How to prove directly sum of non zero divisor and nilpotent is again non zero divisor?
I know that it can be easily proved by extending ring to ring of fraction So that I have a unit as that non zero divisor and then I can prove sum of unit and nilpotent is again unit.
But I was thinking to prove it by a more direct way?
My attempt in this regard:
a is non zero divisor. b is nilpotent of index n
On the contrary, suppose a+b is zero divisor so there is $c\neq 0$
Case 1: $c$ is not a nilpotent element of order $n$
$(a+b)c=ac+bc=0$
$ac=-bc$
multiplying each side n time
$a^nc^n=0 $
But $a$ is not zero divisor this implies $c$ is the nilpotent element of order $n$
Contradictions
CAse 2: If $c$ is the nilpotent element of order $n$
Then I am unable to find the contradiction.
Any Help will be appreciated
In general this is false: In the ring of $2\times 2$ matrices with entries in ... whatever, $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ is not a zero-divisor, and $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ is nilpotent, and their sum $\begin{pmatrix}0&0\\1&0\end{pmatrix}$ is a zero-divisor (in fact, nilpotent).
So, assume $a$ is a zero divisor and $c$ is nilpotent in a commutative ring. Say, $ab=0$ with $b\ne 0$ and $c^n=0$ with $n\in\Bbb N$. Let $k\in\Bbb N_0$ be maximal with $c^kb\ne 0$ (so certainly $0\le k<n$). Then $$(a-c)\underbrace{c^{k}b}_{\ne0}=abc^{k}-c^{k+1}b=0 $$ shows that $a-c$ is a zero-divisor. But if subtracting a nilpotent from a zero-divisor produces a zero-divisor, it follows that adding a nilpotent to a non-zero-divisor produces a non-zero-divisor.