Consider a ring with identity that is the direct sum (as a ring) of non-zero subrings $A = A_1 \oplus ... \oplus A_r$. Suppose that $A$ has exactly $n$ isomorphism types of simple modules. Show that $r \le n$.
This is exercise 3.2 in A Course in Finite Group Representation Theory
I think I should prove it by controdiction: assume that $r \gt n$. If we can find $r$ non-isomorphic maximal left ideals $M_i, i = 1,...,r$, then we get $r (\gt n)$ non-isomorphic simple modules $A/M_i$, a contradiction. To find the maximal ideals, I wonder whether or not $A_2 \oplus ... \oplus A_r$ is a maximal left ideal. And if not, can we find a maximal ideal $N_1$ of $A_1$ so that $N_1 \oplus A_2 \oplus ... \oplus A_r$ becomes a maximal left ideal of $A$? And assume that we have found $r$ maximal left ideals how can we prove that they are non-isomorphic?
So my question is: how to properly solve this exercise? It took me hours and any help is appreciated.
You have already found all the ingredients:
Note that for all ideals $I_i \trianglelefteq A_i$ we have that $I := I_1 \times \dotsb \times I_r$ is an ideal in $A := A_1 \times \dotsb \times A_r$. (It can be shown that every ideal of $A$ is of this form.) Also note that there exists an isomorphism of abelian groups $$ A/I = (A_1 \times \dotsb \times A_r) / (I_1 \times \dotsb \times I_r) \cong (A_1 / I_1) \times \dotsb \times (A_r / I_r) $$ under which the $A$-module structure of $A/I$ corresponds to $$ (a_1, \dotsc, a_r) \cdot ([x_1], \dotsc, [x_r]) = ([a_1 x_1], \dotsc, [a_r x_r]) $$ for all $(a_1, \dotsc, a_r) \in A$ and $([x_1], \dotsc, [x_r]) \in (A_1 / I_1) \times \dotsb \times (A_r / I_r)$.
For every $i = 1, \dotsc, r$ there now exists a maximal (left) ideal $N_i \trianglelefteq A_i$ because $A_i$ has an identity (this is a standard application of Zorn’s Lemma, which I assume you have seen before). The resulting ideal $$ M_i := A_1 \times \dotsb \times A_{i-1} \times N_i \times A_{i+1} \times \dotsb A_r $$ is maximal in $A$: We have that $$ A / M_i \cong 0 \times \dotsb \times 0 \times (A_i / N_i) \times 0 \times \dotsb \times 0 \cong A_i / N_i \,, $$ with the $A$-module structure of $A/M_i$ corresponding to $$ (a_1, \dotsc, a_r) \cdot [x] = [a_i x] = a_i \cdot [x] $$ for all $(a_1, \dotsc, a_r) \in A$, $[x] \in A_i/N_i$. It follows that $A/M_i$ is a simple $A$-module if and only if $A_i/N_i$ is simple as an $A_i$-module (because a subset $P \subseteq A_i/N_i$ is an $A_i$-submodule if and only if it is an $A$-submodule.) But this holds because $N_i$ is a maximal submodule of $A_i$.
To see that the simple modules $A/M_1, \dotsc, A/M_r$ are pairwise non-isomorphic, consider the elements $e_1, \dotsc, e_r \in A$ which are given by $e_i = (0, \dotsc, 0, 1, 0, \dotsc, 0)$ with $1$ at the $i$-th position. Then $e_i \cdot A/M_i = A/M_i$ but $e_i \cdot A/M_j = 0$ for all $j \neq i$.
The idea behind the above approach is the following: Let $A, B$ be rings with identity.
Given an $A$-module $M_1$ and a $B$-module $M_2$, we can endow the (external) direct sum $M_1 \oplus M_2$ with the structure of an $(A \times B)$-module via $$ (a, b) \cdot (m_1, m_2) := (a \cdot m_1, b \cdot m_2) $$ for all $(a,b) \in A \times B$, $(m_1, m_2) \in M_1 \oplus M_2$.
We can also decompose every $(A \times B)$-module $M$ as a(n internal) direct sum $M = M_1 \oplus M_2$ with $M_i = e_i M$ for $e_1 = (1,0)$, $e_2 = (0,1)$. Then $M_1$ carries the structure of an $A$-module, and $M_2$ carries the structure of a $B$-module, via $$ a \cdot m_1 := (a,0) \cdot m_1 \qquad\text{and}\qquad b \cdot m_2 := (0,b) \cdot m_2 $$ for all $a \in A$, $b \in B$, $m_i \in M_i$.
Both of these constructions are inverse to each other (up to isomorphism), and they are compatible with submodules, quotients, homomorphisms, etc. We can therefore think about $(A \times B)$-modules “componentwise”, one $A$-component and one $B$-component. (We actually get an equivalence of categories $$ (A \times B)\mathrm{-Mod} \simeq (A\mathrm{-Mod}) \times (B\mathrm{-Mod}) $$ by the above costruction.)
I would also like to point out that $A_i$ is not really a subring of $A$ (unless $A_j = 0$ for every $j \neq i$) because $1_{A_i} \neq 1_A$ (this is also mentioned at the beginning of section 3.6 of A Course in Finite Group Representation Theory, so I’m not sure why it is written this way in the exercise.)