# Quotient Rings of Polynomials Over Rings that are NOT fields

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.

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

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.

• 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. – Ted Aug 18 '13 at 6:22
• @Ted Yes, you are right. But it seems that $f=x^2-1$ is a counterexample for Alex's problem...? – Willard Zhan Aug 18 '13 at 6:36
• 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. – Ted Aug 18 '13 at 6:42
• 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. – Just Some Old Man Aug 19 '13 at 3:36
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