Proving a commutative ring is isomorphic to Cartesian product of sets I'm having trouble finding a starting point for this problem:
Let $R$ be a commutative ring containing elements $a, b$, both $\neq 0_R$ such that:
$$
a + b = 1_R,\ a^2 = a,\ b^2 = b,\ and\ a · b = 0_R
$$
Show that the ideals $Q := R · a$ and $S := R · b$ are rings, but not subrings of
$R$, and that the ring $R$ is isomorphic to the ring $Q×S$.
Given the $Q×S$ ring follows $(r,s) + (r', s') = (r+r', s+s')$ and $(r,s)·(r',s') = (rr', ss')$
I understand that if $R$ is isomorphic to $Q×S$, then that would mean that $a = (1_R, 0_R)$ and $b = (0_R, 1_R)$ satisfies all of the conditions, and by extension how they are not subrings.
I just can't for the life of me determine how to use that information to prove that $R$ is isomorphic to $Q×S$ without circular reasoning.
 A: Starting with your consideration:
The isomorphism if $R$ is isomorphic to $Q\times S$ with
(i)$a:=(1,0)$ and (ii)$b:=(0,1)$
would be the one that brings $(r_1,r_2)$ in $r_1a+r_2b$ that is $r_1a+r_2b$. So from this particular case you can construct a map that use $a$ and $b$ in a general form by reversing (*) and (**) to pass from an example to a general case:
So let $\phi:Q\times S\longrightarrow R$ defined by $(ar_1,br_2)\mapsto ar_1+br_2$ you can verify that it is an homomorphism of rings and it is surjective and injective and the reason rely on the properties $a^2=a,$ $b^2=b$, $a+b=1$.
For example note that $(a,b)$ is the unit of $Q\times S$ just as $(1,1)$ but the fact is that $a$ and $b$ doesen't need to be invertible, take $\frac{\mathbb{Z}}{(12)}$ with $a=\bar{9}$ and $b=\bar{4}$.
If you know what is an $R$ module the answer is even easier becouse the $\phi$ map is forced to be a map of $R$ module that turns to be an homomorphism of rings.
A: The other answer is enough of a hint in my book, so I'll just mention a little intuition. In linear algebra, a matrix $P$ satisfying $P^2 = P$ is a projection. The idea is that if you've got a vector space $V \oplus W$, then projections $P \colon (v, w) \mapsto (v, 0)$ and $Q \colon (v, w) \mapsto (0, w)$ satisfy $P^2 = P$, $Q^2 = Q$, $P+Q = \mathrm{id}$, and $PQ = QP = 0$. These operators all live in the (non-commutative) endomorphism ring of $V \oplus W$. The algebraic situation you're looking at is different in the particulars, but it's something like an abstraction of this rather intuitive situation.
