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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.

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why not just work in the subring generated by $a$, which is commutative? –  mt_ May 21 '12 at 9:09
    
You're right. The general case may be reduced to the commutative case. I've made an EDIT. –  Martin Brandenburg May 21 '12 at 9:13

2 Answers 2

up vote 3 down vote accepted

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.

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Note that's the same explicit inverse in my answer. The argument there shows how to find it, vs. pulling it out of hat, as you do. –  Bill Dubuque May 21 '12 at 16:43
    
You are right. The explicit inverse is same in your answer. But the OP has asked the second question in the "Edit" part, I am focus on that question. And they are relevant. –  wxu May 21 '12 at 16:49
    
Out of curiosity, did you have some other way to find the explicit inverse? If so, it might prove of interest to elaborate. –  Bill Dubuque May 21 '12 at 16:51
    
Dear @BillDubuque, I have no. My first trying is to do the baby case: $a(1-a)=0$. And after some efforts, I just see that replacing $t$ by $t^-1$ will aslo give a the same property. And multiplying them I see the hope. –  wxu May 21 '12 at 16:55
    
I elaborated in my answer how I computed the inverse by CRT, and how it works more generally. –  Bill Dubuque May 21 '12 at 17:16

A simple local argument works more generally. Note that $\rm\: c = 1 + a(t-1)\:$ is a unit both $\rm\:mod\ a\:$ and $\rm\: mod\ b = a\!-\!1,\:$ since $\rm\: c\equiv 1\pmod a,\:$ and $\rm\:c\equiv t\pmod{a\!-\!1}.\:$ Now apply

Theorem $\rm\ \ ab\:$ nilpotent $\rm\: \Rightarrow\ (U\!+\!aR)\cap(U\!+\!bR) \subseteq U\ $ where $\rm\:U = $ units of $\rm\:R$

Proof $\ $ An element is a unit iff it lies in no maximal ideal, so it suffices to show that every element $\rm\:c\:$ of said intersection lies in no maximal ideal $\rm\:M.\:$ Suppose $\rm\:c\in M.\:$ By $\rm\:(ab)^n = 0\in M\:$ prime, either $\rm\:a\in M\:$ or $\rm\:b\in M.\:$ But if $\rm\:a\in M\,\:$ then $\rm\:c\in U\!+\!aR$ $\Rightarrow$ $\rm\:c = u + a\:r,\:$ for $\rm\:u\in U,\:r\in R,\:$ thus $\rm\:c,a\in M\:$ $\Rightarrow\:$ unit $\rm\:u = c - a\:\!r\in M,\:$ contradiction. Ditto by symmetry if $\rm\:b\in M.\ \ $ QED

This yields an explicit inverse. With above c, note $\rm\:c(t)c(t^{-1}) = 1 + a(1-a)(t-1)^2 t^{-1},\:$ i.e. $\rm\: cc' = 1\!-\!n,\:$ for $\rm\:n\:$ nilpotent, say $\rm\:n^k = 0.\:$ Multiply both sides by $\rm\:d = 1+n+\cdots+n^{k-1}\:$ to get $\rm\:cc'd = 1-n^k = 1,\:$ yielding $\rm\:c^{-1} = c'd$.

This works generally if $\rm\:(a,b) = 1,\:$ viz. since $\rm\:c\:$ is a unit mod $\rm\:a\:$ and $\rm\:b\:$ we can use CRT to find $\rm\:c' = c^{-1}$ mod $\rm\:ab,\:$ so $\rm\:cc' = 1\!+\!abr = 1\!-\!n.\ $ $\rm\:ab\:$ so $\rm\:n\:$ is nilpotent, so $\rm\: c^{-1} = c'd\:$ as above.

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Thanks. But this is even more unconstructive. I would like to write down the inverse element explicitly. And this should work somehow. –  Martin Brandenburg May 21 '12 at 15:08
    
@Martin Indeed, it does yield an explicit inverse - see above. It can probably be simplified, but I haven't had time to try. –  Bill Dubuque May 21 '12 at 15:58
    
+1 Nice! Great! Could I ask what is "viz"? –  wxu May 21 '12 at 17:27
    
@wxu Thanks. For "viz." see here. –  Bill Dubuque May 21 '12 at 17:41

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