilon(c)d). \end{align*} (In the second equality, we have used the $C/i(N)\cong C'$-module structure of $i(N)$.) So $\varphi(cd)=\varphi(c)\varphi(d)$ if and only if $0=cd+s\varepsilon(c)s\varepsilon(d)-s\varepsilon(d)c-s\varepsilon(c)d$. Indeed, \begin{align*} cd+s\varepsilon(c)s\varepsilon(d)-s\varepsilon(d)c-s\varepsilon(c)d &=(c-s\varepsilon(c))d-(c-s\varepsilon(d))s\varepsilon(d)\\ &=-(c-s\varepsilon(c))(d-s\varepsilon(d)), \end{align*} which is a product of two elements of $i(N)$, hence zero, since $i(N)^2=(0)$.