Let $F$ be a field and $A$ be an $F$-vector space which is also a ring with $1$. Suppose for all $c \in F$ and $x,y \in A$ we have $$ (cx)y = c(xy) = x(cy) $$ Then $A$ is called an $F$-algebra. If for an $F$-algebra $A$ we have some finite-dimensional $F$-vector space such that for each $v \in V$ and $x \in A$ there is a unique $vx \in V$ defined such that for $x,y \in A, v, w \in V$ and $c \in F$ we have
$(v+w)x = vx + wx$
$v(x+y) = vx + vy$
$(vx)y = v(xy)$
$(cv)x = c(vx) = v(cx)$
$v1 = v$
then $V$ is an $A$-module. If $A$ is any algebra, then $A$ itself is an $A$-module under right multiplication. This module is called the regular $A$-module and denoted by $A^{\circ}$. For an $A$-module we call $W \subseteq V$ an $A$-submodule if $W$ is a linear subspace that is also $A$-invariant, i.e. $wx \in W$ for $w \in W, x \in A$. An $A$-module is called completely reducible if for every $A$-submodule $W \subseteq V$ there exists another $A$-submodule $U$ such that $V = W + U$, $W \cap U = \{0\}$. An algebra $A$ is called semisimple if its regular module $A^{\circ}$ is completely reducible. A nonzero $A$-module $V$ is called irreducible if $\{0\}$ and $V$ are its only submodules. As it turns out, for an $F$-algebra $A$, every irreducible $A$-module is isomorphic to a factor module of $A^{\circ}$, and further if $A$ is semisimple, then every irreducible $A$-module is isomorphic to a submodule of $A^{\circ}$.
Let $V$ be a completely reducible $A$-module and let $M$ be an irreducible $A$-module. The $M$-homogeneous part of $V$, denotes $M(V)$, is the sum of all those submodules of $V$ which are isomorphic to $M$. If $W \ncong M$ then it follows that $M(A) \cap W(A) = 0$.
Now I have a question on the following proof:
Let $A$ be a semisimple algebra and let $M$ be an irreducible $A$-module. If $W$ is irreducible, then it is annihilated by $M(A)$ unless $W \cong M$.
Proof: If $W$ is an irreducible $A$-module with $W \ncong M$, then $W(A) \cap M(A) = 0$ by the above remark. Since $W(A)$ and $M(A)$ are ideals, we have $W(A)M(A) = 0$. By the above remarks, $A^{\circ}$ has a submodule $W_0 \cong W$ and $W_0 \subseteq W(A)$, so that $M(A)$ annihilates $W_0$. Since $W \cong W_0$, they have the same annihilator in $A$ and the statement follows. $\square$
I do not understand that why we have $W(A)M(A) = 0$ in the above proof. I found the result that for ideals $I, J$ in a ring $R$, that if $I + J = R$ (i.e. both ideals are coprime) then we have $I\cdot J = I \cap J$. But I do not know if that applies here, as I do not see that $M(A)$ and $W(A)$ are coprime. So why does the above hold?
These definitions and statement are taken from I. Martin Isaacs, Character Theory of Finite Groups, Chapter 1. Specifically my question concerns Theorem 1.15 (Wedderburn).
The product of two ideals is always contained in their intersection (even if they are not coprime), so that $W(A)M(A)\subseteq W(A)\cap M(A)=0$.