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 $\{ 0 \}$.