Units and Nilpotents

Question

If $ua = au$, where $u$ is a unit and $a$ is a nilpotent, show that $u+a$ is a unit.

I've been working on this problem for an hour that I tried to construct an element $x \in R$ such that $x(u+a) = 1 = (u+a)x$. After tried several elements and manipulated $ua = au$, I still couldn't find any clue. Can anybody give me a hint?

Answer 1

If $u=1$, then you could do it via the identity $$(1+a)(1-a+a^2-a^3+\cdots + (-1)^{n}a^n) = 1 + (-1)^{n}a^{n+1}$$ by selecting $n$ large enough.

If $uv=vu=1$, does $a$ commute with $v$? Is $va$ nilpotent?

Answer 2

Let $v$ be the inverse of $u$, and suppose $a^2=0$. Note that $$(u+a)\cdot v(1-va)=(1+va)(1-va)=1-v^2a^2=1-0=1.$$ See if you can generalize this.

Answer 3

I think that you do not need to overthink of this problem. The point is to get some idea of the multiplicative inverse.

Firstly, think about if $x$ is a nilpotent, what is the multiplicative inverse of $1+x$. Let's forget about the commutative algebra for a moment. A natural choice is $\frac{1}{1+x}$. However, you need to write $\frac{1}{1+x}$ in terms of $x$, because this is the only thing you know that belongs to the ring. This brings you the Taylor expansion of $$\frac{1}{1+x}= \sum_{k=0}^{\infty}(-1)^{k}x^{k}.$$ The point is that $x^{n}=0$ for some $n>0$, so this sequence is kind of repeating. I do not want to go to details analysis here, but this should give you an idea to try to verify that $$(1+x)\Bigg(\sum_{k=0}^{n}(-1)^{k}x^{k}\Bigg)=1,$$ and this is true.

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Ok, for general case, if $x^{n}=0$ for some $n>0$ and $y$ is a unit, then $y^{-1}$ and all the power of $y^{-1}$ make sense. Now, we use the same idea. The natural choice of the inverse of $(x+y)$ is for sure $\frac{1}{x+y}$, but then again you need to write it as in terms of $x,y$ or $y^{-1}$. This again brings you the Taylor expansion. Treat $y$ as a constant, you expand in terms of $x$. You can feel free to choose the expansion around any point (even around infinty can work I believe), here I expand it around $x=0$, which is $$\frac{1}{x+y}=\sum_{k=0}^{\infty}(-1)^{k}x^{k}y^{-(k+1)}.$$ Again, the $y^{-(k+1)}$ make sense because $y$ is a unit. Hence, what you should consider to prove is $$(x+y)\Bigg(\sum_{k=0}^{n}(-1)^{k}x^{k}y^{-(k+1)}\Bigg)=1.$$ This is again easy to verify.

Answer 4

Given, $R$ is a commutative ring with unity. a be the unit and $b^2=0$. Let $c$ be the multiplicative inverse of $a$.

Case $1$:
If $b=0$
$a.c=1$
$\Rightarrow(a+0).c=1$
$\Rightarrow(a+b).c=1$

$\implies a+b$ is a unit.

Case $2$:
$If b≠0$
$b^2=0$
$\Rightarrow(bc)²=0$
$\Rightarrow1-(bc)²=1$
$\Rightarrow[(1+bc)(1-bc)]=1$. [ Since, $R$ is a commutative ring]
$\Rightarrow[(a+b)(c-bc²)]=1$. [ Since, $c$ is the multiplicative inverse of $a$]

$\implies a+b$ is unit.

Hence proved.