For questions involving prime ideals in commutative or noncommutative rings.

**8**

votes

**0**answers

204 views

### Does this kind of non-noetherian bimodule exist?

Question: Do there exist simple rings $R$ and $S$ (i.e., rings with no proper nonzero ideals) and an $(R,S)$-bimodule $M$ such that
$M$ is finitely generated both as a left $R$-module and a right
...

**2**

votes

**0**answers

1k views

### Prime ideals in polynomial rings over integers

Im trying to find a characterization of the prime ideals in the polynomial ring $R = \mathbb Z[X,Y]$ in two variables over the integers.
Actually I need to find the maximal ideals in quotient rings ...

**2**

votes

**0**answers

292 views

### Orthogonality (wrt. Ext, Tor) in commutative noetherian rings

Hi,
it is a folklore, that:
let $p$, $q$ be two primes of a commutative Gorenstein ring $R$.
$$
\operatorname{Tor}^k(E(R/p), E(R/q)) \neq 0 \iff p = q\mbox{ and }k = \operatorname{height} p.
$$
...

**1**

vote

**0**answers

137 views

### A family of maximal ideals

Let $m_i $, $i \in I,$ be an infinite family of maximal ideals in a commutative ring with identity (it is not supposed to be Noetherian). When does there exist $j \in I$ such that $\cap_{i\not= j} ...

**1**

vote

**0**answers

192 views

### Classification of rings between a PID and its field of fractions?

Let $D$ be a PID and let $\mathrm{Frac}(D)$ be its field of fractions. I want to classify the intermediate rings $D\subseteq R\subseteq \mathrm{Frac}(D)$.
Theorem: Every such ring $R$ is a ...

**0**

votes

**0**answers

53 views

### Monoid prime ideals and prime congruences

I was wondering what the connection is between the notion of "prime congruence" on a monoid, and the notion of "prime ideal" in a monoid. Starting from a prime ideal $P$ in a monoid $M$, one can ...

**0**

votes

**0**answers

101 views

### Can it occur that $q^{ce}$ is a prime ideal (of $S$), while $q^{ce}\neq q $?

Let $R$ and $S$ be commutative rings (with $1$) and $f : R\to S$ be a ring homomorphism. For an ideal $I$ of $R$, set $I^e:=\langle f(I)S\rangle$ (called the extension of $I$ to $S$) and for an ideal ...