admissible submodule and direct summand
The following is a terminology from "Linear Algebra" by K. Hoffman and R. Kunze. Let $R$ be a commutative ring with unity and $V$ be an $R$-module. An $R$-submodule $W\subset V$ is called admissible if for any $v\in V$ and $r\in R$ with $rv\in W$, then there exists an element $w\in W$ such that $rv=rw$.
The above terminology is defined in K. Hoffman and R. Kunze, section 7.2, in the special case when $R=F[T]$, a polynomial ring of one variable over a field $F$, and $V$ is a finite dimensional vector space over $F$. It is proved that in this setting, $W\subset V$ is admissible if and only if $W$ is $R$-module summand of $V$, i.e., iff there exists a $R$-module $W'\subset W$ such that $V=W\oplus W'$.
Consider the following condition for $R$-modules $W\subset V$, $(*)$: there exists a $R$-homomorphism $s:V\to W$ such that $s(w)=w$ for $w\in W$. For commutative ring $R$ with unity, $(*)$ is equivalent to $W$ is a direct summand of $V$.
Apparently, $(*)$ implies that $W\subset V$ is admissible, because if $rv\in W$ for $r\in R,v\in V$, then $rv=r w$ for $w=s(v)\in W$. What is proved in Hoffman and Kunze is, in the situation when $R=F[T]$ and $V$ is finite dimensional, then admissibility and $(*)$ are equivalent.
My question is: are $(*)$ and admissibility equivalent in general? If no, is there any other situation such that they are equivalent?
Thanks.
Any pure submodule is "admissible". In fact, if $R$ is a principal ideal domain, the two notions are equivalent.
One source of pure submodules that are not direct summands is that if $$0\to L\to M\to N\to0$$ is a short exact sequence with $N$ a flat module, then $L$ is a pure submodule of $M$. So, for example, taking $R=\mathbb{Z}$, if $$0\to P_1\to P_0\to\mathbb{Q}\to0$$ is a projective resolution of $\mathbb{Q}$, then $P_1$ is a pure subgroup of $P_0$, but not a direct summand, since $\mathbb{Q}$ is not projective, so the sequence cannot split.
A situation where "pure" is equivalent to $(*)$ is when $M/L$ is a finitely presented module. Then $L$ is a pure submodule of $M$ if and only if it is a direct summand. So if $R$ is a principal ideal domain and $M/L$ is finitely generated, then $L$ is an "admissible" submodule of $M$ if and only if it is a direct summand.