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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.

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

2 Answers 2

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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.

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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.
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