# How to prove directly sum of non zero divisor and nilpotent is again non zero divisor?

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

• 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. – darij grinberg Feb 10 at 8:17
• 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. – SRJ Feb 10 at 8:27
• 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 :) – darij grinberg Feb 10 at 8:49

## 1 Answer

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 $$0\le k). Then $$(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.