non-zero non-invertible polynomial implies zero divisior

Question

Given $F$ is a field and $a \in F[x]$ prove that each non-zero non-invertible element of $F[x]/(a)$ is a zero divisor.

Answer 1

Let $0 \neq \bar{p} \in F[x]/(a).$ Consider any preimage $q \in F[x]$ of $\bar{q}$. (And fix it for the rest of this solution.)

If $p \in F[x]$ and $a$ are coprime, then clearly, $\bar{p}$ is invertible. (There exist polynomial $q, q' \in F[x]$ such that $pq + aq' = 1$ or $\bar{p}\bar{q} = \bar{1}$.)

Thus, if $\bar{p}$ is non-invertible, then $p$ and $a$ are not coprime. Let $d$ be their gcd. (Note that $d$ is not a constant polynomial.)

Consider the polynomial $a' = a/d \in F[x]$. (This is a polynomial as $d \mid a$ in $F[x]$.)
Note that $\deg a' < \deg a$. (Since $d$ is not constant.)
Hence, $a \nmid a'$. In particular, $\bar{a'} \neq 0$.

As $d \mid p$, we can write $p = p'd$ for some $p' \in F[x]$. This gives us $$pa' = p'da' = p'a.$$

Going modulo $a$, we get $$\bar{p}\bar{a'} = 0.$$

As $\bar{a'} \neq 0$, we have the desired result that $\bar{p}$ is a zero divisor.

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Notation: I have used the standard notation that for any $f \in F[x]$, $\bar{f}$ denotes its equivalence class in $F[x]/(a)$.

Answer 2

$V=F[x]/(a)$ is a finite dimensional vector space over $F$ (assuming $a\ne0$).

Let $b \in V$. Then $T: v \mapsto bv$ is a linear transformation $V \to V$. Then $b$ is invertible iff $T$ is invertible:

• If $b$ is invertible with inverse $c$, then $T$ is invertible with inverse $v \mapsto cv$.

• If $T$ is invertible with inverse $S$, then $b$ is invertible with inverse $S(1)$ because $1=TS(v)=bS(1)$.

Therefore, if $b$ is not invertible, then $T$ is not invertible and so $T$ is not injective. Take $v \ne 0$ in $\ker(T)$. Then $0=T(v)=bv$ and so $b$ is a zero divisor.