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Here's what I have so far and I would like to know if I am right or if my proof needs to be edited:

Since $a$ is a unit it means $a1=a$, with $1$ being the unity element

We know $b^2=0$ and this means $b=0$

So $a+b=a+0=a=a1$ hence $a+b$ is a unit.

Alex Wertheim
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cambelot
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  • The question is a duplicate of this: http://math.stackexchange.com/q/119904/29335 but I guess this is a solution verification, hence distinct. – rschwieb Dec 02 '14 at 20:52

2 Answers2

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The element $a$ is unit means that there's $a'$ such that $aa'=1$. Now we have

$$(a+b)(a'-(a')^2b))=aa'-a(a')^2b+ba'+(a')^2b^2=1$$ so $a+b$ is unit.

  • How did you come up with the (a'-$(a')^2$b) part? – cambelot Dec 02 '14 at 20:55
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    By analogy with the real analysis we have $$\frac1{a+b}=\frac{a^{-1}}{1+a^{-1}b}=a^{-1}(1-a^{-1}b+(a^{-1}b)^2+\cdots)$$ and recall that in this ring we have $b^2$ so we have only two terms in the previous sum. –  Dec 02 '14 at 21:00
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Your proof does not make any sense. How do you conclude $a+b = a+0$, for instance? $b^{2} = 0$ does not imply $b = 0$; consider, for instance $2 \in \mathbb{Z}/4\mathbb{Z}$. Also, your definition of a unit is not correct. A unit is an element of a ring with a multiplicative inverse. If $a$ is a unit, then there exists $u \in R$ such that $au = 1$. $a\cdot 1 = a$ is true for every element of $R$ by the definition of the multiplicative identity in a ring!

However, the proposition given is actually a special case of a more general fact: if $R$ is a commutative ring and $a$ is a unit and $b$ is nilpotent, then $a+b$ is a unit.

Here is my hint in your case:

Consider $(a+b)(a-b) = a^{2}-b^{2} = a^{2}$. Since $a$ is a unit, can you use this to construct an inverse to $(a+b)$?

Alex Wertheim
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