Reflections on Idempotent Elements: A Problem Revisited
I encountered the following homework problem in my algebra class. We assume throughout this post that a ring is commutative.
Problem: An idempotent in a ring $R$ is an element $e$ satisfying $e^2 = e$. Prove that $R$ is isomorphic to a product of two nonzero rings if and only if $R$ has an idempotent other than $0_R, 1_R$.
The forward direction is trivial. For the other direction, we need to construct an isomorphism. My original proof was unmotivated. I constructed an isomorphism $i:R\rightarrow R/(e)\times R/(1-e)$. Since the codomain is the product of two quotient rings, I spent a lot of time on checking whether $i$ is well-defined. The proof also did not tell me why this interesting theorem holds, or at least why we should expect this theorem to be true. Recently I come up with a natural proof of this problem.
Before going into the details, let me do some review on a projection map. Let $P:V\rightarrow V$ be a projection defined on a vector space $V$. Geometrically, this often means that $P$ maps the whole space into a lower-dimensional subspace and is equal to identity when it is restricted onto this subspace. Abstractly, or algebraically, we are requiring $P$ to be idempotent. That is, $P^2=P\circ P=P$. In this case, we can prove that
\[V\cong \text{ker }P \oplus \text{im }P\]In fact, this is even true for idempotent homomorphism defined on a module.
Theorem: Let $M$ be an $R$-module, and $P:M\rightarrow M$ an idempotent $R$-module homomorphism. We then have the following isomorphism
\[M\cong \text{ker }P \oplus \text{im }P\]Proof: We use $P$ to define a map $f:M\rightarrow \text{ker }P \oplus \text{im }P$ by the following formula:
\[f(m):=(m-P(m),P(m))\]We leave the details for readers to check that $f$ is indeed an isomorphism. $\square$
Now a perceptive reader may notice that if an element $e\in R$ is idempotent, then the left multiplication on $R$ by $e$ defines an idempotent ring homomorphism! Let’s still use $P$ to denote this map. Then by the lemma above, we have an $R$-module isomorphism
\[\begin{align*} f:R\rightarrow \text{ker }P \oplus \text{im }P\\ \forall r\in R, f(r):=(r-P(r),P(r))=(r-er,er) \end{align*}\]The formula for $f$ also suggests that $\text{ker }P$ is the ideal generated by $1-e$ and $\text{im }P$ is the ideal generated by $r$. Hence, $f$ is an $R$-module isomorphism between $R$ and $(1-e)\times (e)$.
We now turn $f$ into a ring isomorphism. Notice that $(e)$ can be turned into a ring by defining $e=1$ (because $e\cdot (er)=e^2r=er=1\cdot (er)$). $(1-e)\times (e)$ is thus a ring. $f$ is then a ring homomorphism because $(1-e,e)$ is the multiplicative identity in $(1-e)\times (e)$ and
\[\begin{align*} f(1)&=(1-e,e)\\ \forall a,b\in R, f(ab)&=(ab-eab,eab)\\ &=(ab-eab-eab+e^2ab,e^2ab)\\ &=((a-ea)(b-eb),(ea)(eb))\\ &=(a-ea,ea)\cdot (b-eb,eb)\\ &=f(a)\cdot f(b) \end{align*}\]This proof explains why the existence of an idempotent element should lead to a decomposition of $R$—because a projection induces a decomposition of a vector space into the direct sum of its kernel and image.
The proof also explains why my original construction works. It is easy to see that $R/(e)\cong (1-e)$ and $R/(1-e)\cong (e)$.