Atiyah-Macdonald Sum of Submodules Vs Intersection of Submodules On page 19 in Atiyah-Macdonald, if $M$ is an $A$-modules and $(M_i)_{i \in I}$ is a collection of submodules of $M$, then their sum $\sum M_i$ is defined to be the set of all finite sums $\sigma x_i$ where $x_i \in M_i$ for all values of $i$ and all but finitely many values of $i$ are zero.  It goes on to say that $\sum M_i$ is the smallest submodule of $M$ which contains all of the $M_i$.
Immediately after, it states that $\bigcap M_i$ is again a submodule of $M$.
My question is that I am tempted to think of the smallest submodule of $M$ containing all of the $M_i$ as precisely being the intersection $\bigcap M_i$, since it clearly contains all of the $M_i$ but it seems like this is not the case.
Is it not true in general that $\sum M_i = \bigcap M_i$?  I.e., is it not true in general that the intersection of all submodules $M_i$ is the smallest submodule of $M$ which contain all of the $M_i$?
 A: The intersection of submodules is contained in each of the submodules, not the other way around. If $M_i\neq M_j$ then $M_i$ and $M_j$ are both not contained in $M_i\cap M_j$. They both contain $M_i\cap M_j$.
For a very tangible example, you can make explicit what finite sums and intersections of submodules of $\Bbb{Z}$ look like. It is easy and very illustrative.
A: A sum of modules is the smallest module containing the union. So it's like the opposite of the intersection. The intersection is contained inside of each $M_i$ and each $M_i$ is in turn contained in the union. The inclusions go this way:
$$ \bigcap_{i \in I} M_i \subseteq M_i \subseteq \bigcup_{i \in I} M_i \subseteq \sum_{i \in I} M_i. $$
For instance, consider two lines inside a vector space: $V_i = \{t v_i : t \in \mathbb{F}\}$. The intersection of these two lines is just $\{0\}$ and the union is not a subspace. But the sum is $V_1 + V_2 = \{sv_1 + tv_2 : (s, t) \in \mathbb{F}^2\}$ which is the plane containing each of the two lines.
