# Orthogonal idempotents give direct sum?

Say $x$ and $x'$ are orthogonal idempotents so $x+x'=1$ and $xx'=0$ in some commutative ring $A$. Then for any $a\in A$, $a=ax+ax'$, so $A=Ax+Ax'$. Why does $Ax\cap Ax'=0$? If $y=ax=a'x'$, then $y^2=aa'xx'=0$. Maybe I'm missing something, but does $y^2=0$ imply $y=0$? Is there a no zero divisor condition needed for this to work in general?

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No, $y^2=0$ does not imply $y=0$ in general.

But if $ax=a'x'$, then $ax = a(xx) = (ax)x = (a'x')x = a'(x'x) = a'0 = 0$.

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If $ax=a'x'$ then multiply both sides by $x$ alone to get $ax^2=0$. Invoke indempotence.

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Generalizing the well-known fact that $\rm\:lcm(1,n) = 1\cdot n,\:$ an lcm with an idempotent is trivial in any ring;  more precisely $\rm\:x^2 = x\:$ $\Rightarrow$ $\rm\:lcm(x,x') = x\:x'.\:$ Yours is special case $\rm\:xx' = 0\:$ of this

Theorem $\ \ \:$ $\rm\:x,x'\ |\ y\iff xx'\ |\ y\ \ \:$ if $\rm\ \ xx\ |\ x,\$ i.e. $\rm\ axx = x\:$ for some $\rm\:a\in A$

Proof $\rm\,\ \ (\Leftarrow)\,\ \ x,x'\ |\ xx'\ |\ y\ \ \$ $\rm (\Rightarrow)\ \ x'c = y = xb = axxb = axx'c\ \Rightarrow\ xx'\ |\ y\ \$ QED

Note $\ \$ The ring factorization you study is known as the Peirce decomposition.

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