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I was trying the following: Let $R$ be a commutative ring with identity, then $R$ has a unique prime ideal if and only if $R$ has a minimal prime ideal which contains all zero-divisors, and all non-units of $R$ are zero divisors.

While proving necessary part, I stuck at the following place: whether the ideal generated by zero divisors, a proper ideal?

While I am unable to give a rigorous proof for sufficient part. Please somebody explain.

Thank you.

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  • $\begingroup$ do you want to prove "R has a unique prime ideal then R has a minimal prime ideal which contains all zero divisors"? $\endgroup$ – user 1 Apr 5 '15 at 18:57
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The ideal generated by zero divisors is a proper ideal since it has not $1$.


If $R$ has a unique prime ideal then:
1- It is unique maximal, hence have all non-units;
and
2- It is unique minimal prime, hence it is the set of nilpotent elements (which are zero divisors).

So all non-units of $R$ are zero divisors.


conversely


"$R$ has a minimal prime ideal (say $P$) which contains all zero divisors" and since "all non-units of $R$ are zero divisors", $P$ contains all non-units of $R$. Hence $P$ is the unique maximal ideal of $R$. But $P$ is minimal prime. so $R$ has only one prime ideal, namely $P$.

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  • $\begingroup$ In the converse part, I agree that the minimal prime ideal is the unique maximal ideal. But how it is the unique prime ideal as there may be some other prime ideal not containing all the zero divisors. $\endgroup$ – akansha Apr 9 '15 at 15:53
  • $\begingroup$ Sir, P is minimal prime which contains all zero divisors but what I am saying is there may exists a prime ideal which is contained in P but it does not contain all zero divisors. In that case minimality condition of P won't violate. $\endgroup$ – akansha Apr 11 '15 at 15:57
  • $\begingroup$ the maximal ideal is unique because it contains all non-units of R. so every (prime) ideal is contained in it. on the other hand it is minimal. so every (prime) ideal is contained in a minimal prime. so there is only one prime. $\endgroup$ – user 1 Apr 11 '15 at 18:04
  • $\begingroup$ Sir, I think my definition of minimal prime ideal differs from you. What I have read in Hungerford is: A prime ideal P is said to be minimal prime ideal of an ideal I if there does not exist any prime ideal strictly in between I and P. Also, a definition of minimal ideal is given where I is taken to be the trivial ideal. Now I am confused regarding my question that which ideal are they talking about? $\endgroup$ – akansha Apr 15 '15 at 16:11
  • $\begingroup$ Your definition of "minimal prime ideal of an ideal" is correct. BUT, definition of "minimal prime ideal of a ring", which is in question, is different: it is the "minimal prime ideal of the ideal $0$". $\endgroup$ – user 1 Apr 15 '15 at 17:47

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