divisible modules

Question

In surveying LMR of T.Y.Lam, I reached a paragraph stating that "when R is a domain every direct sum or direct product of divisible modules is divisible." My question is that "should R is not a domain, might it be possible to have a divisible direct sum ( or product ) of divisible modules?" Not being able to find any (dis-)proof, I so appreciate any one answering my question.

Answer 1

This answer handles direct products, direct sums, direct factors, and quotients.

Short answer: A direct product of (Lam) divisible modules is divisible. A direct sum of (Lam) divisible modules is divisible. A quotient of a (Lam) divisible module need not be divisible. In particular the partial quote of Lam in the question could be misleading (but the full quote in the book is just fine).

My thanks to Ed Enochs for chatting about this in the hall. Here is the longer answer using relative injectivity to organize the statements:

Definition: Let $R$ be a unital, associative ring; let $A_R$, $B_R$, and $C_R$ be right $R$-modules, and let $i:A \to B$ be a $R$-linear map. We say that $C$ is (relatively) $i$-injective if for every $R$-linear map $f:A_R \to C_R$ there is an $R$-linear map $g:B_R \to C_R$ so that $f(a) = g(i(a))$ for all $a \in A$. In other words, if $\newcommand{\Hom}{\operatorname{Hom}}\Hom(i,C) : \Hom(B,C) \to \Hom(A,C)$ is surjective. $C$ is said to be principally injective iff $C$ is relatively $i$-injective for every inclusion $i:aR \to R$ of principal ideals, that is for each $a \in R$.

The following is a general property of relative injectives:

Proposition: (direct products) If $\{ M_j : j \in J \}$ is a family of right $R$-modules, then $\prod M_j$ is $i$-injective (principally injective, respectively) iff every $M_j$ is $i$-injective (principally injective, respectively).

Proof: Suppose $\{ M_j : j \in J \}$ is a family of $i$-injective modules for $i:A\to B$. Let $f:A \to \prod M_i$ be $R$-linear. Then $f(a) = \left( f_j(a) \right)_{j \in J}$ for $R$-linear maps $f_j :A \to M_j$. Since $M_j$ is $i$-injective, there is some $R$-linear function $g_j : B \to M_j$ with $g_j(i(a)) = f(a)$ for all $a \in A$. Define $g:B \to \prod M_i$ by $g(b) = \left(g_j(b)\right)_{j \in J}$ and notice that $g(i(a)) = \left(g_j(i(a))\right)_{j \in J} = \left(f_j(a)\right)_{j \in J} =f(a)$. Hence $\prod M_i$ is $i$-injective. Conversely, if $\prod M_j$ is $i$-injective and $f_k:A \to M_k$ is an $R$-linear map for some single $k \in J$, then define $f$ as before with $f_j=0$. Extend $f$ to $g:B \to \prod M_j$ and define $g_j:B \to M_j$ as the projection of $g$. Clearly $g_k(i(a)) = f_k(a)$, so $M_k$ is $i$-injective. $\square$

The Bass-Papp characterization of noetherian rings motivates the following:

Definition: An $R$-linear map $i:A \to B$ is said to be noetherian if for every family of right $i$-injective $R$-modules $M_j$, the direct sum $\sum M_j$ is $i$-injective.

Theorem: (Bass-Papp) $R$ is right noetherian (all right ideals are finitely generated) iff every inclusion $i:I_R \to R$ of right ideals into the ring is noetherian.

Unfortunately a strong characterization of noetherian maps seems unrelated, and possibly out of reach (for instance, if $i$ is a split monomorphism, then every module is $i$-injective, so $i$ is noetherian regardless of its $A$ and $B$). However, for the original question asked we get a nice result:

Proposition: (direct sums) If $A$ is finitely generated and $i:A \to B$ is $R$-linear, then $i$ is noetherian. In other words, if $\{ M_j : j \in J \}$ is a family of right $R$-modules and $i:A_R \to B_R$ is $R$-linear with $A$ finitely generated, then $\sum M_j$ is $i$-injective iff every $M_j$ is $i$-injective. In particular, a direct sum of principally injective modules is always principally injective.

Proof: If $\sum M_j$ is $i$-injective, then $M_k$ is $i$-injective, since $\sum M_j = (M_k) \times (\sum_{j\neq k} M_j)$. Conversely, suppose $M_j$ are all $i$-injective, and $f : A \to \sum M_j$. If $\{a _k \}$ is a finite generating set of $A$, then $f(a_k) \in \sum M_j$, so $f_j(a_k) = 0$ for all but finitely many $j$, so $f_j =0$ identically for all but finitely many $j \in J$. If $f_j=0$, then set $g_j=0:B \to M_j$. For the finitely many $f_j \neq 0$, we have $M_j$ is $i$-injective, so set $g_j:B \to M_j$ such that $g_j(i(a))=f_j(a)$ for all $a \in A$. Define $g(b) =\left( g_j(b)\right)_{j \in J}$ and check that $g(i(a))=f(a)$ and $g$ does indeed have codomain $\sum M_j$, not just $\prod M_j$. $\square$

Now we look into quotients:

Definition: If $i:A \to B$ is $R$-linear, then we say $i$ is hereditary iff every quotient $M/N$ of every $i$-injective module $M$ (for $N \leq M$) is $i$-injective. We say a right ideal $I_R$ is hereditary iff the inclusion $i:I\to R$ is hereditary.

The following is the definition of injective module (injective relative to monomorphisms):

Lemma: If $i$ is a monomorphism, then every injective module is $i$-injective.

We then have a classification of when $i$-injective modules are closed under quotient modules:

Proposition: (quotients) If $B$ is projective and $i:A \to B$ is a monomorphism, then $i$ is hereditary iff $A$ is projective. A right ideal is projective iff it is hereditary.

Proof: If $A$ is projective, $M$ is $i$-injective, $N \leq M$, and $f:A \to M/N$, then because $A$ is projective, we can find $g: A \to M$ with $g(a) + N = f(a)$. Since $M$ is $i$-injective, we can find $h:B \to M$ with $h(i(a)) = g(a)$. Hence defining $k:B \to M/N$ by $k(b) = h(b)+N$ we get $k(i(a)) = h(i(a)) + N = g(a) + N = f(a)$, so that $k$ extends $f$. Hence $M/N$ is $i$-injective in this case. Conversely, suppose $i$ is hereditary. Let $E_R$ be an injective right $R$-module, $S \leq E$, and $f:A\to E/S$. Since $E$ is injective and $i$ is a monomorphism, $E$ is $i$-injective, and since $i$ is hereditary, $E/S$ is $i$-injective. Hence we can find $g:B \to E/S$ with $g(i(a))=f(a)$. But then $B$ is projective, so we can find $h:B \to E$ with $h(b)+S = g(b)$. Hence we can define $k:A \to E$ by $k(a) = h(i(a))$. Obviously $k$ extends $f$, and so $\Hom(I,E) \to \Hom(I,E/S) \to 0$ is exact. By the definition of Ext, $\operatorname{Ext}^1(I,S) =0$. Since every right module $S_R$ is contained in some injective module $E$, we get that $I$ is projective. $\square$

Theorem: A ring is right hereditary (in the sense that quotient modules of injective modules are injective) iff all of its right ideals are hereditary = projective.

In particular, the ideal $2\mathbb{Z}/4\mathbb{Z}$ is not projective, so not hereditary. Explicitly, $C=\mathbb{Z}/4\mathbb{Z}$ is $2\mathbb{Z}/4\mathbb{Z}$-injective, but its quotient $C/2C$ is not $2\mathbb{Z}/4\mathbb{Z}$-injective.

Domains

For domains, we have a simpler characterization that immediately handles direct sums, direct products, direct factors, and quotients:

Proposition: If $a$ is not a zero divisor, then $\Hom(R,M) \to \Hom(aR,M)\to0$ is exact iff $Ma=M$.

Proof: If $a$ is not a zero-divisor (that is, if $ar=0$ implies $r=0$, I only need the one side), then $h:R \to aR:r \mapsto ar$ is an isomorphism, so $aR$ is free. Hence for every $m \in M$ there is some $f:aR \to M$ with $f(a) = m$. If $\Hom(R,M) \to \Hom(aR,M)\to0$ is exact, then there is some $g:R \to M$ with $g(ar) = f(ar)$. Setting $n = g(1)$, we get $m = f(a) = g(a) = g(1) \cdot a = n a \in Ma$. Hence $M=Ma$. Conversely, if $M=Ma$, then for any $f:aR \to M$, we have $f(a) = m \in M = Ma$, so there is some $n \in M$ with $na=m$. Define $g:R \to M: r \mapsto nr$. Then $g(ar) = n(ar) = (na)r = mr = f(a)r = f(ar)$. Hence $\Hom(R,M) \to \Hom(aR,M)\to0$ is exact. $\square$

Definition: A right $R$-module $M$ is $a$-divisible for $a \in R$ if $M=Ma$.

Proposition: Let $\{M_i : i \in I\}$ be a set of right $R$-modules and let $a \in R$. Then $\prod M_i$ is $a$-divisible iff $\sum M_i$ is $a$-divisible iff each $M_i$ is $a$-divisible.

Proof: This follows simply from $\left( \prod M_i \right)a = \prod (M_ia)$ and $\left(\sum M_i \right)a = \sum M_i a$. $\square$

Proposition: Let $M$ be an $a$-divisible right $R$-module and let $N \leq M$. Then $M/N$ is $a$-divisible.

Proof: As before $(M/N)a = (aM+N)/N$, so if $M$ is $a$-divisible, then $(M/N)a = (Ma+N)/N = (M+N)/N = M/N$. $\square$

Answer 2

You will have to read the text more closely for the definition of "divisible" that Lam uses. He uses a definition that is not very ordinary, but he makes the case for it very well.

If I'm recalling correctly, given a right $R$ module $M$, we say $M$ is divisible if for every $m\in M$ and $x\in R$ with $r.ann(x)\subseteq ann(m)$, there exists $n\in M$ such that $nx=m$.

The annihilator condition is obviously necessary for this to occur, and of course in a domain $r.ann(x)=0$ for every nonzero $x$, and the definition becomes the usual definition of divisibility for domains.

I can't remember all of the justifications given offhand, but I do remember two nice things that work out with this definition: every injective module is divisible, and a ring whose modules are all divisible is von Neumann regular.

I seem to remember it is somehow dual to another natural condition, but I cannot rightly remember what the statement was at the moment.