Units and Nilpotents 
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?
 A: 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?
A: Here's a rather different argument.  First, suppose that $R$ is commutative.  Suppose $u+a$ is not a unit.  Then it is contained in some maximal ideal $M\subset R$.  Since $a$ is nilpotent, $a\in M$ (since $R/M$ is a field, and any nilpotent element of a field is $0$).  Thus $u=(u+a)-a\in M$ as well.  But $u$ is a unit, so it can't be in any maximal ideal, and this is a contradiction.
If you don't know that $R$ is commutative, let $S\subseteq R$ be the subring generated by $a$, $u$, and $u^{-1}$.  Then $S$ is commutative: the only thing that isn't immediate is that $u^{-1}$ commutes with $a$, and this this can be proven as follows: $$u^{-1}a=u^{-1}auu^{-1}=u^{-1}uau^{-1}=au^{-1}.$$
The argument of the first paragraph now shows that $u+a$ is a unit in $S$, and hence also in $R$.

This argument may seem horribly nonconstructive, due to the use of a maximal ideal (and hence the axiom of choice) and proof by contradiction.  However, it can be made to be constructive and gives an explicit inverse for $u+a$ in terms of an inverse for $u$ and and an $n$ such that $a^n=0$.
First, we observe that all that is actually required of the ideal $M$ is that it is a proper ideal which contains $u+a$ and all nilpotent elements of $R$.  So we may replace $M$ with the ideal $(u+a)+N$ where $N$ is the nilradical of $R$, and use the fact that if $I=(u+a)$ is a proper ideal in a commutative ring then $I+N$ is still a proper ideal.  This is because $R/(I+N)$ is the quotient of $R/I$ by the image of $N$, which is contained in the nilradical of $R/I$.  If $I$ is a proper ideal, then $R/I$ is a nonzero ring, so its nilradical is a proper ideal, so $R/(I+N)$ is a nonzero ring and $I+N$ is a proper ideal.
Next, we recast this argument as a direct proof instead of a proof by contradiction.  Letting $I=(u+a)$, we observe that $I+N$ is not a proper ideal since $u=(u+a)-a\in I+N$ and $u$ is a unit.  That is, a nilpotent element (namely $a$) is a unit in the ring $R/I$, which means $R/I$ is the zero ring, which means $I=R$, which means $u+a$ is a unit.
Finally, we chase through the explicit equations witnessing the statements above.  Letting $v=u^{-1}$, we know that $v((u+a)-a)=1$ so $$-va=1-v(u+a),$$ witnessing that $a$ is a unit mod $u+a$ (with inverse $-v$).  But $a$ is nilpotent, so $a^n=0$ for some $n$, and thus $0$ is also a unit mod $u+a$.  We see this explicitly by raising our previous equation to the $n$th power: $$0=(-v)^na^n=(1-v(u+a))^n=1-nv(u+a)+\binom{n}{2}v^2(u+a)^2+\dots+(-v)^n(u+a)^n,$$ where every term after the first on the right-hand side is divisible by $u+a$.  Factoring out this $u+a$, we find that $$1=(u+a)\left(nv-\binom{n}{2}v^2(u+a)+\dots-(-v)^n(u+a)^{n-1}\right)$$ and so $$-\sum_{k=1}^n \binom{n}{k}(-v)^k(u+a)^{k-1}= nv-\binom{n}{2}v^2(u+a)+\dots-(-v)^n(u+a)^{n-1}$$ is an inverse of $u+a$.
The fact that this complicated formula is hidden in the one-paragraph conceptual argument given at the start of this answer is a nice example of how powerful and convenient the abstract machinery of ring theory can be.
A: 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.
A: Note that since $u$ is a unit and
$ua = au, \tag 1$
we may write
$a = u^{-1}au, \tag 2$
and thus
$au^{-1} = u^{-1}a; \tag 3$
also, since $a$ is nilpotent there is some $0 < n \in \Bbb N$ such that
$a^n = 0, \tag 4$
and thus
$(u^{-1}a)^n = (au^{-1})^n = a^n (u^{-1})^n = (0) (u^{-1})^n = 0; \tag 5$
we observe that
$u + a = u(1 + u^{-1}a), \tag 6$
and that, by virtue of (5),
$(1 + u^{-1}a) \displaystyle \sum_0^{n - 1} (-u^{-1}a)^k = \sum_0^{n - 1} (-u^{-1}a)^k +  u^{-1}a\sum_0^{n - 1} (-u^{-1}a)^k$
$= \displaystyle \sum_0^{n - 1} (-1)^k(u^{-1}a)^k +  \sum_0^{n - 1} (-1)^k(u^{-1}a)^{k + 1}$
$= 1 + \displaystyle \sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \sum_0^{n - 2} (-1)^k(u^{-1}a)^{k + 1} + (-1)^{n - 1}(-u^{-1}a)^n$
$= 1 + \displaystyle \sum_1^{n - 1} (-1)^k (u^{-1}a)^k + \sum_1^{n - 1} (-1)^{k - 1}(u^{-1}a)^k = 1 + \displaystyle \sum_1^{n - 1} ((-1)^k + (-1)^{k - 1})(u^{-1}a)^k = 1; \tag 7$
this shows that
$(1 + u^{-1}a)^{-1} =  \displaystyle \sum_0^{n - 1} (-u^{-1}a)^k, \tag 8$
and we have demonstrated an explicit inverse for $1 + u^{-1}a$.  Thus, by (6),
$(u + a)^{-1} = (u(1 + u^{-1}a))^{-1} = (1 + u^{-1}a)^{-1} u^{-1}, \tag 9$
that is, $u + a$ is a unit.
Nota Bene: The result proved above has an application to this question, which asks to show that $I - T$ is invertible for any nilpotent linear operator $T$.  Taking $T = a$ and $I = u$ in the above immediately yields the existence of $(I - T)^{-1}$.  End of Note.
A: 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.

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.
A: 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.
