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In my commutative algebra course I was asked to solve the following problem:

(i) Prove that in commutative unital ring $R$ every prime ideal $P$ contains a minimal prime ideal.
(ii) Prove that every commutative unital ring $R$ has a minimal prime ideal

Using Zorn's lemma I was able to prove (i). However I do not understand (ii), because if this is true, that would mean that there is some prime ideal $P$ such that for every prime ideal $Q$ we have $P\subset Q$. Of course if $R$ is an integral domain $(0)$ is a minimal prime ideal. But in $\mathbb{Z}_6$ we can prove that $(2)$ and $(3)$ are prime ideals and $2\notin(3)$, $3\notin (2)$, so that would be a counterexample.

It seems that I am missing something, I'll appreciate some help.

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    $\begingroup$ A minimal prime ideal is simply a prime which does not contain any other prime ideal. $\endgroup$ Sep 12, 2021 at 13:33
  • $\begingroup$ Ohhh, I see. Then it suffices to take some prime ideal $P$, apply (i) and the minimal prime ideal obtained is as in (ii), right? $\endgroup$
    – Marcos
    Sep 12, 2021 at 13:42

1 Answer 1

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From your comment in the comments:

Then it suffices to take some prime ideal ๐‘ƒ, apply (i) and the minimal prime ideal obtained is as in (ii), right?

This will work, yes. If you know that in a ring with identity

  1. There exist maximal ideals; and
  2. maximal ideals are prime.

then you can extract a minimal prime ideal contained in that prime ideal, which would necessarily be a minimal prime in the entire ring.

However I do not understand (ii), because if this is true, that would mean that there is some prime ideal ๐‘ƒ such that for every prime ideal ๐‘„ we have ๐‘ƒโŠ‚๐‘„.

As also discussed in the comments, it seems you are interpreting minimal as minimum (meaning "a prime ideal contained in all other prime ideals").

A minimal prime ideal is simply one that does not properly contain any other prime ideal. In your example, $(2)$ and $(3)$ are both minimal prime ideals of $\mathbb Z_6$, and no minimum prime ideal exists in the ring.

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  • $\begingroup$ Thanks, that is exactly what I was looking for. And yeah, I was confusing minimal with minimum. $\endgroup$
    – Marcos
    Sep 14, 2021 at 13:01

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