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

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


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

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.

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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 $a \in N(R)$ ... – user38268 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. – user38268 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
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.

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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.... – user38268 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. – user38268 Sep 28 '11 at 21:50

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 $(1+a)b_n = 1$.

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can you explain the downvote? is this incorrect? – the L 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

$N(R)$ 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.

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Thanks! nice to know the background – Freeman Sep 28 '11 at 13:08

Hint $\ $ A nilpotent $\rm\,n\,$ lies in every prime ideal $\rm\,P,\,$ because $\rm\, n^k = 0\in P\ \Rightarrow\ n\in P.\,$ In particular, $\rm\,n\,$ lies in every maximal ideal. Hence $\rm\,n\!+\!1\,$ is a unit, since it lies in no maximal ideal $\rm\,M\,$ (else $\rm\,n\!+\!1,\,n\in M\, \Rightarrow\, (n\!+\!1)-n = 1\in M),\,$ 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 $\rm\,n\!+\!1\,$ is coprime to all primes, so it must be the unit $1,\,$ so $\rm\,n = 0,\,$ a contradiction.

Remark $ $ 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 $\rm\,J\,$ with $\rm\,1+J \subset U(R)= $ units of $\rm R,\,$ are precisely those ideals contained in $\rm\,Jac(R).\,$ Indeed, we have the following theorem, excerpted 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.

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

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 $\left(\sum_{j=0}^{k-1}(-a)^j\right)(1+a)$.

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This seems a very complicated way of deriving a formula learnt at school... – user38268 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... – user38268 Sep 28 '11 at 13:55
@DavideGiraudo Merci de me dire. – user38268 Sep 28 '11 at 21:50

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