# Questions tagged [maximal-and-prime-ideals]

For questions about prime ideals and maximal ideals in rings.

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### Long exact sequence of intersected prime ideals

I'm considering a commutative local noetherian ring $R$ (in fact, $R$ is even Cohen-Macaulay) along with a number of prime ideals $I_i$. I can then construct the exact sequence 0\...
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### Why natural map $M \rightarrow \prod M_{\mathfrak{p}}$ makes sense?

I am referring to this post where there is a natural map from $A-$module $M \rightarrow \prod M_p$ for $p$ are maximal ideals of $A$. I understand why the map is injective if it makes sense, but what ...
1 vote
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### Criterion for spanning and linear independence of the Zariski cotangent space of $R_p$ that doesn't use localization or quotient.

Let $R$ be a commutative ring, and $p$ be a prime ideal. Let $a_1, ... a_n \in p$. I'm looking for a criterion that the images of $a_1, ... a_n$ in $pR_p/(pR_p)^2$ are spanning or linearly independent,...
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### if R is a commutative ring in which all the maximal ideals are finitely generated then R is Noetherian? [duplicate]

It is well known that if R is a commutative ring in which all the prime ideals are finitely generated then R is Noetherian. Is this also true for maximal ideals instead of primes: if R is a ...
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• 1,506
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### Minimal elements under division in a ring

Let $p$ be an element of a commutative ring with unity. The following definition is natural: $p$ is minimal under division if its only divisors (up to equivalence) are $1$ and itself. That is, for ...
• 45.9k
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### Show that if a ring $R$ satisfies that for every $x\in R$, $x$ or $1-x$ is invertible then $R$ has a unique max ideal

I tried to prove that if $R$ is like that the group of all not invertible in $R$ is ideal and from there to prove the last but that didnt work for me.
1 vote
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### If P is a prime ideal of a ring R and P contains no non-zero zero divisors, then R is an integral domain

Here, $R$ is a commutative ring with unity. I realize there are many proofs of this online, but all of them seem to assume a definition of zero divisors that Dummit and Foote does not. Here is what ...
• 134
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### Why my proof of every maximal ideal being prime is incorrect

I tried to prove that for a commutative ring $R$ with identity, every maximal ideal $I$ is prime. I think my proof was sort of going in the right direction, but this MSE answer was what I now realize ...
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### Reduced finitely generated k-algebras are isomorphic to $k^n$

We let $k$ denote a fixed algebraically closed field of characteristic zero. We let $R$ denote a reduced finitely generated $k$-algebra where $\dim_kR = n$ as a $k$ vector space and $n$ is a positive ...
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### Prove that if $P$ is a prime ideal of a non-commutative von Neumann regular ring $R$, then $P$ is a maximal ideal

Let $R$ be a ring with identity elements (not necessarily commutative). A ring $R$ is called a von Neumann regular ring if for every $a\in R$ there is $b\in R$ such that $a=aba$. How do I prove that ...
1 vote
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### Prime Ideals of $\mathbb{Z}\lbrack x \rbrack$

I am looking for a bit of help on part of this problem. Let $R = \mathbb{Z}\lbrack X \rbrack$. For each prime $p \in \mathbb{Z}$, let $I_p$ denote the ideal of $R$ generated by $p$ and $X^2+1$. (a) Is ...
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### Prime ideals of pullbacks of commutative rings [closed]

Let $A,B,C$ be commutative rings with given ring homomorphisms $f:A\rightarrow C$ and $g: B \rightarrow C$. Let $A\times_C B:=\{(a,b)\in A \times B : f(a)=g(b)\}$ be their pullback (with the subring ...
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1 vote
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### A unital commutative ring having a square-zero maximal ideal is local

I have the following question in my homework. Let $R$ be a commutative ring with identity and $M$ be a maximal ideal of $R$ such that $M^2=\{0\}$. Show that $M$ is the unique maximal ideal of $R$. ...
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### Prime ideals of the product of all countable integral domains

Let's gather up all the countable integral domains into a set $\vec{R}$ such that each countable integral domain appears exactly once up to isomorphism in $\vec{R}$. Let $\vec{R}$ be indexed by $I$. ...
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### If every prime ideal of $A$ is maximal, then the chain ${}\dots \leq x^2 A \leq x A$ stabilizes, for all $x \in A$. [duplicate]
I was given this question, where $A$ is a commutative ring: If every prime ideal of $A$ is maximal, then the chain ${}\dots \leq x^2 A \leq x A$ stabilizes, for all $x \in A$. Also, the clue: ...