2
$\begingroup$

Let R be a commutative ring with identity and $f(x) \in R[x]$ be monic.

In general, I am having trouble with identifying what $R[x] \over (f(x))$ is when R is not a field. What "tools" exist that can help me identify what $R[x] \over (f(x))$ is when R is not a field? In particular, I am unable to solve the following problem.

Let $R= \Bbb Z$. Prove if $\{a \in {R[x] \over (f(x))}:a^2=a\}=\{0,1\}$, then $f(x)=(q(x))^n$ for some monic, irreducible $q(x) \in R[x]$. I tried viewing the quotient in terms of cosets but got nowhere.

$\endgroup$
  • 3
    $\begingroup$ Would you be able to solve the problem if $R$ were a field? $\endgroup$ – Ted Aug 18 '13 at 5:22
  • $\begingroup$ Alex, this problem is incorrect for $R=\mathbb{Z}$. See counterexample in comments below Wild Chan's answer. $\endgroup$ – Ted Aug 18 '13 at 6:44
1
$\begingroup$

First $f$ is monic, then the division with remainder always works on any commutative ring with identity. For any $g\in R[x]$, we have $q,r\in R[x]$ so that $$g=fq+r,\quad \deg r<\deg f$$ This could be proved by induction on the degree of $g$. Using this assertion, we can easily prove that in any equivalent class of $R[x]/\langle f(x)\rangle$, there's a polynomial whose degree is smaller than $f$. This can be the represention element. (However no use for your problem.)


Now for your problem. In order to make $g(x)$ meet the requirements, there should be $$g(x)(g(x)-1)=f(x)h(x)$$ where $h\in R[x]$. If the conclusion is false, then we can decompose $f$ into two relatively prime factors: $f(x)=p(x)q(x)$ (because of unique factorization). $p,q$ are both monic, and neither of them is $1$. Then by Bézout's Identity we have $$p(x)u(x)-q(x)v(x)=1$$ Now we let $g=pu$, and $h=uv$. You can verify that the equivalent class contianing $g$ cannot be $0$ nor $1$, where contradiction gained.

The most important part is Bézout's Identity, and it works in any principal ideal ring, no need for the ring to be a field.

$\endgroup$
  • 2
    $\begingroup$ But $\mathbb{Z}[x]$ is not a principal ideal domain. Bezout's identity doesn't hold. For example, no linear combination of $x+1$ and $x-1$ could be 1 in $\mathbb{Z}[x]$: If $(x+1)f(x) + (x-1)g(x) = 1$ then substituting $x=1$ leads to a contradiction. $\endgroup$ – Ted Aug 18 '13 at 6:22
  • $\begingroup$ @Ted Yes, you are right. But it seems that $f=x^2-1$ is a counterexample for Alex's problem...? $\endgroup$ – Willard Zhan Aug 18 '13 at 6:36
  • $\begingroup$ Yep, looks like it is a counterexample. Letting x=1 gives g(1) is either 0 or 1. Letting x=-1 gives g(-1) is either 0 or 1. If they are both 0 or both 1, then we get the trivial idempotents 0 and 1. (Since f is monic quadratic, we can assume g(x) = a+bx.) The other 2 cases cannot happen since g(1) - g(-1) is always even. $\endgroup$ – Ted Aug 18 '13 at 6:42
  • $\begingroup$ I thought up the exact same counter example when R is a field but, I assumed I was missing something when $R=\Bbb Z$—mainly because I cannot think of the quotient as adjoining roots unless R is a field. This problem was on an exam and was the only problem I could not solve, so bugged me for hours afterwards. $\endgroup$ – Just Some Old Man Aug 19 '13 at 3:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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