In Dummit and Foote's Abstract Algebra, all the main results about representations of finite groups are derived from the Wedderburn-Artin theorem. The proof of the Wedderburn-Artin theorem itself is sketched in a series of exercises for the reader, which I am working through right now.
I'm stuck on the following exercise, which is the second exercise in the series.
Let $R$ be a ring with $1$ (not necessarily commutative). Suppose that every $R$-module is injective. Show that $R = R_1 \times R_2 \times \dots \times R_r$, where $R_j$ is a 2-sided ideal with no proper non-zero 2-sided ideals.
In the previous exercise, we proved that the assumption that every $R$-module is injective implies that $R$ satisfies the descending chain condition. The book suggests we use this observation to tackle the current exercise:
Hint: Use the fact that $R$ satisfies the descending chain condition to show that $R$ has a minimal 2-sided ideal $R_1$. As a left R-module, $R = R_1 \oplus R'$ for some left ideal $R'$. Show that $R'$ is 2-sided ideal and proceed inductively using the descending chain condition.
Okay, so it's easy to see that $R$ must have a minimal 2-sided ideal $R_1$. (If not, then we'd be able to construct an infinite descending chain of 2-sided ideals, which would also be an infinite descending chain of left ideals, violating the descending chain condition.)
The fact that there exists a left ideal $R'$ such that $R = R_1 \oplus R'$ follows immediately from the fact that $R_1$ is an injective left $R$-module.
Now suppose we (somehow) manage to show that $R'$ is actually a 2-sided ideal. Then for any $a \in R_1$, $b \in R_2$, we have $ab \in R_1 R' \subset R_1 \cap R' = 0$ and $ba \in R' R_1 \subset R_1 \cap R' = 0$. Thus the operations of addition and multiplication are "decoupled" between $R_1$ and $R'$, which means that $R$ is the ring product $R_1 \times R'$. Furthermore, the way that $R$ acts on $R'$ (from the left or the right) can be understood as follows: writing $R = R_1 \oplus R'$, the $R_1$ component acts trivially on $R'$, whereas the $R'$ component acts on $R'$ by left or right multiplication. Thus, left/right/two-sided ideals of $R'$ are also left/right/two-sided ideals of $R$. This makes me confident that it will be easy to construct the inductive argument to complete the exercise once we have managed to show that $R'$ is a 2-sided ideal.
But how do we show that $R'$ is a 2-sided ideal? This is where I am stuck.
I noticed that showing that $R'$ is a 2-sided ideal is equivalent to showing that $R'R_1 = 0$. (Indeed, if $R'R_1 = 0$, then certainly $R'R = R'(R_1 + R') \subset R'$. Conversely, if $R'$ is a 2-sided ideal, then $R'R_1 \subset R'$; however $R_1$ is a 2-sided ideal, so $R'R_1 \subset R_1$, and thus $R'R_1 \subset R' \cap R_1 = 0$.)
Furthermore, $R'R_1$ is itself a 2-sided ideal (since $R'$ is a left ideal and $R_1$ is a right ideal). $R'R_1$ is also contained inside $R_1$, which is a minimal 2-sided ideal. So the only two possibilities are (i) $R'R_1 = 0$ and (ii) $R'R_1 = R_1$. Therefore, it suffices to show that $R'R_1 \neq R_1$. In other words, we have to find some element in $R_1$ that is not in $R'R_1$.
But how do we do that?
Following my intuition, I want to decompose $1 \in R$ as $1 = e_1 + e'$ with $e_1 \in R_1$ and $e' \in R'$; then perhaps we might be able to prove that $e_1$ is an element in $R_1$ that is not in $R'R_1$? Unfortunately I wasn't able to follow through on this line of thought.
What am I missing?
You have already shown that $R_1R’\subseteq R_1\cap R’=0$, since $R_1$ is a 2-sided ideal and $R’$ is a left ideal. If we write $1=e+f$ with $e\in R_1$ and $f\in R’$, then multiplying on the left by $e$ gives $ef=0$ and $e=e^2$. Thus $R_1^2\neq0$, and since it is again a 2-sided ideal it must equal $R_1$. Finally, if $R’R_1=R_1$, then $$ R_1 = R_1^2 = R_1R’R_1 = 0, $$ a contradiction. Thus by minimality we must have $R’R_1=0$. It follows, as you already noted, that $R’$ is also a 2-sided ideal.