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 ...
- 50.6k
13
votes
Accepted
Can a Dedekind domain have a power basis over a ring that isn't a Dedekind domain?
If $R\subseteq S$ with $S$ Dedekind and free as an $R$-module, then $R$ is Dedekind because every $R$-ideal $I$ is projective (hence invertible, if non-zero). For $I\otimes_RS$ $\cong$ $IS$ is ...
- 1,388
9
votes
Finding prime ideals for ideal classes in arbitrary Dedekind domains
Some authors add the requirement that a Dedekind domain not be a field.
I will assume that a Dedekind domain is not a field, since otherwise any field is a counter-example.
The answer is no: there are ...
- 7,893
7
votes
Is there a finite extension with a non-trivial class group of any PID?
Counterexample. Let $S$ be an infinite set of primes (of $\mathbf Z$) of density $0$. Let $R$ be the localisation of $\mathbf Z$ away from $S$, i.e. the elements $\tfrac{a}{b}$ with $p \nmid b$ for ...
- 28.7k
6
votes
Accepted
Steinitz isomorphism theorem for non-Dedekind domains
Here is a general fact attributed to Serre. Let $A$ be a Noetherian ring of Krull dimension $d$. If $P$ is a projective module over $A$ of rank larger than $d$, then $P\cong Q\oplus A$. It is easy to ...
- 6,262
5
votes
Accepted
On integral domains over which special kind of modules are projective
An integral domain $R$ such that every proper non-zero $R$-submodule of $\text{Frac}(R)$ is projective is a local principal ideal ring. (The converse is David Handelman's comment above).
Indeed, we ...
- 7,893
4
votes
Finding prime ideals for ideal classes in arbitrary Dedekind domains
Here is a very different sort of example. Let $X$ be a smooth projective curve of genus $\geq 2$ over an algebraically closed field $k$, let $x$ be a single point of $X$, and let $A$ be the coordinate ...
- 156k
4
votes
quotient by ideals and fractional ideals
The reason that you can not find a counterexample is that the statement is true. First notice that the support of both the modules in question is just finitely many primes containing $I$ and thus the ...
- 6,262
4
votes
Accepted
Decomposing Noetherian hereditary rings of Krull dimension $1$ into product of hereditary domains (i.e. Dedekind domains)
Certainly finite products of Dedekind rings with fields work as well. These are the only ones, even without the assumption on the Krull dimension:
Lemma. Let $R$ be a Noetherian (commutative) ...
- 28.7k
3
votes
What is the right level of generality for $(R/a) \times (R/b) \cong (R/\gcd(a,b)) \times (R/\operatorname{lcm}(a,b))$?
Note that unlike the canonical map $R/(I \cap J) \stackrel\sim\to R/I \times R/J$ of the Chinese remainder theorem (whenever $I+J=R$), your isomorphism relies on a choice: for each prime $\mathfrak p$ ...
- 28.7k
3
votes
Algebraic characterization of commutative rings of Krull dimension 1,2, or 3
A remark about the one-dimensional case. By the characterization of Krull dimension given by T. Coquand and H. Lombardi, dim($R$)$\,\leq 1$ iff $\forall_{x,y\in R}\exists_{a,b\in R,m,n\in\mathbb{N}}\,...
- 1,388
3
votes
Accepted
Locally isomorphic algebras over a Dedekind domain
Counterexample. Let $R$ be a Dedekind domain with $\operatorname{Cl}(R) \neq 0$. Let $I \subseteq R$ be an ideal that is not principal (in algebraic geometry language, let $\mathscr L$ be a nontrivial ...
- 28.7k
3
votes
Special submodules over almost Dedekind domains
Without loss of generality, we can assume that $M = R/I$, $x = 1 + I$ where $I$ is a proper non-zero ideal of $R$ and $\mathfrak{m}$ is any maximal ideal of $R$ containing $I$ (recall that $R$ is one-...
- 7,893
3
votes
Accepted
Extension of Dedekind domains and their codifferent
Any $A$-module homomorphism $B/b\to A/a$ gives, by composing with $B\to B/b$, a homomorphism $B\to A/a$. Using the surjection $A\to A/a$ and the fact that $B$ is $A$-projective, the map lifts to give ...
- 6,262
2
votes
Accepted
On $L$-function of permutation representation
It is not easy to give all the details so I'll give a sketch in the case of unramified prime
For $p$ an unramified prime number, $Q\subset O_K$ a prime ideal above $p$, those of $O_k$ are of the form ...
- 3,403
2
votes
Noetherian almost Dedekind domain
In any ring, the ACC on radical ideals is the same as satisfying (i) the ACC on prime ideals and (ii) every element has only finitely many minimal primes. Since almost Dedekind domains are one-...
- 276
2
votes
A question about Dedekind schemes and proper morphisms
Injectivity is because $X$ is separated. The locus where two morphisms $S \to X$ agree is a closed subscheme and if it contains the generic point, it's everything.
For surjectivity, we can "...
- 7,746
1
vote
Steinitz isomorphism theorem for non-Dedekind domains
We answer in the negative the question as to whether there is an $R$-module isomorphism $$I \oplus J \simeq R \oplus IJ$$ whenever $I$ and $J$ are non-invertible ideals of a non-maximal order $R$ of ...
- 7,893
1
vote
An example of a special $1$-dimensional non-Noetherian valuation domain
This isn't possible in any $1$-dimensional quasi-local domain $D$. If $a,b\in D$ and $b$ is not a unit then consider the multiplicative set $S$ generated by $b$. Clearly $S$ is not disjoint from any ...
- 938
1
vote
Accepted
Existence of algebraic integer with absolute value equal to reciprocal of maximum of $1$ and absolute value of a given algebraic number
The answer is already in the comments, but here again for completeness: It indeed follows easily from strong approximation:
Let $S=\{v_1,\dots,v_n\}\cup\{v:|\alpha|_v>1\}$. By the strong ...
- 2,051
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