May 25th, 2017, 01:59 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Nilpotent Element
Let $R$ be a commutative ring with unity and 1!=0. Let a be a nilpotent element, x be a unit. Then A) 1+a is not a unit. B) ax is a nilpotent element. C) x+a is a unit. D) none of the above statements is true. My answer is Option C Please let me know if I'm wrong 
May 31st, 2017, 05:56 AM  #2  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Quote:
Thanks!  
May 31st, 2017, 06:52 AM  #3 
Member Joined: May 2017 From: Russia Posts: 33 Thanks: 4 
Yes, Option C is true. $\displaystyle x$ is a unit, $\displaystyle a$ is a nilpotent element, so $\displaystyle \frac{a}{x}$ is a nilpotent element too. $\displaystyle a+x=x\left(1\left(\frac{a}{x}\right)\right).$ Let $\displaystyle a^n=0$ for some $\displaystyle n\in\mathbb{N}.$ Then $\displaystyle 2^n\gt n$. And $\displaystyle \left(\frac{a}{x}\right)^{2^n}=0.$ $\displaystyle x\left(1\left(\frac{a}{x}\right)\right)\cdot \frac{1}{x}\prod_{j=1}^{n} \left(1+\left(\frac{a}{x}\right)^{2^{j1}}\right)=x\cdot\frac{1}{x}\left(1\frac{a}{x}\right)=x\cdot\frac{1}{x}\left(1\left(\frac{a}{x}\right)^{2^n}\right)=1.$ We can find a multiplicative inverse for $\displaystyle a+x$. Last edited by skipjack; May 31st, 2017 at 09:39 AM. 
May 31st, 2017, 06:56 AM  #4  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Quote:
Last edited by skipjack; May 31st, 2017 at 09:40 AM.  
May 31st, 2017, 09:47 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1383 
Please avoid making a topic unnecessarily lengthy by quoting an entire previous post, especially if it's the immediately preceding post.

May 31st, 2017, 01:43 PM  #6 
Member Joined: May 2017 From: Russia Posts: 33 Thanks: 4 
Sorry, I made an error. I'd better correct it: $\displaystyle x\left(1\left(\frac{a}{x}\right)\right)\cdot \frac{1}{x}\prod_{j=1}^{n} \left(1+\left(\frac{a}{x}\right)^{2^{j1}}\right)=x\cdot\frac{1}{x}\left(1\left(\frac{a}{x}\right)^{2^n}\right)=1.$ 

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