I am trying to prove that if $R$ is not a field, then $R[x]$ is not a PID. I can prove the proposition, assuming that $R$ is a commutative unitary ring and an integral domain, but I cannot prove it without assuming that $R$ is an integral domain.
The way I want to prove this, is by looking at the ideal generated by an element $a \in R$ that is not a unit (and $a \neq 0$),and by $x$. So I looked at the ideal $(a,x)$, which is basically just $$\{a_0a + \sum_{i=1}^{n_a}(a_ia)x^i+ \sum_{i = 1}^{n_b}b_ix^i\ |\ a_i,b_i \in R\}$$
Obviously $1 \not \in (a,x)$, but $x \in (a,x)$. So if I take any element $c \in R[x]$, I have the following cases:
- If $c$ is a unit in $R[x]$ then obviously $1\in (c)$, so $(a,x)$ cannot be generated by $c$
- If $c$ is not a unit in $R[x]$, but constant, then $x\not \in (c)$, since we would have to find an element $r \in R $ with $cr = 1$.
- If $c$ is not a unit in $R[x]$ and not constant: Here is what I struggle with. For integral domains $R$ I can show that $x \not \in (c)$ or $a \not \in (c)$, since multiplying two polynomials (by definition) would give me the equalities $r_0c_0 = 0$ and $r_0c_1+r_1c_0 = 1$ for $c = \sum_{i = 0}^m c_ix^i$ and $r_i \in R$. Then using the definition of an integral domain, I would get $r_0 = 0$ or $c_0 = 0$. If $r_0 = 0$ I get a contradiction since $c_0$ cannot be a unit in $R$ if $c$ is not a unit in $R[x]$. If $c_0 = 0$ then obviously $r_0c_0 \neq a$ so $a \not \in (c)$
How can I generalise my prove to general unitary commutative rings?