I am a math student, in the course of abstract algebra we have shown that in a unitary commutative ring every ideal I possesses at least one minimal prime ideal. I am trying to find an example of ideal contents in a non-commutative (not unitary) ring that does not possess a minimal prime ideal.
Example of a ring with no minimal prime ideal
abstract-algebraidealsmaximal-and-prime-ideals
Related Solutions
Claim: $A$ is a ring such that every primary ideal is prime if and only if $A$ is absolutely flat.
Let us say a ring is a PP(primary is prime) ring if every primary ideal is prime.
Suppose $A$ is absolutely flat, then $A$ is PP. (exercise 3 in page 55 of "Introduction to Commutative Algebra" by Atiyah & Macdonald)
We only need to show that if $A$ is PP then $A$ is absolutely flat.
Notice that every primary ideal of $S^{-1}A$ is of the form $S^{-1}\mathfrak{q}$ where $\mathfrak{q}$ is a $\mathfrak{p}$-primary ideal such that $\mathfrak{p}\cap S=\emptyset$. And if $J/I$ (here $J\supset I$) is a primary ideal of $A/I$ then every zerodivisor in $A/J=(A/I)/(J/I)$ is nilpotent, so $J$ is is primary ideal of $A$.
Now it is easy to show that, if $A$ is PP, then $S^{-1}A$ and $A/I$ is PP for a multiplicatively closed subset $S$ and an ideal $I$ and $\mathfrak{m}^2=\mathfrak{m}$ for every maximal ideal $\mathfrak{m}$.
Based on these properties we are able to prove that PP rings are absolutely flat.
We first show every prime ideal in $A$ is maximal. Suppose not, there are two distinct prime ideals $\mathfrak{p}\subset \mathfrak{m}$ where $\mathfrak{m}$ is maximal. Now $B=(A/\mathfrak{p})_{\mathfrak{m}}$ is a PP, local domain(but not a field) we also use $\mathfrak{m}$ to denote the maximal ideal of $B$. Let $0\neq b\in \mathfrak{m}$, suppose $\mathfrak{q}$ is a minimal prime containing $b$, then $C=B_{\mathfrak{q}}$ is a PP, local domain(not a field) and the only maximal ideal of $C$ is minimal over the ideal $(b)=bC$ so $(b)$ is primary and hence maximal, $(bC)^2=bC$, thus $bC=0$. It is impossible. We have proved that every prime ideal in $A$ is maximal.
Let $\mathfrak{m}$ be any prime ideal in $A$, then $A_{\mathfrak{m}}$ is PP. If $A_{\mathfrak{m}}$ is not a field, pick any nonezero $x\in \mathfrak{m}A_{\mathfrak{m}}$, then $(x)$ is primary hence equals $\mathfrak{m}A_{\mathfrak{m}}$, so $\mathfrak{m}A_m=0$, contradiction.
Hence every prime ideal $\mathfrak{m}$ in $A$ is maximal and $A_{\mathfrak{m}}$ is a field, so $A$ is absolutely flat. We are done.
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
- There exist maximal ideals; and
- 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 𝑃⊂𝑄.
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.
Best Answer
$2\mathbb Z/4\mathbb Z$ does not have any prime ideals.
If you want an $R$ such that $R^2\neq 0$, then $2\mathbb Z/8\mathbb Z$.