Pathologies in module theory

Question

Linear algebra is a very well-behaved part of mathematics. Soon after you have mastered the basics you got a good feeling for what kind of statements should be true -- even if you are not familiar with all major results and counterexamples.

If one replaces the underlying field by a ring, and therefore looks at modules, things become more tricky. Many pathologies occur that one maybe would not expect coming from linear algebra.

I am looking for a list of such pathologies where modules behave differently than vector spaces. This list should not only be a list of statements but all phenomena should be illustrated by an example.

To start the list I will post an answer below with all pathologies that I know from the top of my head. This should also explain better what kind of list I have in mind.

Answer 1

My list of pathologies:

• A submodule of a finitely generated module does not have to be finitely generated.

Example: Let $K$ be a field and $R = K[x_1, x_2, \ldots]$ be the ring of polynomials in infinitely many variables over $K$. $R$ considered as an $R$-module is generated by $f(x_1, x_2, \ldots) = 1$, i.e., it is finitely generated. Now, let $S = \langle x_1, x_2, \ldots \rangle$ be the submodule of all elements with zero constant term. Assume that $S$ is finitely generated, say, $S = \langle f_1, f_2, \ldots, f_n \rangle$. Since we have infinitely many variables, there exits $x_k$ that does not occur as a variable of one of the $f_i$. We get $x_k = \sum_{i=1}^n g_i(X) f_i(X)$. Rewrite $g_i(X)$ as $g_i(X) = x_k q_i(X) + r_i(X)$ where $r_i(X)$ does not involve $x_k$. This gives $x_k = \sum_{i=1}^n (x_k q_i(X) + r_i(X)) f_i(X) = x_k \sum_{i=1}^n q_i(X)f_i(X) + \sum_{i=1}^n r_i(X) f_i(X)$. The last sum does not involve $x_k$ and so it must be $0$. Hence, the first sum must equal $1$, which is not possible since $f_i(X)$ has no constant term.

• The following statement is false for modules: $S$ is a linearly dependent set $\Rightarrow$ some element in $S$ is a linear combination of the other elements in $S$.

Example: Consider $\mathbb{Z}$ as a $\mathbb{Z}$-module. Then $2, 3 \in \mathbb{Z}$ are linearly dependent, since $3 \cdot 2 - 2 \cdot 3 = 0$. But neither is a linear combination of the other.

• Not every module has a basis.

Example: It is even worse. There are modules without any non-empty linearly independent sets. For example, consider $\mathbb{Z}_n$ as a $\mathbb{Z}$-module. Since for every element $a \in \mathbb{Z}_n$ we have $na = 0$, no singleton set is linearly independent.

• A submodule of a free module (module with a basis) does not have to be free.

Example: $\mathbb{Z} \times \mathbb{Z}$ is a free module over itself using componentwise scalar multiplication. It has the basis $(1,1)$ but the submodule $\mathbb{Z} \times \{ 0 \}$ is not free.

• A quotient module of a free module does not have to be free.

Example: $\mathbb{Z}$ as a module over itself is free on the set $\{1\}$. For any $n > 0$, the set $n\mathbb{Z}$ is a free cyclic submodule of $\mathbb{Z}$, but the quotient $\mathbb{Z}$-module $\mathbb{Z}_n$ is not free (see above) unless $n=1$.

• A submodule of a module need not have a complement.

Example: Again, consider $\mathbb{Z}$ as a module over itself. All submodules are of the form $n\mathbb{Z}$. We have $n \mathbb{Z} \cap m \mathbb{Z} = \mbox{lcm}(n,m) \mathbb{Z}$. So the only complemented submodules of $\mathbb{Z}$ are $\mathbb{Z}$ and $\{ 0 \}$.

Answer 2

• A module can have a nonzero annihilator. This overlaps a little with the torsion module example you gave above. If $R/I$ is a non-trivial quotient of a commutative ring, $R$, then it is an $R$ module with annihlator $I\neq 0$.

• A module can be Artinian but not Noetherian, and it can be Noetherian but not Artinian. Any non-Artinian but Noetherian ring considered as a module over itself is an example of the latter, and $\mathbb{Z}(p^\infty)$ for a prime $p$ is Artinian but not Noetherian, since $(1/p)\supset (1/p^2)\supset\dots $. For vector spaces, these two conditions are equivalent.

• Somewhat related to the above point is that modules need not have maximal submodules, nor must they have minimal submodules. Any nonfield domain considered as a module over itself has no minimal submodules. The $\mathbb{Z}$ module $\mathbb{Q}$ has no maximal submodules.

• (This is kind of generalizing a few of your points.) For division rings, every module is free (that's basically saying that every module has a basis) and every module is injective. But in the general case, modules don't have to be projective or injective. For example, any commutative domain which is not a field considered as a module over itself is not an injective module (but yes, it is free :) ) and in such a domain, no quotient by a proper ideal can be a projective module.

• Modules can be directly irreducible without being simple. For example, any commutative domain which is not a field, considered as a module over itself, cannot be written as a direct product of two ideals, but it has proper ideals.

• Over any vector space $V$ with endomorphism $f:V\rightarrow V$, there exists another endomorphism $g$ such that $f=fgf$. This is because the endomorphism ring of $V$ is always a von Neumann regular ring. Modules in general, however, do not have this property in their endomorphism rings. Again, any nonfield domain considered as a module over itself is a module whose endomorphism ring is not von Neumann regular.

• Given any two isomorphic finite dimensional subspaces of a vector space $V$, the isomorphism can be extended to an automorphism of $V$. This example would be hard to explain, but in short, a ring which does this for its modules is quasi-Frobenius, so any non-quasi-Frobenius ring has such a module. Again, nonfield domains are not quasi-Frobenius.

• A maximal linearly independent subset needs not be a basis: consider $2\mathbb Z$ in $\mathbb Z$ as a $\mathbb Z$ module.

• A minimal spanning set needs not be a basis: consider that $2$ and $3$ generate $\mathbb Z$ minimally, but they are not $\mathbb Z$ independent.

Answer 3

I am surprised that it is not mentioned here-

Example of a free module M which has bases having different cardinalities.

Let $V$ be a vector space of countably infinite dimension over a division ring $D$. Let $R=End_D(V)$. We know that $R$ is free over $R$ with basis $\{1\}$. We claim that given a positive integer $n$, there is a $R$-basis $B_n=\{f_1,f_2, \dots f_n\}$ for $R$ having $n$ elements.

Let $B=\{e_k\}_{k=1}^{\infty}$ be a basis of $V$ over $D$. Define $\{f_1, \dots , f_n\}\in R$ by specifying their values on $B$ as in the following table-

\begin{array}{|c| ccccc|} \hline & f_1 & f_2 & f_3 & \dots & f_n \\ \hline e_1& e_1&0 &0 & \dots & 0\\ e_2& 0 & e_1 &0 & \dots & 0\\ \vdots& & & & \ddots\\ e_n & 0 &0 &0 & \dots & e_1 \\ \hline e_{n+1} & e_2 &0 &0 & \dots & 0 \\ e_{n+2} & 0 &e_2 &0 &\dots &0 \\ \vdots & & & &\ddots & \\ e_{2n} & 0 &0 &0 & \dots & e_2 \\ \hline \vdots & \vdots & \vdots & \vdots& \vdots & \vdots \\ \hline e_{kn+1} & e_{k+1} & 0 & 0 & \dots & 0 \\ e_{kn+2} & 0 & e_{k+1}&0& \dots &0 \\ \vdots & &&&\ddots & \\ e_{(k+1)n} &0&0&0& \dots& e_{k+1} \\ \hline \vdots & \vdots & \vdots & \vdots& \vdots & \vdots \\ \end{array}

Now we check that $B_n$ is an $R$- basis of $R$.

• Linearly independent-

If $\sum_{i=1}^{n} \alpha_i f_i=0$ with $\alpha_i \in R,$ then evaluating on the successive blocks of $n$ vectors, namely , $e_{kn+1}, \dots , e_{(k+1)n}, k=0,1,\dots ,$ we get $\alpha_i(e_{k+1})=0\ \forall\ k$ and $1 \le i \le n ;$ i.e. $\alpha_i \equiv 0\ \forall\ i$ showing that $B_n$ is linearly independent over $R$.

• $B_n$ spans $R$-

Let $f\in R$ then $f= \sum_{i=1}^{n} \alpha_i f_i,$ where $\alpha_i \in R$ are defined by their values on $B$ as in the following table-

\begin{array}{|c| ccccc|} \hline & \alpha_1 & \alpha_2 & \alpha_3 & \dots & \alpha_n \\ \hline e_1& f(e_1)&f(e_2) &f(e_3) & \dots & f(e_n)\\ e_2& f(e_{n+1}) & f(e_{n+2}) &f(e_{n+3}) & \dots & f(e_{2n})\\ \vdots& & & & \ddots\\ e_n & f(e_{(n-1)n+1}) & f(e_{(n-1)n+2}) &f(e_{(n-1)n+3}) & \dots & f(e_{n^2}) \\ \hline e_{n+1} & f(e_{n^2+1}) &f(e_{n^2+2}) &f(e_{n^2+3}) & \dots & f(e_{n^2+n}) \\ e_{n+2} & . & . &. &\dots &f(e_{n^2+2n}) \\ \vdots & & & &\ddots & \\ e_{2n} & f(e_{2n^2-n+1}) &f(e_{2n^2-n+2}) &f(e_{2n^2-n+3}) & \dots & f(e_{2n^2}) \\ \hline \vdots & \vdots & \vdots & \vdots& \vdots & \vdots \\ \hline e_{kn+1} & f(e_{kn^2+1}) & f(e_{kn^2+2}) & f(e_{kn^2+3}) & \dots & f(e_{kn^2+n}) \\ e_{kn+2} & . & .&.& \dots &f(e_{kn^2+2n}) \\ \vdots & &&&\ddots & \\ e_{(k+1)n} &.&.&.& \dots& f(e_{(k+1)n^2}) \\ \hline \vdots & \vdots & \vdots & \vdots& \vdots & \vdots \\ \end{array}

This shows that $B_n$ spans $R$.

So for each $n > 0$, $B_n= \{f_n\}$ is a basis of cardinality $n$

Answer 4

All short exact sequences $0\to U\to V\to W\to 0$ of vector spaces split, which can be proven using the fact that all vector spaces have bases. The same is not true for modules in general. Over $\mathbb{Z}$, the exact sequence $0\to\mathbb{Z}\overset{\times n}{\to}\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}\to 0$ does not split since any homomorphism $\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}$ is zero.