# $R/Rg$ is a field iff $g\in R$ is irreducible.

Let $R$ be a PID and $g\in R$. I want to show:

$R/Rg$ is a field iff $g\in R$ is irreducible.

I.e. I want to show that all $a\notin Rg$ are invertible modulo $g$ iff $g$ is irreducible.

So if I take $a\notin Rg$, how do I use irreducibility of $g$ to find an inverse of $a$, modulo $g$?

This should follow straight from the definition but I am utterly confused.

• This is not true unless you assume more things about $R$. Jul 22, 2016 at 14:27
• What is $R$? Any commutative ring? Jul 22, 2016 at 14:27
• It may be easier to show that $Rg$ is a maximal ideal (I'm assuming $R$ is commutative?) Jul 22, 2016 at 14:27
• @mynameis Far from true. For instance, $R=\Bbb C[x,y]$, $g=x$.
– user228113
Jul 22, 2016 at 14:29
• $R$ is a commutative ring, yes. Actually sorry, $R$ is a PID!! My bad for omitting that vital information. Jul 22, 2016 at 14:29

Hint  Note that for principal ideals: $$\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$$,  i.e. $$(a)\supset (b)\iff a\mid b,\,$$ thus, since generally $$\,R/M\,$$ is a field $$\iff M\,$$ is a maximal ideal, we have
$$\qquad\quad\begin{eqnarray} R/(p)\,\text{ is a field} &\iff& (p)\,\text{ is maximal} \\ &\iff&\!\!\ (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (a)\\ &\iff&\ p\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ a\\ &\iff&\ p\ \ \text{ is irreducible}\\ &\iff&\!(p)\ \text{ is prime,}\ \ \text{by PID} \Rightarrow\text{UFD, so ireducible = prime } \end{eqnarray}$$
Remark $$\$$ PIDs are the UFDs of dimension $$\le 1,\,$$ i.e. where all prime ideals $$\ne 0\,$$ are maximal.