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Tensoring with descending chain of modules

Let $A \to B$ be a ring homomorphism. Let $M_1 \supseteq M_2\supseteq \ldots$ be an infinite chain of $A$-modules ($M_i$ not necessarily finite free). Suppose that the limit $\cap_{i=1}^{\infty} M_i$ ...
ringq's user avatar
  • 11
1 vote
0 answers
1k views

Algebraic Independence of Polynomials in n Variables with Real Coefficients

I am considering the problem of determining the algebraic independence of $n$ polynomials in $m$ variables with real coefficients, where $m \geq n$. The variables will be denoted by $a_{1}, a_{2}, ... ...
mnh1364's user avatar
  • 11
7 votes
2 answers
942 views

Hilbert's Nullstellensatz on polynomials with integer coefficients

Let $f_1, f_2, \ldots, f_m \in \mathbb{Z}[x_1, \ldots, x_n]$. Assume $f_1(X) = f_2(X) = \ldots = f_m(X) = 0$ have no solutions over $\mathbb{C}^n$, then by Hilbert's Nullstellensatz, there exists ...
jsliyuan's user avatar
  • 651
38 votes
2 answers
6k views

Over which fields are symmetric matrices diagonalizable ?

The question is motivated by this one real symmetric matrix has real eigenvalues - elementary proof: Are there other fields $F$ than $\mathbb{R}$ (maybe some valued fields or real closed fields) ...
tomasz 's user avatar
  • 567
2 votes
1 answer
284 views

Dimension of An Ultraproduct Field as a Vector Space over Another Ultraproduct Field

Suppose $F \subseteq K$ are fields with $G$ an ultrafilter on an infinite set $X$. If $F^{\ast}$ and $K^{\ast}$ represent the ultraproducts respectively of $F$ and $K$, it is easy to see that $[K : F]...
Shankman's user avatar
6 votes
1 answer
276 views

Linear algebra over principal rings

Consider an extension $R\subseteq S$ of commutative rings, and suppose that $R$ is principal (i.e., $0$ is the only zero-divisor of $R$ and every ideal of $R$ has a generating set of cardinality $1$). ...
Fred Rohrer's user avatar
  • 6,700
20 votes
2 answers
1k views

a determinantal identity

Dusan Pokorny and Jan Rataj have just posted a paper (http://arxiv.org/abs/1209.2305) in which they prove the identity $$ \det (A-B) = \frac 1{d!} \sum_{k=0}^d (-1)^k \binom dk \det((d-k)A + kB) $$ ...
Joe Fu's user avatar
  • 340
2 votes
1 answer
118 views

Dual of a semilinear morphism

Let $R$ be a commutative ring and let $M$ and $N$ be $R$-modules. Let $\sigma:R\rightarrow R$ be a ring automorphism. Let $f: M\rightarrow N$ be a $\sigma$-semilinear map, i.e. a map of abelian ...
HenrikRüping's user avatar
2 votes
1 answer
196 views

Relations between a set of inner products of vectors

Suppose we have n normalized vectors on an arbitrarily large Hilbert space $|A_1\rangle,\dots,|A_n\rangle$, $\langle A_i|A_i\rangle=1$ for every i. And there're $\frac{n(n-1)}{2}$ inner products $\...
YN Wang's user avatar
  • 23
1 vote
1 answer
253 views

Chain of ideals in a complex algebra

Suppose $\mathfrak{A}$ is an unital algebra over complex numbers and $\mathfrak{J}$ is chain of left-ideals in $\mathfrak{A}$ ordered by inclusion such that none of its elements is countably generated....
GiroCont's user avatar
1 vote
2 answers
313 views

How many DISTINCT vectors we get from pairs v_i + v_j for some set of given vectors v_i ?

Consider some set of vectors v_i i=1...N , v_i \in Z^k. e.g. N = 10^4; k = 10 Consider all possible sums: v_i + v_j. Is it possible to estimate how many DISTINCT vectors we get in advance without ...
Alexander Chervov's user avatar
30 votes
2 answers
2k views

When is $SL(n,R) \rightarrow SL(n,R/q)$ surjective?

Let $R$ be a commutative ring with unit and let $q$ be an ideal of $R$. There is thus a natural map $SL(n,R) \rightarrow SL(n,R/q)$ for all $n$. This map is surjective if $SL(n,R/q)$ is generated by ...
Ira L's user avatar
  • 418
0 votes
1 answer
1k views

Conjugate Matrix

Let $A$ be a nilpotent square matrix, $J$ be the antidiagonal matrix with 1's on the secondary diagonal (i.e. $J^{2}=E$) and let $B=J A J.$ Suppose we conjugate the matrices $A,B$ by a matrix $...
Melania's user avatar
  • 301
4 votes
2 answers
818 views

Double orthogonal complement of a finite module

Crossposted from math.stackexchange since I'm not getting any answer. Let $W$ be the finite $\mathbb{Z}$-module obtained from $\mathbb{Z}_q^n$ with addition componentwise where $\mathbb{Z}_q$ is the ...
Carl's user avatar
  • 141
5 votes
3 answers
1k views

adjoint of multiplication operator in a commutative algebra

Dan Popescu asked me the following question, and since I'm not an expert I'm throwing his question on MO. Suppose that $A$ is a finite-dimensional vector space over an ordered field $k$ with $\...
Tom De Medts's user avatar
  • 6,614
6 votes
2 answers
5k views

Periodic matrices

A square matrix $M$ such that $M^{k+1}=M$, for some positive integer $k$, is called a periodic matrix. Can we characterize the periodic matrices in $\mathcal{M}_n(\mathbb{Z})$? If we replace $\mathbb{...
Portland's user avatar
  • 2,829
16 votes
3 answers
2k views

Hom(A,C) ⊗ Hom(B,D) injects into Hom(A⊗B,C⊗D): when? why?

Sorry for asking a linear algebra question on a research forum, but this seems to be either a case of extreme blindness on my side, or a case of a result lying much deeper than it seems. The ...
darij grinberg's user avatar
3 votes
3 answers
461 views

Multiplicity of eigenvalues in 2-dim families of symmetric matrices

Say you have 2 symmetric matrices, $A$ and $B$, and you know that every linear combination $xA+yB$ ($x,\\,y\in \mathbb{R}$) has an eigenvalue of multiplicity at least $m>1$. Such a situation can of ...
CuriousUser's user avatar
  • 1,452
3 votes
1 answer
321 views

spurious torsion under compositions of linear maps

Say we have a PID $R$, integers $1 \leq a \leq b$, and $R$-homomorphisms $R^a \stackrel f\to R^b \stackrel g\to R^a$ with $g \circ f$ of full rank. For $h = f, g, g \circ f$, let $c(h)$ be the ...
Jay's user avatar
  • 735
13 votes
2 answers
1k views

Is the Characteristic of a Field Detectable from the Topology of a Topological Vector Space?

Motivation A topological vector space is a vector space over a (topological) field, K, that carries a topology such that addition and scalar multiplication are continuous maps, e.g., all normed vector ...
Charlie Cunningham's user avatar
3 votes
0 answers
474 views

Jacobson-Bourbaki correspondence

The Jacobson-Bourbaki correspondence induces the traditional, finite Galois correspondence by suitable restriction; I've been pondering two things: 1. Are there any (other) interesting applications of ...
Stephan F. Kroneck's user avatar
3 votes
0 answers
473 views

Infinite Galois correspondence "according to Artin"

Ever since Artin's lectures on Galois Theory one knows how to set up and derive the usual Galois correspondence in the finite(-dimensional) case using just a bit of elementary Linear Algebra, and ...
Stephan F. Kroneck's user avatar
3 votes
2 answers
344 views

Pseudo-idempotent matrix generating a free module

Let $R$ be a commutative ring with $1$. Let $n$ and $k$ be nonnegative integers, and let $A\in\mathrm{M}_n\left(R\right)$ be a matrix such that $A\cdot R^n\cong R^k$ as $R$-modules. Assume that $A^2=\...
darij grinberg's user avatar
4 votes
4 answers
596 views

Generalization of Jordan Decomposition for Several Commuting Operators

Recently I became curious about the following question: Let $V$ be a finite dimensional vector space over $k$ and let $A_1, \cdots, A_n: V \rightarrow V$ be a set of commuting maps. Question: ...
Mikhail Gudim's user avatar
53 votes
9 answers
13k views

Is there a preferable convention for defining the wedge product?

There are different conventions for defininig the wedge product $\wedge$. In Kobayashi-Nomizu, there is $\alpha\wedge\beta:=Alt(\alpha\otimes\beta)$, in Spivak, we find $\alpha\wedge\beta:=\frac{(k+l)...
agt's user avatar
  • 4,306
3 votes
3 answers
2k views

How to define the orientation of a vector space over an arbitrary field?

I know the construction of the Hodge star operator in the context of (pseudo-)euclidean real vector spaces. Apart from the scalar product it involves a orientation of the vector space, which one has ...
Marc Nieper-Wißkirchen's user avatar
5 votes
2 answers
2k views

Iterated calculation of determinants

Given a $4 \times 4$ matrix $S$ over a commutative ring $R$. I want to consider it as a $2\times 2$ matrix over $M_2(R)$. Lets say $S=\left(\begin{array}{cc} A&B \\\ C&D\end{array}\right)$ ...
HenrikRüping's user avatar
1 vote
0 answers
169 views

Sum of two free o-submodules in a vector space over a local field

Let $V$ be a countably infinite dimensional $K$-vector space over a local field $K$ (nontrivially discretely valued with finite residue field). Let $o$ be the ring of integers of $K$. Given two free ...
Tiffy's user avatar
  • 107
16 votes
1 answer
2k views

Commuting Matrices and the Weak Nullstellensatz

In the Wikipedia article on Hilbert's Nullstensatz, http://en.wikipedia.org/wiki/Hilbert%27s_Nullstellensatz the following application of the Weak Nullstensatz is mentioned: Commuting matrices ...
Holger Partsch's user avatar
6 votes
1 answer
301 views

Orbits in commutative groups.

Let A be finite commutative group say $(Z_m)^h$. I will say that $S \subset A$ is an orbit if exist group $H$ which acts on A such that $S$ is an orbit of $H$. Can one give a simple characterization ...
Klim Efremenko's user avatar
2 votes
2 answers
369 views

vectors with entries from a finite ring

I've been working recently with vectors over finite fields, but I was hoping to work in a more general setting and consider vectors over finite commutative rings. The question I had is as follows: if ...
Sarah's user avatar
  • 21
7 votes
3 answers
2k views

Is there a field which is the union of finitely many proper subfields?

Is there a field which is the union of finitely many proper subfields?
heiko's user avatar
  • 79
6 votes
0 answers
267 views

Is there a straightforward way to solve unmixed, homogeneous systems of polynomials?

I came across this problem in my research. It might just be an easy algebraic geometry question, but I don't know much algebraic geometry. Suppose we have a system of $k\leq n$ polynomials in $\...
Jeffrey Doker's user avatar
14 votes
4 answers
6k views

When is an algebra of commuting matrices (contained in one) generated by a single matrix?

Let C be an nxn matrix, then the polynomials in C (with appropriate coefficients) form an algebra of commuting matrices. I feel that I should know if the converse is true but I do not. So my first ...
Aaron Meyerowitz's user avatar
9 votes
3 answers
1k views

Vector spaces with natural bases

Sergeib's question asks about vector spaces without a natural basis. Actually, I would claim (apparently in accord with many comments and answers to Sergeib's question ) that this is the default ...
28 votes
6 answers
5k views

Expressing $-\operatorname{adj}(A)$ as a polynomial in $A$?

Suppose $A\in R^{n\times n}$, where $R$ is a commutative ring. Let $p_i \in R$ be the coefficients of the characteristic polynomial of $A$: $\operatorname{det}(A-xI) = p_0 + p_1x + \dots + p_n x^n$. I ...
Laurent Lessard's user avatar
28 votes
5 answers
4k views

Does Smith normal form imply PID?

Let $R$ be a nonzero commutative ring with $1$, such that all finite matrices over $R$ have a Smith normal form. Does it follow that $R$ is a principal ideal domain? If this fails, suppose we ...
user avatar
9 votes
2 answers
2k views

Jordan Form Over a Polynomial Ring

Let $X$ be the set of $k\times k$ matrix with entries in $\mathbb{C}$, and let $M\in X$. The group $GL(k,\mathbb{C})$ acts on $X$ by conjugation, and according to the Jordan decomposition theorem (...
Clay Cordova's user avatar
  • 2,087
12 votes
4 answers
752 views

Additive commutators and trace over a PID

I would like to find an example of principal ideal domain $R$, such that there exists a square matrix $A\in \mathfrak{M}_n(R)$ with zero trace that is not a commutator (i.e. for all $B,C \in \mathfrak{...
Portland's user avatar
  • 2,829
38 votes
1 answer
10k views

Infinite tensor products

Let $A$ be a commutative ring and $M_i, i \in I$ be a infinite family of $A$-modules. Define their tensor product $\bigotimes_{i \in I} M_i$ to be a representing object of the functor of multilinear ...
Martin Brandenburg's user avatar

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