# Pathologies in module theory

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.

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See math.stackexchange.com/questions/196203/… for an example where two notions of dimension give answers of 1 and uncountably-infinite for the same module. – Jack Schmidt Sep 24 '12 at 11:48
Should this be CW? – kahen Sep 24 '12 at 12:30
@kahen: I thought I did this already when posting the question. Now I cannot find the checkbox anymore. How do I make it community wiki? – Gregor Sep 24 '12 at 12:47
I would suggest every write one pathology in each answer. – Fredrik Meyer Sep 24 '12 at 12:57
@FredrikMeyer What would the utility be? There would be at least thirteen answers now... – rschwieb Sep 24 '12 at 13:41

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 $\{ 0 \}$.

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For what it's worth, I think this answer is deserving of reputation and, as such, shouldn't be CW. On the other hand, do whatever you feel comfortable with. – Jason DeVito Sep 24 '12 at 15:10

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 $n$

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This example is taken from "Rings and modules" by C. Musili – Bhaskar Vashishth Jun 24 '15 at 6:37

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.

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Just as an observation, this is saying that not every module is projective, or not every module is injective, which is the fourth point in rschwieb's answer. In your specific example, $\mathbb{Z}$ is not injective and $\mathbb{Z}/n\mathbb{Z}$ is not projective. – lentic catachresis Sep 24 '12 at 14:11
@BrunoStonek: Good point! I'm not an algebraist and I never learned that terminology, but wikipedia agrees that it means exactly what I said. – Noah Stein Sep 24 '12 at 15:00
@NoahStein If you were looking at the projective module article, I'm really going to chuckle: I was just editing it yesterday. – rschwieb Sep 24 '12 at 19:24
@rschwieb: Yes I was, but it's not quite a coincidence since I clicked through from the link in your answer. – Noah Stein Sep 24 '12 at 19:55
• 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.

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I'm pretty amazed at how "badly behaved" domains are :) I got a lot of mileage out of them for counterexamples, here! – rschwieb Sep 24 '12 at 12:55