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?

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?

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.

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.

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.