2
$\begingroup$

I am looking at some old exams, and one thing is to prove that in a quotient ring $R = F[x]/\langle p(x)\rangle$, where $F$ is a field and $\langle p(x)\rangle$ is the ideal generated by a polynomial $p(x) \in F[x]$, that all elements are either units or zero divisors. I have tried looking this up and read several documents on quotient rings, but I haven't seen a mention of this fact.

Here's my attempt at a proof. Suppose $a(x) + \langle p(x)\rangle$ is not a unit. Then this element generates a proper ideal in $R$. This corresponds to a proper ideal in $F[x]$ which contains $a(x)$ and $p(x)$. $F[x]$ is a PID, so this ideal is generated by some other polynomial $f(x)$, which cannot be a unit in $F[x]$. Honestly, I'm not sure where to proceed, and I'm a little tired and scared preparing for this exam.

Any hints would be appreciated, more than a full solution.

Thanks

$\endgroup$

1 Answer 1

3
$\begingroup$

Say you take an element $\overline{q(x)} \in {F(x)/\langle p(x)\rangle}$. You have two cases:

Case I: $q(x)$ does not share any common factor with $p(x)$ i.e. they are coprime to each other. In this case, the Euclidean algorithm gives you that there are $a(x), b(x)$ such that $a(x) p(x) + b(x) q(x) = 1$. This means that

$b(x) q(x) \equiv 1 ($ mod $p(x))$ which exactly means $\overline{q(x)}$ is a unit.

Case II: $q(x)$ shares a factor $c(x)$ with $p(x)$ i.e. $q(x) = c(x) d(x)$ and $p(x) = c(x) e(x)$ for some $d(x), e(x)$.

In this case, you see that $q(x) e(x) \equiv 0 ($ mod $p(x) )$ which means that $\overline{q(x)}$ is a zero-divisor.

$\endgroup$
1
  • $\begingroup$ Thanks! Thread can be closed $\endgroup$
    – Andrea B.
    Commented Jan 10, 2021 at 5:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .