Tagged Questions

0
votes
1answer
47 views

Show that every ideal of the matrix ring $M_n(R)$ is of the form $M_n(I)$ where $I$ is an ideal of $R$ [duplicate]

Suppose $R$ is a commutative ring. Show that every ideal of $M_n(R)$ is of the form $M_n(I)$ where $I$ is an ideal of $R$. I have spent 30 minutes on this question and I still got nowhere. Can ...
1
vote
1answer
35 views

Conjugacy class in matrix ring

Let $M_{2}(\mathbb{R})$ be the ring of $2\times 2$ matrices over the reals and $M_{2}(\mathbb{R})^*$ the set of invertible such matrices. Consider any $A \in M_{2}(\mathbb{R})$ such that $ A^{2}=-I$, ...
3
votes
1answer
63 views

Definition of $\text{GL}(n,R)$

How do one usually define the general linear group over a ring $R$, denoted by $\text{GL}(n,R)$. I was told in a paper that $\text{GL}(n,R)$ is a group, and I presumed that $$\text{GL}(n,R)=\{A\in ...
1
vote
1answer
58 views

Problem about a matrix on the ring $\mathbb Z$.

Suppose $A$ is a $m\times n$ ($n\geq m$) matrix on the ring $\mathbb Z$ of integers and the greatest common divisor of its $m\times m$ minor determinants is $1$. Prove that there is a $n\times m$ ...
4
votes
1answer
44 views

Finding an idempotent that satisfies certain conditions in a matrix ring.

I've been stuck on a problem, and I was wondering if anyone could help me out. The problem is: Let $R$ be the $2 \times 2$ matrix ring over the reals $\mathbb{R}$ of the form $$ \begin{bmatrix}a ...
5
votes
1answer
165 views

Polynomials in matrices with integer entries

I'm looking for references, if there is any, for this problem: Characterize all elements $a \in M_n(\mathbb{Z})$ for which we have $\mathbb{Q}[a] \cap M_n(\mathbb{Z})=\mathbb{Z}[a].$ Here, by ...
5
votes
3answers
156 views

How to construct a $2\times 2$ real matrix A not equal to Identity such that $A^3$=I?

How to construct a $2\times 2$ real matrix A not equal to Identity such that $A^3$=I? There is a correspondence between the ring of complex numbers and the ring of $2\times2$ matrices (0 matrix is ...
2
votes
1answer
43 views

Is there an explicit way to determine $\mathrm{Mat}_n(R[X_1,\dots,X_m])\simeq\mathrm{Mat}_n(R)[X_1,\dots,X_m]$?

For a commutative ring $R$, let $\mathrm{Mat}_n(R[X_1,\dots,X_m])$ denotes the matrix ring with entries from $R[X_1,\dots,X_m]$, and let $\mathrm{Mat}_n(R)[X_1,\dots,X_m]$ denotes the polynomial ring ...
1
vote
1answer
53 views

Are there rngs whose rngs of matrices are commutative?

If $R$ is a unital ring and $M_{2\times 2}(R)$ is a commutative ring, then $R$ is a trivial ring because if $$\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}=\begin{pmatrix}1 & 0 \\ 0 & ...
2
votes
0answers
124 views

Why is $M_{mn}(R)\simeq M_m(M_n(R))$? [duplicate]

Intuitively, it's not hard to believe that for a ring $R$, the matrix ring $M_{mn}(R)$ is isomorphic to $M_m(M_n(R))$. Taking a matrix in $M_{mn}(R)$ and turning the $n\times n$ blocks into single ...
4
votes
3answers
167 views

Rings with isomorphic proper subrings

Rings will be unital here but I don't require that subrings share the identity elements with superrings. I just accidentally came up with an example of a ring $R$ with a proper subring $S$ such that ...
4
votes
3answers
123 views

Is this action of $\mathbb F[[x]]$ on $\bigoplus_{i=0}^{\infty}\mathbb F$ natural?

The title of my question has a field $\mathbb F$ in it, but to make sure I'm not losing anything, I would like to introduce my question in full generality. But still, I will be happy with an answer in ...
0
votes
1answer
370 views

Smith Normal Form

Would the Smith Normal Form of the following matrix over $\mathbb Q[x]$ $$\begin{pmatrix}   (x+a)(x+b) & 0 & 0 &0 \\  0 & (x+c)(x+d) & 0 & 0 \\   0 ...
3
votes
1answer
208 views

Matrices over a ring

How might I find $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ where $a,b,c,d \in \mathbb Z[x]$ such that there does not exist $B, C \in M_2(\mathbb Z[x])$ such that $B^{-1}AC$ is ...
3
votes
1answer
163 views

Note on Ring Homomorphisms of Matrices Rings

Assume that $\mathbb{F}$ is a field, and let $\mathbb{M}_{t}\left( \mathbb{F}\right) $ be the ring of matrices of order $t$ over $\mathbb{F}$. Does there exist a non-trivial ring homomorphism from ...
12
votes
2answers
773 views

necessary and sufficient condition for trivial kernel of a matrix over a commutative ring

In answering Do these matrix rings have non-zero elements that are neither units nor zero divisors? I was surprised how hard it was to find anything on the Web about the generalization of the ...
9
votes
1answer
452 views

Do these matrix rings have non-zero elements that are neither units nor zero divisors?

Let $R$ be a commutative ring (with $1$) and $R^{n \times n}$ be the ring of $n \times n$ matrices with entries in $R$. In addition, suppose that $R$ is a ring in which every non-zero element is ...
5
votes
2answers
338 views

Is the determinant of a zero divisor zero?

Suppose that $A$ is a zero divisor in the ring of $(n\times n)$-matrices over the ring $R$. Is $\det(A) =0$ if $R$ is a field? Is $\det(A) =0$ if $R$ is an integral domain? It's not necessarily ...
3
votes
2answers
120 views

What is useful about commutativity/non-commutativity in practice

How is having commutativity or not having it useful for in practice? (whether it is for linear algebra, rings, basic airthmetic etc)
1
vote
1answer
147 views

How to find matrices with given commutator

Consider $M_2(\mathbb{Z})$. Is it possible to find two matrices A,B such that their commutator AB - BA equals a given matrix C? Is there any chance to characterize all possible occuring commutators in ...
8
votes
4answers
492 views

“weird” ring with 4 elements - how does it arise?

For no real reason, I did a classification of (associative, with a 1) rings with less than 8 elements (they all happen to be commutative). Most of the rings I got were of a type I knew - namely: ...
5
votes
2answers
146 views

Rings of matrices

Let $ A\in {\mathbb{F} }^{n\times n} $ be a fixed matrix. The set of all matrices that commute with A forms a subring of ${\mathbb{F} }^{n\times n}$. Is any subring of ${\mathbb{F} }^{n\times n }$ ...