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 ring, let $I$ be an ideal of $R$, and let $u\in R$ be idempotent modulo $I$ (that is, $u^{2}-u\in I$). Then $u$ can be lifted to an idempotent in $R$ in case there is an idempotent $e$ in $R$ with $e-u \in I$.

I want to show that :

Let $n > 1$ in $\mathbb{N}$. Prove that if $n$ is not a power of a prime, then there exist idempotents modulo $n\mathbb{Z}$ in $\mathbb{Z}$ that cannot be lifted to idempotents in $\mathbb{Z}$

share|improve this question
add comment

2 Answers 2

up vote 1 down vote accepted

If $n$ is not a prime power, then you can write $n=ab$ with $a,b\gt 1$, $\gcd(a,b)=1$.

By the Chinese Remainder Theorem, there exists $x\in\mathbb{Z}$ such that $x\equiv 0\pmod{a}$ and $x\equiv 1\pmod{b}$. Since $x^2\equiv 0\equiv x\pmod{a}$ and $x^2\equiv 1\equiv x\pmod{b}$, it follows that $a|x^2-x$ and $b|x^2-x$, and since $\gcd(a,b)=1$, then $n|x^2-x$. Thus, $x$ is an idempotent modulo $n\mathbb{Z}$.

Now show that neither $0-x$ nor $1-x$ are in $n\mathbb{Z}$.

share|improve this answer
    
@Magidin; Can you also prove that if $R$ is the ring of $n\times n$ upper triangular matrice over a field $\mathbb{Q}$ and if $J$ is the ideal of matrices having zero on the diagonal, then every idempotent modulo $J$ can be lifted to an idempotent in $R$. –  Vahid Dec 14 '11 at 21:10
    
@Vahid: As that is a completely different question, you should post it in a completely different post. You might also want to add why you are considering these questions (homework? assignment?). –  Arturo Magidin Dec 14 '11 at 21:25
add comment

Suppose $n=p^rq$ with $p$ prime not dividing $q$ and $q\gt 1$ (i.e. $q$ divisible by some prime $l\neq p$).
Write a Bézout relation $1=ap^r+bq$ and define $e:=ap^r=1-bq \; $. Then:

The element $\bar e\in \mathbb Z /n\mathbb Z$ is idempotent
Indeed in the isomorphism given by the Chinese remainder theorem $\mathbb Z /n\mathbb Z\simeq \mathbb Z /p^r\mathbb Z \times \mathbb Z /q \mathbb Z$ the element $\bar e =\overline {ap^r}= \overline {1-vq}$ is sent to $(0,1)$

The element $\bar e$ cannot be lifted to $1$ or $0$, the only idempotents in $\mathbb Z$
Indeed it is clearly impossible to have $1= e+xn=ap^r+xp^rq$ .
And it is also impossible to have $0=e+xn = 1-bq+xp^rq $ (since $q$ does not divide $1$).

share|improve this answer
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.