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 May 25th, 2017, 01:59 AM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 154 Thanks: 1 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) a-x 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
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 Originally Posted by Lalitha183 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) a-x 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

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 May 31st, 2017, 06:52 AM #3 Newbie   Joined: May 2017 From: Russia Posts: 19 Thanks: 1 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 $\displaystyle \left(-\frac{a}{x}\right)^{2^n}=0.$ $\displaystyle x\left(1-\left(-\frac{a}{x}\right)\right)\cdot \frac{1}{x}\prod_{j=1}^{n} \left(1+\left(-\frac{a}{x}\right)^{2^{j-1}}\right)=x\cdot\frac{1}{x}\left(1-\frac{a}{x}\right)=x\cdot\frac{1}{x}\left(1-\left(-\frac{a}{x}\right)^{2^n}\right)=1.$ 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
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 Originally Posted by ABVictor 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 $\displaystyle \left(-\frac{a}{x}\right)^{2^n}=0.$ $\displaystyle x\left(1-\left(-\frac{a}{x}\right)\right)\cdot \frac{1}{x}\prod_{j=1}^{n} \left(1+\left(-\frac{a}{x}\right)^{2^{j-1}}\right)=x\cdot\frac{1}{x}\left(1-\frac{a}{x}\right)=x\cdot\frac{1}{x}\left(1-\left(-\frac{a}{x}\right)^{2^n}\right)=1.$ We can find a multiplicative inverse for $\displaystyle a+x$.
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Last edited by skipjack; May 31st, 2017 at 09:40 AM.

 May 31st, 2017, 09:47 AM #5 Global Moderator   Joined: Dec 2006 Posts: 17,520 Thanks: 1318 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 Newbie   Joined: May 2017 From: Russia Posts: 19 Thanks: 1 Sorry, I made an error. I'd better correct it: $\displaystyle x\left(1-\left(-\frac{a}{x}\right)\right)\cdot \frac{1}{x}\prod_{j=1}^{n} \left(1+\left(-\frac{a}{x}\right)^{2^{j-1}}\right)=x\cdot\frac{1}{x}\left(1-\left(-\frac{a}{x}\right)^{2^n}\right)=1.$

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