User hagen - MathOverflow

Answer by Hagen for Which clustering algorithm could I use to group 2D points that are arranged over a time series?

2013-01-15T12:17:34Z

Hello Julius.

The proposal of Felix Goldberg will only work, if the data are sampled in such a way, that the time intervals between successive (in time) data are correlated with the spatial data. Essentially short time intervals should imply spatially small movements. In particular the proposal will not work (without further processing) if the data are equidistantly sampled.

You may want to take a look at the so called QT-algorithm (Quality threshold) for clustering (wikipedia describes it). You can apply it without prior specification of the number of clusters but specifying for example the cluster diameter. You could try to cluster the spatial data first. Then you could analyse the various clusters separately with respect to time: points with subsequent time stamps form one cluster, gaps in time signal the passage to a new cluster at the same location.

Of course whether this approach works or not again depends on the sampling structure ... If you have enough data and the clusters you search for are typically big you could force an equidistant sampling by leaving out some samples. If the movement in space is not too erratic you could also try to replace points by interpolated points thus again forcing equidistant time intervals.

Moreover the QT-algorithm has problems with large data sets. Maybe you have to take some kind of stochastic speed up into account.

Answer by Hagen for Henselization of valued field

2013-01-10T08:11:08Z

In general the answer is no: let $(K,v)$ be non-henselian and let $v_x$ be the Gauss extension of $v$ to the rational function field $K(x)$:

$v_x(a_nx^n+\ldots +a_0):=\min (va_i : i=0,\ldots ,n)$.

Then $(L,v_x)$ is non-henselian for every subfield $L\subseteq K(x)$.

Proof: by assumption there exists a monic polynomial $f\in O_v[z]$ such that the reduced polynomial $\overline{f}\in k[z]$, $k$ the residue field of $O_v$, has a simple root $\bar{\alpha}$ in $k$ but $f$ has no root in $K$ that reduces to $\bar{\alpha}$. This property of $f$ does not change when considering $f$ as a polynomial in $(O_{v_x}\cap L)[z]$: $K$ is algebraically closed in $K(x)$ and thus in $L$. The residue field of $O_{v_x}$ is $k(x)$, hence $k$ is algebraically closed in the residue field of $O_{v_x}\cap L$.

Answer by Hagen for Criterions for Reflexiveness of sheaves and a special case

2012-01-30T13:41:11Z

Let $O$ be the valuation ring of a valuation $v$ having a subgroup of the reals different from the integers as its value group. Let $v_1>v_2>\ldots >0$ be a monotonously decreasing sequence of values converging to $0$ and chose elements $x_i\in O$ such that $v(x_i)=v_i$. Then $O x_1\subset O x_2\subset\ldots \subset M$, where $M$ is the maximal ideal of $O$. Moreover

$M=\bigcup\limits_{i}Ox_i$.

But $M$ is not reflexive: $(O:M)=O$, hence $(O:(O:M))=(O:O)=O$.

However: this may only tell us that we should define reflexivity for non-coherent modules / sheaves in a different way.

Answer by Hagen for When is the tensor product of two fields a field?

2011-11-29T11:37:59Z

A remark concerning Georges general question: if both $K$ and $L$ are separable algebraic extensions, and $k\subseteq K$ is finite, then $K\otimes_k L = L[x]/fL[x]$, where $f\in k[x]$ is the minimal polynomial of a primitive element of $K$. So to decide whether the tensor product is a field amounts to deciding whether $f\in L[x]$ is irreducible. This seems to indicate that an answer to Georges questions highly depends on the nature of the field $k$, more precisely on its Galois theory.

Note also that if $k\subseteq K$ is separable algebraic and $k\subseteq L$ is purely inseparable, then regardless of the embeddings into the algebraic closure of $k$, the resulting extensions are linearly disjoint over $k$. Hence the tensor product is a field in this case. I guess that one can exploit this fact to reduce the whole problem to considering two separable algebraic or two purely inseparable extensions.

Answer by Hagen for an easy example of valuation ring which is not noetherian？

2011-11-29T09:26:20Z

Construction of valuation domains of Krull dimension $>1$:

Let $O\neq K:=\mathrm{Frac}(O)$ be a valuation domain. Consider the natural map $h:O\rightarrow k$, where $k$ is the residue field of $O$. Let $\overline{O}$ be a valuation domain of $k$. Then $O^\prime:=h^{-1}(\overline{O})\subseteq O$ is a valuation domain of $K$ with the following properties:

$\mathrm{Spec}(O)\subset\mathrm{Spec} (O^\prime)$,
$O=O^\prime_M$, where $M$ is the maximal ideal of $O$,
$O^\prime/M\cong\overline{O}$.

In particular: $O^\prime$ is never noetherian.

Answer by Hagen for Is every poset the poset of prime ideals of a ring?

2011-04-28T15:53:23Z

Beside considering the article by Hochster you could also take a look at

WILLIAM J. LEWIS, The Spectrum of a Ring as a Partially Ordered Set, JOURNAL OF ALGEBRA 25, 419-434 (1973).

It is online:

www.maths.manchester.ac.uk/.../Lewis-Thespectrumofaringasapartiallyorderedset.pdf

Answer by Hagen for Dimensionality of a map -- distance

2011-04-28T14:03:08Z

I guess what you want to say is this: one defines a distance function $d$ on the set ${ A,B,C,D }$ of four elements through setting $d(A,A)=0$, $d(A,B)=1$, $d(A,C)=2$ and so on. Of course one can put the various values of the distance function into a matrix as you did, but if I understaand your problem correctly this is not the point.

What you call "one-dimensional" is the fact that one can find a map $f: {A,B,C,D}\rightarrow\mathbb{R}$ that preserves the distances using the ordinary absolute value in the reals to define the distance between to numbers. So this map satisfies for example $d(A,B)=|f(A)-f(B)|$, and similar for the other point combinations. Of course this is a particular property of the map $d$: if you would define a distance map $e$ through $e(A,B)=1$, $e(B,C)=1$ and $e(A,C)=1$ then you could not find such a map $f$.

One way to generalize this problem is this one: consider a finite metric space $(X,d)$, that is a finite set equipped with a function $d:X\times X\rightarrow\mathbb{R}^{\geq 0}$ satisfying the following properties: $d(x,x)=0$ for all $x\in X$, $d(x,y)=d(y,x)$ for all $x,y\in X$ and $d(x,y)\leq d(x,z)+d(z,y)$ for all $x,y,z\in X$. Does there exist a map $f:X\rightarrow\mathbb{R}^n$ for some $n\in\mathbb{N}$ that preserves the distances taking the euclidean distance in the space $\mathbb{R}^n$. Such problems are treated in the theory of finite metric spaces. They occur naturally in Data Mining or theoretical computer science.

Answer by Hagen for When is the degree of the pull-back of a Weil divisor a constant multiple of its degree?

2011-04-26T11:17:45Z

In the situation you consider the degree $\mathrm{deg}(f)$ is equal to the degree $[K(X):K(Y)]$ of the extension of the function fields of $X$ and $Y$. The precise requirements are that $X$ and $Y$ are normal varieties and that $f$ is finite. The base field is not required to be algebraically closed.

For varieties of dimension $\geq 2$ there is no degree function for Weil divisors comparable to the one in the case of curves. Your definition(?) of $\mathrm{deg}(f^\ast Q)$ looks at least strange to me. How would you define the degree of an arbitrary Weil (prime) divisor of $X$? There should then be no reference to $f$ ...

Answer by Hagen for Algebraic applications of Hurwitz' theorem

2011-04-18T12:13:11Z

As far as I remember the Riemann-Hurwitz-formula is used to prove the inequality

$|\mathrm{Aut}(F|K)|\leq 84(g-1)$

for the number of automorphisms of an algebraic function field $F$ of one variable over $K$, where $K$ has characteristic $0$ and $g\geq 2$ holds for the genus of $F|K$.

Answer by Hagen for divisorial part of an ideal

2011-03-29T09:00:32Z

Suppose that $X$ is affine: $X=\mathrm{Spec}(A)$. Then $I$ is an ideal of the noetherian, integrally closed ring $A$ (this property in fact is all you need, smoothness is not required). The ring $A$ can be written as

$A=\bigcap\limits_{p\in\mathrm{Spec} (A):\mathrm{height}(p)=1}A_p$.

The divisorial hull of $I$ is defined as

$\widehat{I}=\bigcap\limits_{p\in\mathrm{Spec} (A):\mathrm{height}(p)=1}IA_p$.

The ideal $\widehat{I}$ then defines a Weil divisor $D$ on $X$ in the usual way: namely

$v_p(D):=v_p(f_p) $ for $p\in\mathrm{Spec} (A):\mathrm{height}(p)=1$,

where $v_p$ is the discrete valuation associated to the valuation ring $A_p$ and $\widehat{I}A_p=f_pA_p$. This divisor is the one you are searching for.

Answer by Hagen for Maximal separable extensions of residue fields

2011-03-17T12:17:59Z

No answer, just a simple starting point: forget about the rings and look at two finite extensions $l_1/k$ and $l_2/k$. Then the separable closure of $k$ in the compositum $l_1.l_2$ indeed is the compositum of the separable closures $l_1^s$ and $l_2^s$ of $k$ in $l_1$ and $l_2$: by definition the extension $l_1/l_1^s$ is purely inseparable, hence the extension $(l_1.l_2^s)/(l_1^s.l_2^s)$ is purely inseparable too. The same holds for $(l_2.l_1^s)/(l_1^s.l_2^s)$ hence for $(l_1.l_2)/(l_1^s.l_2^s)$.

So one has to search for counterexamples in the case $k(m_D )\neq k(m_B ).k(m_C )$. For example one could start with a case in which the ring compositum $B_{m_B}\cdot C_{m_C}$ is not normal.

Answer by Hagen for Valuations and separable extensions

2011-02-22T15:45:29Z

What about this: we have to prove that $K$ and every purely inseparable extension $l/k$ are linearly disjoint over $k$.

Let $x_1,\ldots ,x_r\in l$ be $k$-linearly independent elements and assume $0=a_1x_1+\ldots +a_rx_r$ for some elements $a_i\in K$. We can divide by the coefficient $a_j$ with the least value and thus assume $a_1,\ldots ,a_r\in R$ with at least one coefficient being a unit of $R$.

There is a unique valuation ring $S$ of $K.l$ dominating $R$. Taking residues modulo the maximal ideal of $S$ yields a linear combination $0=\overline{a_1}x_1+\ldots +\overline{a_r}x_r$ in the field $F.l$. The separability of $F/k$ yields $\overline{a_i}=0$ for all $i$, a contradiction.

Answer by Hagen for Two questions about finiteness of ideal classes in abstract number rings

2010-10-21T09:48:50Z

In the article R. C. HEITMANN, PID’S WITH SPECIFIED RESIDUE FIELDS, Duke Math. J. Volume 41, Number 3 (1974), 565-582 the author shows (Thm A):

For every countable set $F$ of countable fields with the property that for every prime $p$ the set $F$ contains only finitely many fields of characteristic $p$, there exists a countable principal ideal domain $R$ of characteristic $0$ such that the set $F$ consists precisely of all residue fields (with respect to maximal ideals) of $R$.

The construction he uses to prove the theorem seems to give domains $R$ such that the extension $K/\mathbb{Q}$ of the field of fractions $K$ of $R$ over the rationals is finitely generated but not necessarily algebraic. (Unfortunately I do not have access to the complete article :c

Answer by Hagen for How badly can Krull's Hauptidealsatz fail for non-Noetherian rings?

2010-10-18T08:59:19Z

Valuation rings demonstrate quite clearly the failure of Krull's principal ideal theorem: take a valuation ring O of finite dimension. The prime ideals then form a chain

$p_0:=0\subset p_1\subset\ldots\subset p_d$

so that for every $i\in{1,\ldots ,d}$ there exists $r_i\in p_i\setminus p_{i-1}$. Obviously $p_i$ is a minimal prime over $r_iO$.

For valuation domains of infinite dimension one has to consider the so-called limit-primes: a prime ideal $p$ of a commutative ring $R$ is called limit-prime if

$p=\bigcup\limits_{q\in\mathrm{Spec} (R): q\subset p}q$.

There exist valuation domains $O$ of infinite Krull dimension such that the maximal ideal $m$ of $O$ is no limit-prime. For example take a valuation ring such that the corresponding value group is

$\mathbb{Z}\times\mathbb{Z}\times\ldots$ (countably many factors ordered lexigraphically).

Then one can find $r\in m$ such that $m$ is minimal over $rO$.

Answer by Hagen for The Polynomial Kernel

2010-10-07T07:33:00Z

The precise definition of a kernel function on a set $X$ is this:

The function $K:X\times X\rightarrow\mathbb{R}$ is a kernel function if it has the following two properties:

$K(x,y)=K(y,x)$.
For all $(x_1,...,x_r )\in X^r$ the matrix $(K(x_i,x_j))_{i,j\in{1,...,r}}$ is positiv semi-definite.

Using basic linear algebra one can prove: the set $K_X$ of all kernel functions on $X$ is a commutative ring with identity taking pointwise addition and multiplication as ring operations. Moreover the product of a kernel function with a non-negative real is a kernel function. In particular it follows that for a polynomial $p(X)$ with non-negative coefficients and every kernel function $K$ on $X$ the function $p(K)$ is a kernel function on $X$. Applying this to the scalar product, which is a kernel function on $\mathbb{R}^n$, one can see that the "polynomial kernel" actually is a kernel function.

The ring $K_X$ has much more structure: one can look at limits of kernel functions, power series, orderings etc.

Personal remark / opinion: according to my experience the people in the maschine learning community tend to ignore the rigorous theory in favor of a more computational / pragmatic point of view. One can learn the theory of kernels much better from publications in functional analysis for example, where kernels are arising in the theory of functional Hilbert spaces.

Hagen

Answer by Hagen for Why is an absolute value generated by a simple subvariety of a variety V well-behaved?

2010-10-04T11:28:47Z

According to Lang the valuation $v$ of the field $K$ is well-behaved, if for every finite extension $E/K$ the equation

$ [E:K] =\sum\limits_{w|v} [E_w:K_v] $

holds, where the summation runs over all extension $w$ of $v$ to $E$, and $K_v$, $E_w$ are the completions of the fields $K$, $E$ with respect to the valuations $v$, $w$ respectively.

For a discrete valuation $v$ the completion $K_v$ is equal to the field of fractions of the $M_v$-adical completion $\widehat{O_v}$ of the local ring $O_v$, where $M_v$ is the maximal ideal of $O_v$.

The discrete valuation $v$ we are discussing here by assumption is a localization of an integral, finitely generated $k$-algebra($k$ the field over which the variety $V$ lives). Such an algebra has a finite normalisation in every finite extension of their field of fractions. This property is inherited by localisations of the algebra, thus $O_v$ has this property too: the normalization $O_v(E)$ of $O_v$ in a finite extension $E$ of the field of fractions $K$ is a finitely generated, torsion-free $O_v$-module. It is well-known that such modules over a factorial ring are free - and $O_v$ is factorial. The rank of $O_v(E)$ msut be $[E:K]$ - just localise at $0$.

There is a bijection between the valuation rings $O_w$ of the extensions $w$ of $v$ to $E$ and the localisations of $O_v(E)$ at maximal ideals $M$.

The product of all these maximal ideals is some ideal $I$. The $I$-adical completion $\widehat{O_v(E)}$ of $O(E)$ satisfies:

$ \widehat{O_v(E)} = \prod\limits_{w|v}\widehat{O_w} $

(Matsumura, Thm. 8.15).

Since a power of $I$ lies in the ideal $M_vO_v(E)$ and a power of $M_vO_v(E)$ lies in $I$, the completions of $O_v(E)$ with respect to these two ideals coincide.

The completion of $O_v(E)$ with respect to $M_vO_v(E)$ on the other hand equals the tensor product $ O(E)\otimes_{O_v}\widehat{O_v} $. Since the extension $\widehat{O_v}/O_v$ is faithfully flat this tensor product is a free $\widehat{O_v}$-module of rank $[E:K]$.

Altogether we see now:

$ \prod\limits_{w|v}\widehat{O_w} $

is a free $\widehat{O_v}$-module of rank $[E:K]$, from which we get

$ \prod\limits_{w|v}E_w $

is a free $K_v$-module of rank $[E:K]$.

Hagen

Answer by Hagen for behavior of places of a function field under automorphism

2010-07-20T13:37:16Z

No, not in general, that is not without particular requirements for $x$:

take $F=\mathbb{R}(y)$, the rational function field in one variable over the reals. Then the equation $\sigma (y)=y+1$ determines an automorphism of $F/\mathbb{R}$.

Let $P_1$ be the place associated to the polynomial $y^2+1$; then $\deg (P_1)=2$.

Let $P_2 := \sigma (P_1)$; then $P_2$ is associated to the polynomial $y^2+2y+2$ and (automatically) $\deg (P_2)=2$.

Let $x := y^2+1$; then $[F:\mathbb{R}(x)]=2$ and $P_1|_{\mathbb{R}(x)}$ has degree $1$.

On the other hand $yP_2 $ either equals $i-1$ or $-i-1$. In both cases $xP_2$ is non-real and thus $\deg (P_2)=2$.

Answer by Hagen for on the computation of decomposition groups

2010-06-08T14:44:19Z

A particularly well-studied case is that of cyclic extensions $L/K$, where $K=k(x)$ is a rational function field, the degree $n:=[L:K]$ is not divisible by the characteristic of $k$ and $k$ contains the $n$-th roots of unity. In this case there exists a kind of normal form $L=K(y)$ for generating $L$: $y^n=f(x)$, where the prime factorization of $f\in k[x]$ satisfies some requirements. One can then compute the ramification indices and inertia degrees using only the multiplicities and degrees of the prime factors of $f$. You can find the result in an article by Helmut Hasse: "Theorie der relativ-zyklischen algebraischen Funktionenkörper, insbesondere bei endlichem Konstantenkörper.", Journal für die reine und angewandte Mathematik 172 (1935).

Similar things can be done for Artin-Schreier-extensions of a rational function field.

Hagen.

Answer by Hagen for Reflexive modules over a 2-dimensional regular local ring

2010-04-29T11:07:39Z

Hello,

I guess that $\mathrm{codh}$ actually means $\mathrm{depth}$, that is the length of a maximal regular sequence on a module. Then $\mathrm{codh}(N/M)\geq 1$ is a consequence of the fact that the maximal ideal of $A$ does not annihilate the module due to reflexivity. The next inequality is Auslander-Buchsbaum. Finally $\mathrm{dh}(M)=\mathrm{dh}(N/M)-1$ is a standard fact from homological algebra: if for some module $Q$ $\mathrm{dh}(Q)\leq d$ and $0\rightarrow S\rightarrow P_{d-1}\rightarrow\ldots\rightarrow P_0\rightarrow Q\rightarrow 0$ is exact with projective $P_k$, then $S$ is projective too. Apply this to $0\rightarrow M\rightarrow N\rightarrow N/M\rightarrow 0$.

Answer by Hagen for Generic fiber of morphism between non-singular curves

2010-01-26T12:37:10Z

Example of a finite morphism with inert points:

Define $C_1 := \mathrm{Spec}(\mathbb{R}[x,y])$, where $\mathbb{R}[x,y]:=\mathbb{R}[X,Y]/(X^2+Y^2+1)$ and $C_2 := \mathbb{A}^1_\mathbb{R}$.

Let the morphism $\phi$ be given by the ring extension $\mathbb{R}[x,y] / \mathbb{R}[x]$.

Then the fibre above every rational point of $C_2$ consists of one element only, because the equation $X^2+Y^2+1=0$ has no real solutions. However $\phi$ has degree $2$.

Comment by Hagen

2013-01-10T07:54:22Z

@ pravnak: the normalization is defined in a general setting in definition 24.48.3 of the Stacks Project ( stacks.math.columbia.edu/tag/035E ). As you see no reducedness assumption is necessary.

Comment by Hagen

2013-01-09T08:38:10Z

A good, general presentation of Stein factorization and related topics covering even the non-noetherian case can be found in the Stacks Project (stacks.math.columbia.edu). To me the most remarkable fact is, that the finite morphism appearing in the Stein factorization is the normalization of the base scheme $Y$ in $X$. Thus the obstructions to getting geometrically connected fibres lie in the behavior of the normalization. This is particularly useful if for example $X$ is itself normal and $Y$ is affine.

Comment by Hagen

2011-07-11T14:21:42Z

A remark concerning your "definitions": Definition: a RKHS over $\mathbb{C}$ is a Hilbert space $H$ that as a vector space is a subspace of the vector space $\mathbb{C}^X$ and that has the property, that for every $x\in X$ the evaluation functional $H\rightarrow\mathbb{C},\; h\mapsto h(x)$ is continous. In particular you cannot define the scalar product: it is given, $H$ is a Hilbert space. Similarly the kernel $k$ is determined by the structure of $H$ via Riesz representation. Against that background I do not understand the two definitions you are talking about.

Comment by Hagen

2011-05-23T15:11:11Z

Your counterexample is correct. The statement in the book you mention is only true, if the maximal ideal of the valuation ring is the union of all prime ideals properly contained in it - such a prime ideal is called a limit prime. H

Comment by Hagen

2011-05-20T09:46:16Z

Actually one does not need the dimension formula in the case of one variable: let $C$ be an integral $k$-algebra such that the fraction field of $C$ has transcendence degree $1$ over $k$, and $k$ is algebraically closed in $C$. Then $C$ is irreducible: it is clear that in a factorization $C\cong A\otimes B$ one of the factors, $A$ say, must contain an element $x$ transcendental over $k$. It remains transcendental over $B$. Hence $B$ must be algebraic over $k$. The natural map $B\rightarrow A\otimes B$ is injective, thus $B=k$ by assumption.

Comment by Hagen

2011-04-27T10:29:14Z

If $f$ is a Galois cover, then for all $P\in Y$ of codimension $1$ we have $r_P e_P f_P =\mathrm{deg}(f)$, where $r_P =|f^{-1}(P)$, $e_P$ is the ramification index of some (and thus all) point $Q\inf^{-1}(P)$ and $f_P =[k(Q):k(P)]$ for some (and thus all) $Q\inf^{-1}(P)$. Your requirement then is equivalent to $f_P$ as a function of $P$ is constant. If in addition $\mathrm{deg}(f)$ is a prime $p$, then either $f_P=p$ for all $P$ or $f_P=1$ for all $P$. Statements like this are treated in algebraic number theory, but I have'nt seen something like that for varieties.

Comment by Hagen

2011-04-18T13:00:15Z

Well, is the genus of a function field as it appears in Felipe Voloch's answer a disguised version of the geometric one? One needs an algebraic theory of algebraic functions which is realized in the form of the field theory of extensions of transcendence degree 1 and their valuation/divisor theory. Of course to prove Hurwitz full result one needs Weierstrass points (valuations) and thus the theory of higher derivatives in function fields, which is an algebraization of parts of analysis. Is it pure algebra then?

Comment by Hagen

2011-04-12T07:28:13Z

Remark: assume $A$ fullfills all the requirements and let $m$ be a maximal ideal of $A$. There exists a maximal ideal $n$ of $B:=A[x]$ lying over $m$. Then $B/n$ is algebraically closed. On the other hand $B/n = A/m [x+n]$, hence by Schreier's theorem either $A/m = B/n$ or $A/m$ is real closed.

Comment by Hagen

2011-04-06T08:32:07Z

There is such a counter example on pages 206 - 207.

Comment by Hagen

2011-04-06T08:08:41Z

A discrete valuation ring $R$ has finite integral closure $S$ in the finite extension $L$ of the fraction field $K=\mathrm{Frac}(R)$ if and only if: \[ \sum\limits_{w|v}e(w|v)f(w|v)=[L:K] \] where $v$ is the valuation associated to $R$ and the sum is taken over all extensions of $v$ to $L$. For $L=K(x)$ with $x\in\widehat{R}\setminus R$ and $x^p\in R$ the following facts are known: there is only one extension $w$ of $v$ to $L$ ($L/K$ is purely inseparable). The equations $e(w|v)=f(w|v)=1$ holds, because the completion is unramified over $R$ and has the same residue field as $R$.

Comment by Hagen

2011-03-29T08:45:38Z

$\mathcal{O}_X /I$ is a sheaf of rings. Its support consists of those points $x\in X$ for which the stalk $(\mathcal{O}_X/I)_x$ is non-zero.

Comment by Hagen

2011-03-22T12:09:05Z

Felipe Voloch's class of examples works equally well in algebraic geometry: replace $\mathrm{Spec}\mathbb{Z}$ by an (affine) algebraic curve.

Comment by Hagen

2010-10-21T11:26:39Z

Of course. My remark was not meant to be offensive. H

Comment by Hagen

2010-10-20T08:08:25Z

Valuation domains are frequently studied via the associated Krull valuations, so that any book on valuation theory can be used. Personally I learned a lot from Otto Endler's Valuation Theory and Zariski-Samuel, Commutative Algebra Vol. 2. Moreover Robert Gilmer's book on Multiplicative Ideal Theory and Fuchs-Salce, Modules over valuation domains. H

Comment by Hagen

2010-10-05T13:13:05Z

Your argument seems correct to me. However you are hiding the work to be done under the notion of excellence, which in particular includes the property of having finite normalisations in finite extensions. And this latter property is the only thing one really needs. Hagen