I have a doubt about one part of this following proposition:
Proposition 4.20: Every local regular ring of dimension one is a principal ideal domain.
Proof: Let $A$ be such a ring, and let $\mathfrak{m}=(\pi)$ be its maximal ideal.
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By assumption, there exists a prime ideal $\mathfrak{p}$ properly contained in $\mathfrak{m}$. ....
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And then the proof goes on, but my doubt is with that part.
If $A$ is a local regular ring of dimension one, as $\mathfrak{m}$ is prime also, it would be the only proper ideal, so if its Krull's dimension is equal to $1$, then the only possible prime ideals are $0$ and $\mathfrak{m}$.
Therefore, when in the proof it is said that exists a prime ideal $\mathfrak{p}$ properly contained in $\mathfrak{m}$... then is necessarily $\mathfrak{p}=0\ \ ?$
You can continue the proof using a general $\mathfrak{p}$, but the proof becomes easier if you use that $\mathfrak{p}=0$. But I am not sure if I am understanding it correctly. So the question would be... is necessarily $\mathfrak{p}=0\ \ ?$
Thanks for your help.
It is not assumed explicitly in the hypotheses of the result that $A$ is a domain. The assumption that $A$ is local implies that every proper (e.g., prime) ideal of $A$ is contained in $\mathfrak{m}$, and the assumption that $A$ has dimension one implies that there must be a prime ideal of $A$ distinct from $\mathfrak{m}$. That the only such ideal is in fact $0$ (which implies that $A$ is a domain) is deduced from the additional hypothesis that $\mathfrak{m}$ is generated by a regular element of $A$.