If $A$ is a semisimple algebra, and $M_1 \ncong M_2$ as irreducible $A$-modules, why we have that every ideal of $M_1(A)$ is an ideal of $A$

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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 $M(A) \cap W(A) = 0$, hence $M(A)W(A) = 0$.

Now let $A$ be a semisimple algebra. Now I have a question on the following statement:

Since $M_1(A)M_2(A) = 0$ for $M_1 \ncong M_2$, it follows that every ideal of the algebra $M(A)$ is in fact an ideal of $A$.

Why does $M_1(A)M_2(A) = 0$ implies that every ideal of $M(A)$ is an ideal of $A$?

These definitions and statements are taken from I. Martin Isaacs, Character Theory of Finite Groups, Chapter 1.