15 votes
Accepted

Quotients of number fields by certain prime powers

I remember working this out 25 years ago. The main idea is to view both rings as quotient rings of completions: $\mathcal O_K/\mathfrak p^e \cong \widehat{\mathcal O_{\mathfrak p}}/\widehat{\mathfrak ...
KConrad's user avatar
  • 49.5k
12 votes
Accepted

Automorphisms of rings fixing all prime ideals

The condition just says that for each prime $\mathfrak{q}$ of $B$, both composite maps $A\rightrightarrows B\to B/\mathfrak{q}$ are equal. Hence for each $x\in A$, $f(x)-g(x)$ belongs to every prime ...
Laurent Moret-Bailly's user avatar
11 votes
Accepted

Constructive proof that a kernel consists of nilpotent elements

This answer provides a scheme how to construct a constructive proof, though I'm still working to actually explicitly extract the constructive proof, so please don't accept the answer just yet. (Update:...
Ingo Blechschmidt's user avatar
10 votes
Accepted

Is it true that this ideal must be principal? (proof verification)

I claim that under the given circumstances $\mathfrak{P}$ is not necessarily principal, i.e., the statement claimed in the question is wrong. Here is a counterexample. Consider $K = \mathbb{Q}$ and $L ...
AlexIvanov's user avatar
10 votes
Accepted

Density of prime ideals of a given degree

This is fairly classical, and the answer is zero for any $n>1$, and $1$ for $n=1$. The reason basically comes down to the fact that asymptotic-wise, almost all prime powers are prime. Here is the ...
Wojowu's user avatar
  • 27.3k
9 votes
Accepted

What is the effect of imaginary quadratic extension on a quaternion algebra's ramified primes

The base change of your quaternion algebra will be ramified at $\mathfrak{P}$ if and only the degree of the extension of completions $K_{\mathfrak{P}}/F_{\mathfrak{p}}$ has odd degree, i.e. case 3. A ...
Aurel's user avatar
  • 4,878
8 votes
Accepted

When $C (X) $ is zero dimensional

I had written this as a comment, but since the discussion is now a bit confused, it is best to write it as an answer. The completely regular spaces $X$ such that the ring $C(X)$ is zero-dimensional (...
Gro-Tsen's user avatar
  • 29.8k
8 votes
Accepted

Rings with all non-prime ideals finitely generated

No. If $\mathbb Z_{p^{\infty}}$ is a Prufer group for prime $p$, then its endomorphism ring is isomorphic to the ring $\mathbb Z_p$ of $p$-adic integers. Hence $\mathbb Z_{p^{\infty}}$ is a $\mathbb ...
Keith Kearnes's user avatar
7 votes

When does prime elements remain prime in certain integral extension

This is true when $R$ is reasonable. The properties that I use are: $R$ is Noetherian; $\tilde R$ is Noetherian; $\tilde R$ is catenary and equidimensional (i.e. every maximal chain $0 = \mathfrak ...
R. van Dobben de Bruyn's user avatar
7 votes
Accepted

For every prime ideal $P$ of any Cohen-Macaulay ring $R$, is the sequence $\operatorname{depth}(R/P^n)$ eventually constant?

Yes, for any ideal in a Noetherian local ring. See: this paper.
Hailong Dao's user avatar
  • 30.3k
6 votes
Accepted

Closed prime ideal in $C[0, 1]$

No. Note that an ideal in a commutative ring with identity is prime if and only if the quotient ring is an integral domain. Now consider $C[0,1]$. It is known that the closed ideals in this Banach ...
Yemon Choi's user avatar
  • 25.5k
5 votes

A property of minimal prime ideals in commutative reduced ring

This is an interesting question. Quentel's Example provides an example. Let $R$ be a reduced ring, $Min(R)$ its space of minimal prime ideals, and $q(R)$ its classical quotient ring. Theorem The ...
W.McGovern's user avatar
5 votes
Accepted

Is the annihilator of a minimal prime ideal principal?

This is false. To see why, consider the following lemma. Lemma. Let $R$ be a Noetherian ring with exactly two minimal primes $\mathfrak p$ and $\mathfrak q$ such that $\mathfrak p \mathfrak q = 0$. ...
R. van Dobben de Bruyn's user avatar
4 votes

When an intersection is contained in a minimal prime ideal

I would extend Jason's comment to say that this condition trivially holds in any Artinian ring since in those rings every intersection of ideals is a finite intersection. It seems that indeed this is ...
Sándor Kovács's user avatar
4 votes
Accepted

On maximal ideals of $k [X_i : i \in I ] $ where $k$ is a field , $I$ is an infinite set with $|k| > |I|$

You are right that the algebraicity of $R/\mathfrak m$ is important. Indeed, suppose $\mathfrak m \cap k[X_i] = 0$. That means that the map $k[X_i] \to R/\mathfrak m$ is injective. But then (the image ...
R. van Dobben de Bruyn's user avatar
4 votes
Accepted

GCD and LCM of elements in Prufer domain

In fact, for any given nonzero $a$ and $b$, if $Ra\cap Rb$ is principal so is $Ra+Rb$. Here is one way to see it (surely there must be a more down-to-earth proof). Without assuming $Ra\cap Rb$ ...
Laurent Moret-Bailly's user avatar
4 votes
Accepted

prime ideals minimal over a zerodivisor

For the first question, consider $R = \mathbb{Z}[X]/(X^3 - 1)$, $P = (x+ 1)$ and $a = (x + 1)(1 + x + x^2)$ where $x$ denotes the image of $X$ in $R$. In order to see that it provides us with a ...
Luc Guyot's user avatar
  • 7,353
4 votes
Accepted

Is every universally catenary ring a going-between ring?

OK, let $R \subset S$ be an integral ring extension with $R$ universally catenary. Let $\mathfrak q \subset \mathfrak q'$ be primes in $S$ such that there is no prime strictly in between them. We have ...
darx's user avatar
  • 56
4 votes

$G_{\mathbb Q}$ and primes of $\overline{\mathbb{Z}}$

There are infinitely many prime ideals $\mathfrak p$ in $\overline{\mathbf Z}$ that lie over $p$ since you can find an arbitrarily large (finite) number of prime ideals lying over $p$ in suitable ...
KConrad's user avatar
  • 49.5k
4 votes
Accepted

Irreducibility of an explicit complex projective variety

Let me explain how to show that the projective surface $\Sigma$ is geometrically irreducible (see also the comments above). First, we know that $\Sigma$ is irreducible (this was checked by the OP). ...
Ariyan Javanpeykar's user avatar
3 votes

Commutative rings with unity over which every non-zero module has an associated prime

There exists a non-noetherian ring $R$ such that every non-zero $R$-module has an associated prime. This in proven in Example 2.3 in P. J. Cahen, Ascending chain conditions and associated primes, ...
Fred Rohrer's user avatar
  • 6,650
3 votes

Constructive proof that a kernel consists of nilpotent elements

Let us use $k_\min$. The references we will give are in the book Commutative Algebra. Constructive methods (Lombardi-Quitté) (arXiv:1605.04832v1). We have to prove the following. Let $k \to A$ ...
user42641's user avatar
3 votes

Constructive proof that a kernel consists of nilpotent elements

(Edit: This proof is incorrect, but see the comments below.) I will use Lemma 6.4 of Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry (used in the proof of the equational ...
Minseon Shin's user avatar
  • 1,987
3 votes

Prime ideals of formal power series ring that are above the same prime ideal

In general, $(P, X)$ is not the only prime containing $P[[X]]$ and contracting to $P$. I don't have anything to say about the problem of characterizing such primes, but in general it seems extremely ...
Badam Baplan's user avatar
2 votes

When a finitely generated ideal is contained in a union of maximal ideals

In the example in my comment, the ideal $I$ is not finitely generated. Here is an example with $I$ finitely generated (this example also illustrates why the Prime Avoidance Theorem only holds for ...
2 votes

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

The question is elementary and was already answered in the comments. I'm posting a cw answer so that the question can be ticked as answer (otherwise it remains regularly bumped). Let $I_D$ be the set ...
2 votes
Accepted

How bad does a ring have to be for a failure of "going-in-between"?

This is a partial answer, giving a two dimensional Noetherian counterexample. It starts from your observation that one has to consider the case where $A/\mathfrak{p}_0$ is not integrally closed. ...
pinaki's user avatar
  • 4,982
2 votes

Zero -dimensional commutative semiprimitive rings

Several nice characterizations for these rings are worked out as Exercise 4.15 in Lam's book "Exercises in Classical Ring Theory." These include (for commutative rings): (A) $R$ is reduced and $K$-...
Pace Nielsen's user avatar
2 votes

Zero -dimensional commutative semiprimitive rings

These are exactly the zero-dimensional reduced commutative rings (a.k.a. "absolutely flat rings"). Clearly semiprimitive rings are reduced. [EDIT: what follows is correct but much too complicated. ...
Laurent Moret-Bailly's user avatar
2 votes
Accepted

Is $Hom_R(S_X^{-1}R, E)$ the minimal injective cogenerator of $S_X^{-1}R$?

I give a counter-example. Let $R$ be a semi-local domain with two maximal ideals $\mathfrak{m}$ and $\mathfrak{n}$. So the minimal injective generator module is $E = E(R/\mathfrak{m}) \oplus E(R/\...
Pham Hung Quy's user avatar

Only top scored, non community-wiki answers of a minimum length are eligible