4
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ A minimal prime ideal is simply a prime which does not contain any other prime ideal. $\endgroup$ Commented 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
    Commented Sep 12, 2021 at 13:42

1 Answer 1

1
$\begingroup$

From your comment in the comments:

Then it suffices to take some prime ideal 𝑃, 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 𝑃 such that for every prime ideal 𝑄 we have π‘ƒβŠ‚π‘„.

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.

$\endgroup$
1
  • $\begingroup$ Thanks, that is exactly what I was looking for. And yeah, I was confusing minimal with minimum. $\endgroup$
    – Marcos
    Commented Sep 14, 2021 at 13:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .