In my lecture, the Prof proved the following statement:
Assume $R$ has maximal ideals $\mathfrak{m}_1,\dots,\mathfrak{m}_r$, such that $\mathfrak{m}_1\cdots\mathfrak{m}_r = (0)$. Then $R$ is noetherian iff $R$ is artinian iff $\text{length}(R)$ is finite.
His proof starts as follows: Define $\mathfrak{a}_0 := R, \mathfrak{a}_1 := \mathfrak{m}_1,\dots, \mathfrak{a}_i := \mathfrak{m}_1\cdots\mathfrak{m}_i$., etc. Then $\mathfrak{a}_r = 0$. This gives us a descending chain of ideals $R = \mathfrak{a}_0 \supset \mathfrak{a}_1\supset\cdots\supset \mathfrak{a}_r = 0$.
Here is the statement that I don't understand: He says that each $\mathfrak{a}_i/\mathfrak{a}_{i+1}$ is an $R/\mathfrak{m}_{i+1}$ vector space. Of course $R/\mathfrak{m}_{i+1}$ is a field, as $\mathfrak{m}_{i+1}$ is maximal. However, I still don't see how or why we can see $\mathfrak{a}_i/\mathfrak{a}_{i+1}$ as a vector space over that field.
Notice that $\mathfrak{m}_{i+1}\mathfrak{a_i} = \mathfrak{a}_{i+1} $ so that $\mathfrak{m}_{i+1} \frac{\mathfrak{a}_i}{\mathfrak{a}_{i+1}} = 0$.
Whenever you have an $R$ - module, $M$, that is annihilated by an ideal $I$ of $R$, you can view $M$ as an $\frac{R}{I}$ module where the scalar multiplication is $(r + I)m = rm$ for all $r+I \in \frac{R}{I}$, $m \in M$. It is straightforward to check that this is a well-defined scalar multiplication.
Therefore since $\frac{\mathfrak{a}_i}{\mathfrak{a}_{i+1}}$ annihilated by $\mathfrak{m}_{i+1}$, you can view $\frac{\mathfrak{a}_i}{\mathfrak{a}_{i+1}}$ as a $\frac{R}{\mathfrak{m}_{i+1}}$ module. That is, it is a vector space.