Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $P$ be a noncommutative ring with units, $M_n(P)$ be a ring of matrices with coefficients in $P$. Is there an two-sided ideal $J$ in $M_n(P)$ that is not of the form $J=M_n(I)$, where $I$ is some two-sided ideal in $P$?

Thanks.

share|improve this question
    
I've added the cases where the ideals need not be two-sided. The examples are very general and trivial. –  Patrick Da Silva Feb 24 '12 at 22:23
add comment

4 Answers

up vote 6 down vote accepted

You said $P$ has units, so I'll assume it has a $1$ (the proper way of saying that $P$ has units is by saying that it is a unitary ring).

Consider the matrix $E_{ij}$ to be the matrix with zeros everywhere except at the $ij^{\text{th}}$ position, where it is worth $1$. It is a quick calculation exercise to show that $E_{ij}AE_{k\ell}$ is a matrix whose $i\ell^{\text{th}}$ coefficient is $A_{jk}$.

Assume $J$ is a two-sided-ideal of $M_n(P)$ that is non-zero. Let

$$I = \{ x \in P \, | \, \exists A \in J \text{ such that } x = A_{ij} \text{ for some } i,j\in\{1,\cdots,n\}\},$$

i.e. $I$ is the set of all possible coefficients of elements in $J$. We show that $J = M_n(I)$. One containment is clear : $J \subseteq M_n(I)$.

Next consider an element $A \in M_n(I).$ For each $a_{ij}$ coefficient of $A$, there exist a matrix $J_{ij} \in J$ whose $k\ell^{\text{th}}$ coefficient is precisely $a_{ij}$, for some $k,\ell\in\{1,\cdots,n\}.$ But then $E_{ik} J_{ij} E_{\ell j}$ is a matrix with zeros everywhere except at $(i,j)$, where it is worth $(J_{ij})_{k\ell} = a_{ij}$. Therefore $$ A = \sum_{i=1}^n \sum_{j=1}^n E_{ik} J_{ij} E_{\ell j} \in J $$ which gives you the reverse containment. In conclusion, there is no two-sided ideal of the form you request.

ADDED : I thought Matt's answer was too particular and not very understandable by everyone so I thought of another one. The set of matrices of the form $$ \begin{bmatrix} a_1 & \dots & a_n \\ 0 & \dots & 0 \\ \vdots & & \vdots \\ 0 & \dots & 0 \\ \end{bmatrix} $$ form a right ideal of $M_n(P)$ but not a left ideal. Similarly, the set of all square matrices of the form $$ \begin{bmatrix} a_1 & 0 & \dots & 0 \\ \vdots & \vdots & & \vdots \\ a_n & 0 & \dots & 0 \\ \end{bmatrix} $$ form a left ideal but not a right ideal. This only works for the cases $n > 1$, but for $n=1$, everything depends on $P$, so there's not much to say there.

Hope that helps,

share|improve this answer
1  
Thanks a lot for answer. –  Richard Feb 20 '12 at 21:25
    
Perhaps you should "accept" the best answer ; if it doesn't do me any good it will at least do some good to others. It's your way of thanking us for working out answers =) –  Patrick Da Silva Feb 20 '12 at 21:26
add comment

Yes, there is. There's probably a simpler answer than this, but the first one that I thought of was if you take $P=KQ$ to be the path algebra of a quiver $Q$, and take:

$$J=\begin{pmatrix}KQ&KQ\\e_1KQ&e_1KQ\end{pmatrix}$$

where $e_1$ is one of the vertices. Then elements of $KQ$ are linear combinations of paths in the quiver (directed graph) $Q$ with coefficients in $K$, and $e_1KQ$ consists of those paths that start from the vertex $e_1$. The multiplication is by concatenation of the paths where possible, so unless I've made a silly error, this is a right ideal of $M_n(KQ)$, but not of the form you describe for any $Q$ with at least one path not starting at $e_1$.

share|improve this answer
    
The question is (I assume) about two-sided ideals. –  Qiaochu Yuan Feb 20 '12 at 21:10
    
Well Matt's answer still remains interesting, now is it? Anyway OP could've made that precision if he wanted to. –  Patrick Da Silva Feb 20 '12 at 21:11
    
Well, in that case this doesn't work! It does at least show that a two-sided hypothesis is necessary though. –  Matt Pressland Feb 20 '12 at 21:11
add comment

If by ideal you mean a two-sided ideal then no. Every ideal $J$ of $M_n(P)$ is of the form $M_n(I)$ for some ideal $I$ of $P$.

share|improve this answer
    
Thanks. Is it not important whether $P$ is commutative or not? Can you provide references? –  Richard Feb 20 '12 at 21:08
    
No it is not. Check my answer. –  Patrick Da Silva Feb 20 '12 at 21:10
add comment

Let $E_{i,j}$ be the matrix with $1$ at the $(i,j)$ place and $0$ elsewhere.

Note that for any matrix $A$, we have that $E_{(i,j)}A E_{(k,l)}$ is a matrix with $0$ everywhere, except at the $(i,l)$ place, where it is equal to $A_{(j,k)}$. It follows that if $I$ is any (Two-sided!) ideal in $M_n(R)$, and $J$ is the set of elements of $R$ which are part of matrices in $I$, then $I$ contains $M_n(J)$. Of course, $I$ is contained in $M_n(J)$, so there is an equality.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.