Hints to prove Existence of Ring Isomorphism $A \to A/(e_{1}) \times A/(e_{2})$

Question

Let A be a commutative ring, $e_{1}, e_{2} \in A$ whereby $e_{1}^{2}=e_{1}$ , $e_{2}^{2}=e_{2}$ , $e_{1}+e_{2}=1$, and $e_{1}e_{2}=0$.

Prove that there is an Isomorphism $\phi$, whereby

$\phi: A \to A/(e_{1}) \times A/(e_{2})$

My first question is: How would quotient rings $A/(e_{1})$ or $A/(e_{2})$ actually look since I have no idea what $e_{1}$ or $e_{2}$ actually are?

I'd suggest a map that for $a \in A$ we define $\phi(a):=(\lambda_{1}[e_{1}],\lambda_{2}[e_{2}])$ so that $\phi$ basically splits any element $a \in A$ into its respective components in terms of the linearly independent $\{e_{1},e_{2}\}$. Am I on the right track?

Any help is greatly appreciated.

Answer 1

Hint 1

For $a \in A$ we have $a = a \cdot 1 = a \cdot e_{1} + a \cdot e_{2}$.

Hint 2

Consider the map $f_{1} : A \to A$ given by $a \mapsto a e_{1}$.

Hint 3

Prove that $f_{1}$ is a ring homomorphism with image $(e_{1})$.

Hint 4

$a \in \ker(f_{1})$ iff $0 = a e_{1}$ iff by Hint 1 $a = a e_{2} \in (e_{2})$. Therefore $\ker(f_{1}) = (e_{2})$ and $A/(e_{2}) \cong (e_{1})$.

Hint 5

Similarly $A/(e_{1}) \cong (e_{2})$.

Hint 6

Consider the ring homomorphism $f : A \to (e_{1}) \times (e_{2})$ given by $a \mapsto (a e_{1}, a e_{2})$. It is injective by Hint 1. It is surjective because for any $b, c \in A$ one has $(b e_{1}, c e_{2}) = ((b e_{1} + c e_{2}) e_{1}, (b e_{1} + c e_{2}) e_{2}) = f(b e_{1} + c e_{2})$.

Answer 2

I'll write $e=e_1$ and $f=e_2$ for simplicity. There is an obvious candidate for the isomorphism: $$ \phi\colon A\to A/(e)\times A/(f) \qquad \phi(a)=\bigl(a+(e),a+(f)\bigr) $$ This is obviously a homomorphism. If $a\in\ker\phi$, then $a\in(e)$ and $a\in(f)$, so $$ a=be=cf $$ for some $b,c\in A$. Then $ae=cfe=0$ and $af=bef=0$. Therefore $$ a=a(e+f)=ae+af=0 $$ Hence $\phi$ is injective.

Suppose $b,c\in A$. You need to find $a\in A$ with $$ a+(e)=b+(e),\qquad a+(f)=c+(f) $$ Hint: try $a=bf+ce$.

For instance, \begin{align}(a-b)e&=(bf+ce-b)e=(ce-be)e=(c-b)e \\ (a-b)f&=(bf+ce-b)f=(ce-be)f=0 \end{align} so $a-b=(c-b)e\in(e)$. Similarly, $a-c\in(f)$.

How do we find such an expression?

If $a-b\in(e)$ and $a-c\in(f)$, then $a-b=(a-b)e$ and $a-c=(a-c)f$ so $a-ae=b-be$ and $af=bf$; similarly $ae=ce$, so $a=ae+af=bf+ce$.