Linked Questions

1
vote
1answer
167 views

The relation between minimal ideals and zero divisors [duplicate]

How we can prove this Theorem. Let $R$ be a reduced ring. Then $a\in ZD(R)$ (the zero divisors of $R$) if and only if $a\in P$ for some minimal prime ideal $P$.
1
vote
1answer
115 views

Minimal prime ideals consist of zerodivisors [duplicate]

I don't find the proof for this little demonstration ... Let $P$ be a minimal prime ideal of $A$. Show that $P$ is contained in the set of zero divisors of $A$.
0
votes
0answers
100 views

Minimal prime ideals are made of zero-divisors [duplicate]

Let $R$ be a commutative ring with unity which is not an integral domain. Let $P$ be any minimal prime ideal of $R$. How can I show that $P⊆Z(R)$, where $Z(R)$ denotes the set of zero-divisors of $R.$
31
votes
5answers
7k views

Showing the set of zero-divisors is a union of prime ideals

I'm working on an exercise from Atiyah and MacDonald's Commutative Algebra, and have hit a bump on Exercise 14 of Chapter 1. In a ring $A$, let $\Sigma$ be the set of all ideals in which every ...
18
votes
3answers
4k views

Complement of maximal multiplicative set is a prime ideal

Let $R$ be a commutative ring with identity. I've been trying to prove the following: If $S \subset R$ is a maximal multiplicative set, then $R \setminus S$ is a prime ideal of $R$. I have spent ...
5
votes
3answers
2k views

In a reduced ring the set of zero divisors equals the union of minimal prime ideals.

If $R$ is a reduced commutative ring with identity, why is the set $Z$ of zero divisors the union of minimal prime ideals? I know that $Z$ is a union of associated primes, and that the intersection ...
7
votes
2answers
1k views

The set consisting of all zero divisors in a commutative ring with unity contains at least one prime ideal

I'm asked to prove that the set consisting of all zero divisors in a commutative ring with unity contains at least one prime ideal. I can't even start in the proof, I've just defined my set but cant ...
5
votes
3answers
4k views

Not a Zero Divisor

Let $R$ be a commutative ring. Then we say $a \in R$ is a zero divisor if there exists $b \neq 0$ such that $ab = 0$. I want to know what it means to not be a zero divisor. So I tried to negate the ...
6
votes
1answer
364 views

A problem about localization of $\mathbb{Z}/6\mathbb{Z}$ at prime ideal $2\mathbb{Z}/6\mathbb{Z}$

We know that Given a prime ideal $P$ of a commutative ring, there is a one-to-one correspondence between $\lbrace\text{prime ideals }Q\subset P\rbrace$ and $\lbrace\text{prime ideals of } S^{-1}R \...
0
votes
1answer
633 views

show if $P$ is minimal prime ideal of $R$ then every element of $PR_P$ is nilpotent.

Show if $P$ is minimal prime ideal of $R$ then every element of $PR_P$ is nilpotent. The only idea that I come to mind is, we know $PR_P$ is the maximal ideal of $R_P$. Since $P$ is a prime ideal of $...
0
votes
1answer
441 views

What is a minimal prime ideal of a ring

From Wikipedia: A prime ideal P is said to be a minimal prime ideal over an ideal I if it is minimal among all prime ideals containing I. (Note that we do not exclude I even if it is a prime ideal....
-1
votes
2answers
241 views

Why is a non-zero-divisor a unit in zero dimensional ring? [closed]

Let $R$ be a commutative ring of Krull dimension $0$. Let $x\in R\setminus\{0\}$ such that $x$ is non-zero-divisor. Then, how do I prove that $x$ is a unit?
2
votes
3answers
110 views

A proof about prime ideals

Assume $R$ is commutative. Prove that if $P$ is a prime ideal of $R$ and $P$ contains no zero-divisors then $R$ is an integral domain. Proof: let $ab \in P$ where $ab \not= 0$. that means $a \in P$ ...
0
votes
2answers
92 views

Why the dimension of $R/(a)$ is $0$?

How do I see the following fact? If $R$ has dimension $1$, and $a$ is a non-zerodivisor and non-unit, then $R/(a)$ has dimension $0$. That is saying if $P_1\supset P_2\supset (a)$ are two prime ...
1
vote
1answer
351 views

Height of prime ideal containing the variable of a polynomial ring

I have a ring $R$ and a prime ideal $P$ of $S=R[t]$ with $t \in P$. I'm trying to prove that if $\mathrm{ht}(P/tS)$ is finite then $\mathrm{ht}(P) > \mathrm{ht}(P/tS)$. Here $\mathrm{ht}(P)$ ...

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