Show that if $S$ is a PID and $t\in S$ irreducible then $S/(t)$ is a field

Question

Let $S$ be a PID and $t\in S$ irreducible. I'm looking at a question that says show that $S/(t)$ is a field, so I need to show that for any $x\in S/(t)$, $x$ has a multiplicative inverse.

They start by saying that if $x\in S/(t)$ then $t\nmid x$ and $\gcd(t,x)=1$, and then since $S$ is a PID, there exist $a,b\in S$ such that $1=at+bx$ and so $x^{-1}=b$.

I don't understand why $t\nmid x$ and $\gcd(t,x)=1$ or why $b$ is the inverse of $x$

Answer 1

Let $0 \neq \overline{x} \in S/(t)$. We must show that $\overline{x}$ has an inverse.

Since $0 \neq \overline{x}$, we have $x \notin (t)$, i.e. $t \not| x$. But then $gcd(t,x) = 1$. This easily follows because $t$ is irreducible.

By Bézout, there are $a,b \in S$ with $at + bx = 1$. Thus

$$\overline{1}= \overline{at+bx} = \overline{b}\overline{x}$$

and thus $\overline{b}$ is an inverse for $\overline{x}$, as $S$ is commutative.

Answer 2

Can you show in a PID, irreducible elements generate maximal ideal?

Use properties of Principal ideal and definition of irreducible element to prove it.

Now coming to your question, $x \in S/(t)$ and $x\neq 0$ implies $x \notin (t)$. Now, $(t,x)=(d)$ for some $d\in S$ so, $t=du $ for some $u$ in $S$. If $u$ is a unit then $(d)=(t) \Rightarrow x\in (t)$. So, d must be a unit. So, $(d)=(1)$ so you get the last equation.