Undergraduate Algebra by Matej Bresar, Exercise 2.104: Prove that a prime ring with zero divisors contains a non-zero nilpotent.
Attempts:
By the definitions,
$\forall a, b \in R-\{0\}, \exists x \in R\; s.t. axb \ne 0$
Let $c, d \in R-\{0\}, c^2, d^2 \ne 0, cd=0, c^2d^2=0 $
$\exists y \in R \;s.t. \;c^2yd^2 \ne 0, c^2y, yd^2 \in R$
$(c^2yd^2)^2=c^2yd^2c^2yd^2=c^2(yd^2c^2y)d^2$
$yd^2c^2y=1, d^2c^2=y^{-2}, y^2=c^{-2}d^{-2} $ then $c^2(yd^2c^2y)d^2=c^2d^2=0 $
$(cd)^{-1}=d^{-1}c^{-1}, (dc)^{-1}=c^{-1}d^{-1}, c^{-1}(dc)^{-1}d^{-1}=c^{-2}d^{-2}$
Also,
$\exists y' \in R \;s.t. \;cy'd \ne 0, cy', y'd \in R$
$(cy'd)^2=cy'dcy'd=c(y'dcy')d$
$y'dcy'=1, dc=y'^{-2}, y'^2=c^{-1}d^{-1} $ then $c(y'dcy')d=cd=0 $
$dcy'=y'dc, dcy'^2dc=dc$
I couldn't proceed further from here.
Best Answer
You are way off in the weeds. This is especially the case because there is no justification for $c$ or $d$ to have an inverse.
The argument should simply be:
If $ab=0$ for nonzero $a$ and $b$, then by definition of primeness, $bRa\neq \{0\}$, so $bxa\neq 0$ for some $x$.
Then $bxa$ is a nonzero element whose square is zero.