7 typo

It seems to me the references in this Mathematics - Stack Exchange answer contain the requested information.

EDIT 1. Here is an excerpt from Hyman Bass's book Algebraic K-Theory, W. A. Benjamin (1968), p. 54:

Exercise.

(a) Show that a module $P$ is finitely generated if and only if the union of a totally ordered family of proper submodules of $P$ is a proper submodule.

(b) Show that $\text{Hom}_A(P,\bullet)$ preserves coproducts if and only if the union of every (countable) chain of proper submodules is a proper submodule.

(c) Show that the conditions in (a) and (b) are not equivalent. (Examples are not easy to find.)

EDIT 2. Here is a solution to Exercise (a) above. Let $R$ be an associative ring with $1$, and $A$ an $R$-module. If $A$ is finitely generated, then the union of a totally ordered set of proper submodules is clearly a proper submodule. Let's prove the converse:

Assume that $A$ is not finitely generated. Let $Z$ be the set of those submodules $B$ of $A$ such that $A/B$ is is not finitely generated. The poset $Z$ is nonempty and has no maximal element. By Zorn's Lemma, there is a nonempty totally ordered subset $T$ of $Z$ which has no upper bound. Letting $U$ be the union of $T$, we see that $A/U$ is finitely generated. There is thus a finitely generated submodule $F$ of $A$ which generates $A$ modulo $U$. Then the $B+F$, where $B$ runs over $T$, form a totally ordered set of proper submodules whose union is $A$. QED

I'd be most grateful to whoever would post a solution to the other exercises in Bass's list. (I haven't been able to do them.) The following references might help, but I haven't been able to find them online:

• R. Rentschler, Sur les modules M tels que $\text{Hom}(M,-)$ commute avec les sommes directes, C. R. Acad. Sci. Paris Sér. A-B 268 (1969), 930-933. [Update: see Edit 3 below.]

• P.C. Eklof, K.R. Goodearl and J. Trlifaj, Dually slender modules and steady rings, Forum Math. 9 (1997), 61-74.

This paper is available online, but I don't understand it:

• Jan Zemlicka, Classes of dually slender modules, Proc. Algebra Symposium Cluj 2005, 129-137.

EDIT 3.

$\bullet$ Rentschler's paper

R. Rentschler, Sur les modules M tels que $\text{Hom}(M,-)$ commute avec les sommes directes, C. R. Acad. Sci. Paris Sér. A-B 268 (1969), 930-933

is available here in one click, and there in a few clicks. [I'm also giving the second option because it's a trick worth knowing.] Thanks to Stéphanie Jourdan for having found this link!

$\bullet$ Exercise (b) in Bass's list is in fact the easiest. [Sorry for not having realized that earlier.] Here is a solution. --- Let $R$ be an associative ring with $1$, let $A$ be an $R$-module, and let "map" mean "$R$-linear map".

If $A_0\subset A_1\subset\cdots$ is a sequence of proper submodules of $A$ whose union is $A$, then the natural map from $A$ to the direct product of the $A/A_n$ induces a map from $A$ to the direct sum of the $A/A_n$ whose components are all nonzero.

Conversely, let $f$ be a map from $A$ to a direct sum $\oplus_{i\in I}B_i$ of $R$-modules such that the set $S$ of those $i$ in $I$ satisfying $f_i\neq0$ [obvious notation] is is infinite. By choosing a countable subset of $S$ we get a map $g$ from $A$ to a direct sum $\oplus_{n\in \mathbb N}C_n$ of $R$-modules such that $g_n\neq0$ for all $n$. It is easy to check that the $$A_n:=\bigcap_{k > n}\ \ker(g_k),$$ form an increasing sequence of proper submodules of $A$ whose union is $A$.

EDIT 4. [Version of Nov. 26, 2011, UTC.] The following result is implicit in Rentschler's paper, and solves Bass's Exercise (c):

Theorem. Let $T$ be a nonempty ordered set $( * )$ with no maximum. Then there is a domain $A$ which has the following property. If $P$ denotes the poset of proper sub-$A$-modules of the field of fractions of $A$, then there is an increasing $( * )$ map $f:T\to P$ such that $f(T)$ is cofinal in $P$.

$( * )$ Since I'm using references written in French while writing in English (or at least trying to), I adhere strictly to linguistic conventions. In particular:

ordered set = ensemble totalement ordonné,

poset = ensemble ordonné,

increasing = strictement croisssant.

Proof. Let $T_0$ be the ordered set opposite to $T$, let $\mathbb Z^{(T_0)}$ be the free $\mathbb Z$-module over $T_0$ equipped with the lexicographic order. Then $\mathbb Z^{(T_0)}$ is an abelian ordered group (groupe abélien totalement ordonné). By Example 6 in Section V.3.4 of Bourbaki's Algèbre commutative, there is a field $K$ and a surjective valuation $$v:K\to\mathbb Z^{(T_0)}\cup { \infty }.$$ Say that a subset $F$ of $\mathbb Z^{(T_0)}$ is a final segment if $$F\ni x < y\in\mathbb Z^{(T_0)}$$ implies $y\in F$. Attach to each such $F$ the subset $$S(F):=v^{-1}(F)\cup { 0 }$$ of $K$. Then $A:=S(F_0)$, where $F_0$ is the set of nonnegative elements of $\mathbb Z^{(T_0)}$, is a subring of $K$. Moreover, by Proposition 7 in Section V.3.5 of the book quoted above, $F\mapsto S(F)$ is an increasing bijection from the final segments of $\mathbb Z^{(T_0)}$ to the sub-$A$-modules of $K$.

Write $e_{t_0}$ for the basis element of $\mathbb Z^{(T_0)}$ corresponding to $t_0\in T_0$. Then the intervals $$I_{t_0}:=[-e_{t_0},\infty)$$ are cofinal in the set of all proper final segements segments of $\mathbb Z^{(T_0)}$, and we have $I_{t_0}\subset I_{u_0}$ if and only if $t\le u$. [We denote an element $t$ of $T$ by $t_0$ when we view it as an element of $T_0$.]

6 rewrote edit 4

EDIT 4. [Version of Nov. 26, 2011, UTC.] The following result is implicit in Rentschler's paper: , and solves Bass's Exercise (c):

Theorem. Let $T$ be an a nonempty ordered set $( * )$ with no maximum. Then there is a domain $A$ and an increasing $( * )$ map from which has the following property. If $T$ to P$denotes the poset$ ( * ) $of proper sub-$A$-modules of the field of fractions of$A$. This follows immediately from Example 6, Section V.3.4, and Proposition 7, Section V.3.5, in Bourbaki's Algèbre commutative, coupled with Section 5 of Rentschler's paper. --- This also solves Bass's Exercise A$, then there is an increasing $( c). * )$ map $f:T\to P$ such that $f(T)$ is cofinal in $P$.

$( * )$ Since I'm using references written in French while writing in English (or at least trying to), I adhere strictly to linguistic conventions. In particular:

increasing = strictement croisssant.

Proof. Let $T_0$ be the ordered set opposite to $T$, let $\mathbb Z^{(T_0)}$ be the free $\mathbb Z$-module over $T_0$ equipped with the lexicographic order. Then $\mathbb Z^{(T_0)}$ is an abelian ordered group (groupe abélien totalement ordonné). By Example 6 in Section V.3.4 of Bourbaki's Algèbre commutative, there is a field $K$ and a surjective valuation $$v:K\to\mathbb Z^{(T_0)}\cup { \infty }.$$ Say that a subset $F$ of $\mathbb Z^{(T_0)}$ is a final segment if $$F\ni x < y\in\mathbb Z^{(T_0)}$$ implies $y\in F$. Attach to each such $F$ the subset S(F):=v^{-1}(F)\cup { 0 }$$of K. Then A:=S(F_0), where F_0 is the set of nonnegative elements of \mathbb Z^{(T_0)}, is a subring of K. Moreover, by Proposition 7 in Section V.3.5 of the book quoted above, F\mapsto S(F) is an increasing bijection from the final segments of \mathbb Z^{(T_0)} to the sub-A-modules of K. Write e_{t_0} for the basis element of \mathbb Z^{(T_0)} corresponding to t_0\in T_0. Then the intervals$$ are cofinal in the set of all proper final segements of $\mathbb Z^{(T_0)}$, and we have $I_{t_0}\subset I_{u_0}$ if and only if $t\le u$. [We denote an element $t$ of $T$ by $t_0$ when we view it as an element of $T_0$.]

5 typo

EDIT 4. The following result is implicit in Rentschler's paper:

Theorem. Let $T$ be an ordered set $( * )$. Then there is a domain $A$ and an increasing $( * )$ map from $T$ to the poset $( * )$ of sub-$A$-modules of the field of fractions of $A$.

This follows immediately from Example 6, Section V.3.4, and Proposition 7, Section V.3.5, in Bourbaki's Algèbre commutative, coupled with Section 5 of Rentschler's paper. --- This also solves Bass's Exercise (c).

$( * )$ Since I'm using references written in French while writing in English (or at least trying to), I adhere strictly to linguistic conventions. In particular:

ordered set = ensemble totalement ordonné,

poset = ensemble ordonné,

increasing = strictement croisssant.

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