I've been looking at a couple of ring theory exercises and there's this one I don't know how to do it. It goes like this.
$A$ is a commutative unital ring, and $e$ an element of $A$, $e \neq 0,1$, with $e=e^{2}$. Given the ideals $I_{1}=(e)$ and $I_{2}=(1-e)$, show that:
a) $I_{1}+I_{2}=A$ and $I_{1} \cap I_{2} = \{0\}$.
b) $A$ and $A/I_{1} \times A/I_{2}$ are isomorphic.
So, for the first one, it is clear that $e$ is idempotent. Moreover, if I consider $e$ and $1-e$, they are orthogonal (their product is commutative and gives 0). But I don't know if this is of any help to prove that their sum is the total and the intersection is $\{0\}$...
The second one, I can't even visualize what $A/I_{1} \times A/I_{2}$ is, so I'd welcome a little help!
a) Let $a \in A$. Then $a = ea + (1-e)a \in I_1 + I_2$, so $A = I_1 + I_2$. As for the intersection, assume that we have an element $f$ in the intersection, i.e. there exists an $a$ and a $b$ such that $f = (1-e)a = eb$. We then multiply by $e$ to get $$ ef = e(1-e)a = e^2b\\ ef = 0 = eb = f $$ so $0$ is the only element in the intersection.
b) There is a natural map $A \to A/I_1 \times A/I_2$, by $a \mapsto (a+I_1, a+I_2)$. We need to show that this map is bijective.
Injectivity. Let's say $a_0 \mapsto (0 + I_1, 0+I_2)$. That means that $a_0$ is both in $I_1$ and in $I_2$, and as detailed above, the only element in the intersection is $0$, so $a_0 = 0$, and the map is injective.
Surjectivity. Take $a, b \in A$. We need to show that there is a $c\in A$ such that $c \mapsto (a + I_1, b + I_2)$. But this is solved by letting $c = (1-e)a + eb = (1-e)a + (1-(1-e))b$, since $$ c + I_1 = a + e(b-a) + I_1 = a + I_1\\ c + I_2 = b + (1-e)(a - b) + I_2 = b + I_2 $$