If for $x\in R$, either $x$ or $x-1$ is regular then for (prime) ideals $I$ and $J$ of $R$ consisting of zero divisors, $I+J\ne R$. Suppose that $R$ is a commutative ring.
Show that if $x\in R$, either $x$ or $x-1$ is regular then for (prime) ideals $I$ and $J$ of $R$ consisting of zero divisors, $I+J\ne R$.
Prove:
Suppose that $I$ and $J$ are ideals consisting of zero divisors and that $I+J=R$. So $1=i+j$ where $i\in I$ and $j\in J$. Then $i$ and $i−1=−j$ are both zero divisors, a contradiction.
Question: How does the contradiction happen? If assume that $i$ is a zero divisor, then $i-1$ should be regular but it was zero divisor also, isn't?
 A: Since $i$ is a zero divisor, the hypothesis says $1-i$ is regular.  Since $j$ is a zero divisor, and equal to $1-i$, you have that it is a zero divisor and a regular element at the same time.  That's not possible.
A: It's simply $\,\lnot (1)\,\Rightarrow\,\lnot (3)\,$ below (which doesn't use $\,zR\subseteq \cal Z),\,$ with $\,{\cal Z} = $ zero-divisors $\,\cup\, \{0\}$.
Lemma $ $ TFAE in a commutative ring $\,R\,$ where $\,{\cal Z}\subseteq R\,$ satisfies $\,zR\subseteq\cal Z\,$ for all $\,z\in \cal Z$
$\begin{align}&(1)\ \ \exists\ x\!:\ x\in {\cal Z}\ \ \&\ \ 1\!-\!x\in \cal Z\\[.2em]  
&(2)\ \ \exists\ x,\,y\,\in\,{\cal Z}\!:\:\! \ \ x+y\,=\,1\\[.2em] 
&(3)\ \ \exists\, X,Y\subseteq {\cal Z}\!:\  X + Y = 1,\, \ {\rm for\ ideals}\ X,Y\end{align}$
Proof $\ \ (1\Rightarrow 2)\ \ $ Let $\,y = 1\!-\!x$.
$(2\Rightarrow 3)^{\phantom{|^|}}\  $ By hypothesis $\,x,y\in{\cal Z}\,\Rightarrow\, xR,yR\subseteq {\cal Z}\,$ and $\, x+y=1\,\Rightarrow\,xR + yR = 1$.
$(3\Rightarrow 1)^{\phantom{|^|}}\  X+Y=1\,\Rightarrow\, x+y = 1\,$ for $\,x,y\in\! X,Y\subseteq{\cal Z}\,$ so $\,x\in{\cal Z},\,$ $ y\!=\!1\!-\!x\in\cal Z$.
