TITLE: Pathologies in module theory
QUESTION [46 upvotes]: 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.

REPLY [5 votes]: 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 $n$