TITLE: Why is $1+a(t-1)$ a unit?
QUESTION [6 upvotes]: Let $R$ be a (noncommutative) ring and $a \in R$ such that $a(1-a)$ is nilpotent. Why is $1+a(t-1)$ a unit in $R[t,t^{-1}]$? Probably one just has to write down an inverse element, but I could not find it. Perhaps there is a trick related to the geometric series which motivates the choice of the inverse element, which can be actually made into a formal proof because the series is finite since $a(1-a)$ is nilpotent?
Here is a proof that the two-sided ideal generated by $1+a(t-1)$ is $R[t,t^{-1}]$ (which already finishes the proof when $R$ is commutative): Let $A$ be the quotient, we have to show $A=0$. Now, $A$ contains elements $a,t$ such that $a(1-a)$ is nilpotent, $t$ is invertible and $(1-a) + ta = 0$. If we multiply this equation by $a^u (1-a)^v$, we see that $a^{u+1}(1-a)^v=0 \Rightarrow a^u (1-a)^v = 0$ as well as $a^u (1-a)^{v+1} = 0 \Rightarrow a^u (1-a)^v = 0$ for all $u,v \geq 0$. Since $a(1-a)$ is nilpotent, we get by induction that $a$ and $1-a$ are nilpotent. But $a$ nilpotent implies that $1-a$ is a unit, so it can only be nilpotent when $A=0$.
As mt_ mentions below, the general case may be reduced to the commutative case by working with the commutative subring $\mathbb{Z}[a] \subseteq R$. Anyway, I would like to know if there is a short proof which just writes down the inverse.
EDIT: I would also like to know why the converse is true: When $1+a(t-1)$ is a unit, why is $a(1-a)$ nilpotent? Rosenberg claims this in his book without proof (he only says "by the same reasoning as in ...", but this doesn't work the same!). Here, we cannot assume that $R$ is commutative, since the inverse may contain coefficients which do not lie in $\mathbb{Z}[a]$.
Background: This is needed in the proof of the Bass-Heller-Swan Theorem.

REPLY [2 votes]: Should be read first: If $R$ is commutative, then both side is easy. Pick any  prime ideal $\mathfrak{p}$ of $R$, $1+a(t-1)=1-a+at$ is a unit in $R/\mathfrak{p}[t,t^{-1}]$ if and only if either $1-a=0$ or $a=0$ in $R/\mathfrak{p}$. So if and only if $(1-a)a\in \mathfrak{p}$.

Since $(1+a(t-1))(1+a(t^{-1}-1))=1+a(1-a)(t+t^{-1}-2)$ and $a(1-a)$ is nilpotent, it follows $1+a(t-1)$ is a unit.
Conversely, if $1+a(t-1)$ is a unit, then $1+a(t^{-1}-1)$ is also a unit. (mapping $t\to t^{-1}$ gives a ring map of itself). So their product $1+a(1-a)(t+t^{-1}-2)$ is a unit. Now, $R[t,t^{-1}]\to R[t,t^{-1}]$ sending $t$ to $t^2$, we know $1+a(1-a)(t^2+t^{-2}-2)=1+a(1-a)(t-\frac{1}{t})^2$ is a unit. We want to show $x=a(1-a)$ is nilpotent.
Just write down the inverse of $1+a(1-a)(t-\frac{1}{t})^2$, say $s=b_{-k}t^{-k}+\ldots+b_0+\ldots+b_nt^n$ is the inverse. Then mapping $t\to t^{-1}$, we know that the inverse has very good property: $b_{-i}=b_i$ for all $i$ and no odd terms, i.e., $s$ indeed polynomial of $(t-1/t)$. 
OK, Consider the ring map $R[Y]\to R[t,t^{-1}]$ sending $Y$ to $t-1/t$. We come to the case: $1+xY^2$ is a unit of $R[Y]$, we want to show $x$ is a nilpotent element. This is true of course.

Edit: It seems I have made it comlicated. Go to the step: $1+x(t+t^{-1}-2)=1-2x+x(t+t^{-1})$ is a unit. Mapping $t$ to $t^{-1}$, we see the inverse of $1-2x+x(t+t^{-1})$ is a polynomial of $t+t^{-1}$. Now $R[Y]\to R[t,t^{-1}]$ sending $Y\to t+t^{-1}$ is injective. Come to the case $1-2x+xY$ is a unit in $R[Y]$. 
Finally, if we are in the world of commutative rings, then modulo every prime ideal of $R$, to see $x$ is nilpotent. If we are not in the world of commutative rings, just write down the inverse and compare the coefficients to see $x$ is nilpotent.