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 $n$



multiplying each side n time

$a^nc^n=0 $

But $a$ is not zero divisor this implies $c$ is the nilpotent element of order $n$


CAse 2: If $c$ is the nilpotent element of order $n$

Then I am unable to find the contradiction.

Any Help will be appreciated

  • 1
    $\begingroup$ If $a$ is a regular element (= non-zero-divisor) and $n$ is nilpotent with $n^k = 0$, then $\left(a-n\right)\sum_{i=0}^{k-1}a^in^{k-i} = a^k - n^k = a^k$. The right hand side is regular; thus, $a-n$ is regular. Apply this to $-n$ instead of $n$ to conclude that $a+n$ is regular. $\endgroup$ – darij grinberg Feb 10 at 8:17
  • $\begingroup$ Thanks. Sir I had first time read regular as term for nonzero divisor. I like that term. Is there any reason to call it regular? Just asking for information, please don't mind. $\endgroup$ – SRJ Feb 10 at 8:27
  • $\begingroup$ In don't know of a reason, but I have seen this word being used a few times, and even know someone who believes it so common that he doesn't define it :) $\endgroup$ – darij grinberg Feb 10 at 8:49

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.


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