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May 25th, 2017, 01:59 AM   #1
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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
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May 31st, 2017, 05:56 AM   #2
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Quote:
Originally Posted by Lalitha183 View Post
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
Please somebody check

Thanks!
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May 31st, 2017, 06:52 AM   #3
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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.
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May 31st, 2017, 06:56 AM   #4
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Quote:
Originally Posted by ABVictor View Post
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 $\displaystyle a+x$.
Thank you.

Last edited by skipjack; May 31st, 2017 at 09:40 AM.
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May 31st, 2017, 09:47 AM   #5
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Please avoid making a topic unnecessarily lengthy by quoting an entire previous post, especially if it's the immediately preceding post.
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May 31st, 2017, 01:43 PM   #6
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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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