Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $R$ be a commutative ring with 1, we define

$$N(R):=\{ a\in R \mid \exists k\in \mathbb{N}:a^k=0\}$$

and

$$U(R):=\{ a\in R \mid a\mbox{ is invertible} \}.$$

Could anyone help me prove that if $a\in N(R) \Rightarrow 1+a\in U(R)$?

I've been trying to construst a $b$ such that $ab=1$ rather than doing it by contradiction, as I don't see how you could go about doing that.

share|improve this question
4  
If $a\in N(R)$, then we can find $k\in\mathbb N^*$ such that $a^k=0$. What about $\sum_{j=0}^{k-1}(-a)^j$? –  Davide Giraudo Sep 28 '11 at 10:44
    
@LHS You may want to change your notation: I would not use $a$ to describe $U(R)$ and then say if $a \in N(R)$ ... –  fpqc Sep 28 '11 at 11:36
    
@DBLim Do not know him sorry, but I appear to have 5 mutual friends with him, Wadham i'm guessing? –  Freeman Sep 28 '11 at 13:13
    
I know Adam from a few years ago when I met him at Warwick. Then he was doing his A-levels. I am no longer friends with him on the internet, but a last check reveals we have 16 mutual friends. –  fpqc Sep 28 '11 at 13:50
    
@DBLim Small world ;) I'll keep a look out for him in my lectures! –  Freeman Sep 28 '11 at 14:20
add comment

5 Answers

up vote 5 down vote accepted

This really is a problem in disguise: How did you derive the formula for the sum of the geometric series in year (something) at school??

$(a + 1)(a^{k-1} - a^{k-2} + \ldots 1) = 1-(-a)^n$

but as $a^n = 0$, we have (it does no matter whether $n$ is even or odd) that $(a+1)$ is invertible. Prove the following analogous problem, it may strengthen your understanding:

Let $A$ be a square matrix. If $A^2 = 0$, show that $I - A$ is invertible.

If $A^3 = 0$, show that $I - A$ is invertible.

Hence in general show that if $A^n = 0$ for some positive integer $n$, then $I -A$ is invertible.

share|improve this answer
    
Ah this is excellent, I don't think I would have spotted that! Need to get my game back I think ;) Thanks! –  Freeman Sep 28 '11 at 13:09
    
@LHS Sometimes maths gives the illusion that everything is so complicated; rings, nilradicals, commutative rings, etc and the simple ideas get lost under a heap of terminology.... –  fpqc Sep 28 '11 at 13:48
    
Indeed, just an extra thing, how'd you think this lets you deduce that is $u\in U(R)$ and $a\in N(R) \Rightarrow u+a\in U(R)$? What i'm currently working on! –  Freeman Sep 28 '11 at 14:19
    
As U(R) forms a group with multiplication i've been trying to multiply $a+u$ by things to get $u(a+1)=au+u \in U(R)$ or as a factor.. just use that fact in some way –  Freeman Sep 28 '11 at 14:23
    
@LHS When $n$ is odd, $a^n + b^n = (a+b)(a^{n-1} + \ldots b^{n-1})$. There might be some alternating signs in there. –  fpqc Sep 28 '11 at 21:50
show 2 more comments

Note the following:

$(1+a)(1-a) = 1-a^2$.

$(1-a^2)(1+a^2) = 1-a^4$

$(1-a^4)(1+a^4) = 1-a^8$

...

Thus, continuing in this way, we may find some $b_n$ such that $(1+a)b_n = 1-a^{2^n}$

For large enough $n$, this will give us $(1+a)b_n = 1$.

share|improve this answer
    
can you explain the downvote? is this incorrect? –  anonymous Sep 28 '11 at 11:25
    
No, this is correct, just not obviously how so. $(1+a)$ is a factor of $(1-a^{2^n})$ for $n\geq 1$. If we pick the smallest $n$ such that $a^{2^n} = 0$, taht will yield your result. –  Arthur Sep 28 '11 at 11:54
add comment

$N(R)$ is at least in some texts referred to as the nilradical of $R$. It is contained in all prime ideals (in fact, it is the intersection of all prime ideals, Atiyah, MacDonald prop. 1.8), so taking a nilpotent element $a$, since it is contained in all maximal ideals, $a+1$ is not in any maximal ideal. Then the ideal generated by $a+1$ must neccesarily be the whole ring, which means that it specifically generates 1 at some point.

share|improve this answer
    
Thanks! nice to know the background –  Freeman Sep 28 '11 at 13:08
add comment

HINT $\ $ Note that a nilpotent element $\rm\:n\:$ lies in every prime ideal $\rm\:P\:$, since $\rm\: n^k = 0\in P\ \Rightarrow\ n\in P\:.\:$ In particular, $\rm\:n\:$ lies in every maximal ideal. So $\rm\:n+1\:$ is a unit, since it lies in no maximal ideal $\rm\:M\:$ (else $\rm\:n,n+1\in M\ \Rightarrow (n+1)-n = 1\in M\:$); i.e. elements coprime to every prime are units.

You may recognize a hint of this in proofs of Euclid's theorem that that are infinitely many primes. Namely, if there are only finitely many primes then their product $\rm\:n\:$ is divisible by every prime, so $\rm\:1 + n\:$ is coprime to all primes, so it must be the unit $1\:,\:$ so $\rm\:n = 0\:,\:$ a contradiction.

You'll meet related results later when you study the structure theory of rings. There the intersection of all maximal ideals of a ring $\rm\:R\:$ is known as the Jacobson radical $\rm\:Jac(R)\:.\:$ The ideals $\rm\:J\:$ such that $\rm\:1+J \subset U(R)\:,\:$ i.e are all units, are precisely those ideals contained in $\rm\:Jac(R)\:.\:$ Indeed, we have the following theorem, from my post on the fewunit ring theoretic generalization of Euclid's proof of infinitely many primes.

THEOREM $\ $ TFAE in ring $\rm\:R\:$ with units $\rm\:U,\:$ ideal $\rm\:J,\:$ and Jacobson radical $\rm\:Jac(R)\:.$

$\rm(1)\quad J \subseteq Jac(R),\quad $ i.e. $\rm\:J\:$ lies in every max ideal $\rm\:M\:$ of $\rm\:R\:.$

$\rm(2)\quad 1+J \subseteq U,\quad\ \ $ i.e. $\rm\: 1 + j\:$ is a unit for every $\rm\: j \in J\:.$

$\rm(3)\quad I\neq 1\ \Rightarrow\ I+J \neq 1,\qquad\ $ i.e. proper ideals survive in $\rm\:R/J\:.$

$\rm(4)\quad M\:$ max $\rm\:\Rightarrow M+J \ne 1,\quad $ i.e. max ideals survive in $\rm\:R/J\:.$

Proof $\: $ (sketch) $\ $ With $\rm\:i \in I,\ j \in J,\:$ and max ideal $\rm\:M,$

$\rm(1\Rightarrow 2)\quad j \in all\ M\ \Rightarrow\ 1+j \in no\ M\ \Rightarrow\ 1+j\:$ unit.

$\rm(2\Rightarrow 3)\quad i+j = 1\ \Rightarrow\ 1-j = i\:$ unit $\rm\:\Rightarrow I = 1\:.$

$\rm(3\Rightarrow 4)\ \:$ Let $\rm\:I = M\:$ max.

$\rm(4\Rightarrow 1)\quad M+J \ne 1 \Rightarrow\ J \subseteq M\:$ by $\rm\:M\:$ max.

share|improve this answer
    
A lot to think about there! I am definitely only scratching the surface of this area! –  Freeman Sep 28 '11 at 21:21
add comment

Let $a\in N(R)$ and $k$ such that $a^k=0$. We have \begin{align*}(1+a)\left(\sum_{j=0}^{k-1}(-a)^j \right)&=\sum_{j=0}^{k-1}(-a)^j+\sum_{j=0}^{k-1}-(-a)^{j+1}\\ & =\sum_{j=0}^{k-1}(-a)^j-\sum_{j=1}^k(-a)^j\\ &=1-(-a)^k=1+(-1)^ka^k=1, \end{align*} and we have $\left(\sum_{j=0}^{k-1}(-a)^j\right)(1+a)=1$ by the same computation (it's true even if the ring is not commutative), hence $1+a$ is invertible, and it's inverse is $\left(\sum_{j=0}^{k-1}(-a)^j\right)(1+a)$.

share|improve this answer
1  
This seems a very complicated way of deriving a formula learnt at school... –  fpqc Sep 28 '11 at 11:34
    
@DBLim: Indeed, yours does explain the thought process needed to come to the conclusion, I guess this is another way of expressing the same idea. Thanks though, this is helpful! –  Freeman Sep 28 '11 at 13:10
    
@DavideGiraudo Cette question n'a rien à voir, vous étudiez à quelle université maintenant? Vous avez dit que l'anglais n'est pas votre première langue, le même que le franćais n'est pas le mien. On peut se discuter en anglais/franćais pour que... –  fpqc Sep 28 '11 at 13:55
    
@DavideGiraudo Merci de me dire. –  fpqc Sep 28 '11 at 21:50
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.