$R/Rg$ 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.

Solution 1:

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.

Solution 2:

Use (or prove) the following facts, in order.

• In an integral domain, an element is irreducible iff the ideal it generates is maximal among principal ideals;
• Therefore, in a PID, an element is irreducible iff the ideal it generates is maximal;
• An ideal in a commutative ring is maximal iff the quotient of the ring by that ideal is a field.