Let D be a principal ideal domain. Prove that every non-zero prime ideal in D is a maximal ideal in D.

So I'm think I need to use the fact that all PID's are UFD's. If it is a UFD I can infer that every irreducible element in D is prime. Does that mean it is itself irreducible though? If so can I conclude thus a field, and thus must be maximal?


Hint $\ $ For principal ideals: $\,\ \rm\color{#0a0}{contains} = \color{#c00}{divides},\,$ i.e. $\,(a)\supset (b)\iff a\mid b,\,$ hence

$$\begin{align} (p)\text{ is prime}\ \Rightarrow&\,\ \ p\ \text{ is prime}\\ \Rightarrow&\,\ \ p\ \text{ is irreducible}\\ \Rightarrow&\ \ \,p\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ a\\ \Rightarrow&\ (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (a)\\ \Rightarrow&\ (p)\, \text{ is maximal} \end{align}\qquad$$


Note that for the statement to be true, you need to require the ideal to be non-zero.

Showing this statement is straightforward from the definitions. If you use the fact that every PID is a UFD, your reasoning might be circular.

Let $(x)$ be a non-zero prime ideal of a PID $R$. Suppose $(x) \subset (y)$ for some $y \in R$. Then $x = ay$ for some $a \in R$. Since $(x)$ is prime, either $a \in (x)$ or $y \in (x)$.

If $a \in (x)$, then $a = bx$ for some $b \in R$ and $x = bxy$. Since $x \ne 0$, if follows that $y$ is a unit. Thus, $(y) = R$.

If $y \in (x)$, then $(x) = (y)$. We conclude that $(x)$ must be maximal.


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