User miguel - MathOverflow

Chain Condition on Rings

2012-11-23T18:14:12Z

Let $R$ be a noncommutative ring. The ring $R$ has descending chain condition on two-sided ideals (D.C.C.), if for a chain of two-sided ideals $J_1\supset J_2\supset \cdots$, then there exists an $N\in\mathbb N$ such that $J_n = J_N$ for $n\geq N$.

Is there any example of a ring $R$ with D.C.C. on two-sided ideals but without D.C.C on left ideals and also without D.C.C on right ideals?

Commuting Linear Operators In Hilbert Spaces

2012-10-10T23:05:28Z

Let $V$ be a finite dimensional vector space over the complex field $\mathbb C$. Let $L:V\rightarrow V$ be a linear operator. Using the matrix of $L$ and the Jordan canonical form it is easy to find all the linear operators that commute with $L$.

Now suppose that $H$ is a Hilbert space and let $L:H\rightarrow H$ be a continuous linear operator. There is some method to determine all the continuous linear operatores that commute with $L$?

Inner Automorphisms of Matrix Algebras

2012-10-09T14:55:21Z

Let $\mathbb R$ be the field of real numbers and $\mathbb C$ the field of complex numbers. It is well known that that $\mathbb C$ can be embedded in $M_2(\mathbb R)$.

This embedding can be extended in the obvious way to an embedding function $\varphi : M_n(\mathbb C)\rightarrow M_{2n}(\mathbb R)$.

My question is: consider the embedding $\varphi :M_2(\mathbb C)\rightarrow M_{4}(\mathbb R)$, it is possible to find an inner automorphism $\Psi : M_4(\mathbb R)\rightarrow M_{4}(\mathbb R)$ such that the intersection $$\varphi (M_2(\mathbb C))\cap \Psi(\varphi (M_2(\mathbb C)))$$

is the scalar matrices in $M_4(\mathbb R)$?

Decomposition of Matrices in Semisimple and Nilpotent Parts

2012-09-29T11:56:20Z

I asked this question in http://math.stackexchange.com/posts/204115/edit but remains unanswered.

For any matrix $A\in M_n(\mathbb F)$, where $\mathbb F$ is an algebraically closed field, there is a matrix $S\in M_n(\mathbb F)$ such that

$$SAS^{-1}=D+N,$$ where $D$ is diagonal and $N$ nilpotent. Moreover, this decomposition is unique.

Suppose now that $A\in M_n(\mathbb K)$, but $\mathbb K$ is not necessarily algebraically closed. It is also true that there is a matrix $L\in M_n(\mathbb K)$ such that

$$LAL^{-1}=R+M,$$

where $M$ is nilpotent and $R$ is diagonalizable in the algebraic closure of $\mathbb K$? Moreover when we consider the decomposition in $\mathbb K$ and in the algebraic closure of $\mathbb K$ the nilpotent part is the same?

Annihilators in algebras

2012-06-13T13:45:47Z

Let $R$ be an artinian commutative ring and $A$ an algebra over $R$ with basis ${a_1, a_2, \ldots , a_n}$ where each $a_i$ is a unit and

$a_ia_j=u_{ij}a_ja_i,$

where each $u_{ij}\in R$ and is a unit. is true that for every element $x\in A$ the left annihilator of $x$ denoted by $ann_l{x}$ equals the right annhilator $ann_r{x}$?

I think that if exist an element $z$ with $ann_l{z}\neq ann_r{z}$, this lack of symmetry must necessarily be reflected in the ring $R$ or in the elements of the base. Since, for the ring $R$ and for the elements of the base there is symmetry, this symmetry extends to all elements of $A$. This is my perception, but how to prove?

Finite local rings

2012-06-05T15:38:04Z

There is some classification of finite commutative local rings. For example how many not isomorphic finite local rings with the same order $p^k$ and the same residue field $\mathbb F_p$ exist?

Annihilators in artinian ring

2012-05-29T12:44:01Z

Let $A$ be an artinian ring. It is well known that for an element $x$ in $R$ the right annihilator $Ann_r(x)$ is non trivial (i.e. contains a nonzero element ) if and only if the left annihilator $Ann_l(x)$ is non trivial.

If we assume that the ring $A$ is finite is true that the cardinality of $Ann_r(x)$ equals the cardinality of $Ann_l(x)$?

Note: If the ring $A$ is finite and semisimple, them it is a direct sum of full matrix rings over fields and in this case it is true. I think it may help to think first in full matrix rings over finite commutative rings.

Permanent function over finite commutative rings

2012-03-10T15:20:26Z

Let $R$ be a finite commutative ring with identity and $R=R_1+...+R_k$ its decomposition in a direct sum of finite local rings. Considere the matrix ring $M_n(R)$ of $n\times n$ matrices with elements in $R$. It is easy to see that a matrix $A$ decomposes in a unique way as

$$A=A_1+ .... +A_k,$$

with $A_i\in R_i$. Let $per(A)$ denote the permanent of the matrix $A$.

Is true that $per(A)$ is a unit in $R$ if and only if each $per(A_i)$ is a unit in $A_i$?

Determinants over commutative rings

2012-03-03T22:42:22Z

Hello, I am preparing a paper on determinants in commutative rings. Someone can give me examples of applications of determinants in commutative rings to other areas of mathematics or physics. Thank you

Answer by Miguel for Online free copy of the article "What is an answer?" (Amer. Math. Monthly)

2012-02-19T12:57:59Z

Hello. send me your email and I send you a copy of the article "What is an Answer? Amer. Math. Monthly, 89 (5), pp. 289-292"

Tensor product of two algebras

2012-01-20T15:26:24Z

Let $A$ and $B$ be algebras over a field $K$. The ideals of the tensor product $A\bigotimes_K B$ are of the form $I\bigotimes_K J$ where $I$ is an ideal of $A$ and $J$ is an ideal of $B$?

Maximal ideals in a polynomial ring over the real numbers.

2012-01-17T16:23:30Z

Let $\mathbf{R}$ be the field of real numbers. What are the generators of the maximal ideals of the polynomial ring $\mathbf{R}[x_1, ... , x_n]$? If instead of $\mathbf{R}$ one considers the field $\mathbf{C}$ of complex numbers, then Hilbert's Nullstellensatz implies that each maximal ideal $\mathfrak{m}$ of $\mathbf{C}[x_1, ... , x_n]$ is generated by $n$ generators of the form $x_i-a_i$.

Noetherian ring

2012-01-16T19:25:16Z

Let $R$ be a noetherian ring. By the Hilbert Basis Theorem the polinomial ring $R[x_1, ... , x_n]$ is also a noetherian ring. What can we say about the number of generators of an ideal $I$ of $R[x_1, ... , x_n]$? (We can suppose that every ideal in $R$ is principal)

subalgebra of a matrix algebra

2011-12-30T14:33:29Z

Let $K$ be an algebraic closed field and $M_n(K)$ the $K$-algebra of all matrices $n\times n$ over $K$. If $L$ and $M$ are two commutative isomorphic subalgebras of $M_n(K)$ it is true that there exista a regular matrix $S\in M_n(K)$ such that $SLS^{-1}=M$. That is the isomorphism is inner?

Comment by Miguel

2012-11-23T18:46:54Z

Muchas gracias Fernado.

Comment by Miguel

2012-06-13T18:35:07Z

Thank you. It is a nice example to prove that we have not $ann_l x=ann_rs$. But, probably, their cardinality is the same. I'm trying to find a proof of this.

Comment by Miguel

2012-06-06T09:41:39Z

Martin, I do not understand your comment, you want to explain?

Comment by Miguel

2012-06-06T07:37:56Z

Thanks, your answer is very useful for me.

Comment by Miguel

2012-06-05T17:10:29Z

Gracias Mariano

Comment by Miguel

2012-05-29T17:05:06Z

thanks, is a very clever example.

Comment by Miguel

2012-03-04T00:17:23Z

The paper is on the convertion the determinant into the permanent. Yes i want applications where the ring has zero divisors.

Comment by Miguel

2012-01-17T17:47:05Z

Thank you Francesco Polizzi

Comment by Miguel

2011-12-30T18:05:13Z

thank you Denis Serre. Your example responds to my question. Your answer made me think of another question: two unital subalgebras (that is, containing the scalar matrices) are conjugated?

Comment by Miguel

2011-12-30T15:23:12Z

Dear darij grinberg thanks for your suggestion. But I think it can not be applied directly the Skolem–Noether theorem since the subalgebra $A$ is not necessarily simple. -Miguel