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.

Denote by $M_{n \times n}(k)$ the ring of $n$ by $n$ matrices with coefficients in the field $k$. Then why does this ring not contain any two-sided ideal?

Thanks for any clarification, and this is an exercise from the notes of Commutative Algebra by Pete L Clark, of which I thought as simple but I cannot figure it out now.

share|improve this question
    
Sorry for posting an elementary question, but I am really stuck. –  awllower Feb 18 '11 at 5:28
3  
Amusingly, a student in my course asked me about this exercise during a problem session. My reply: "What? That's not a commutative algebra question: how did that get in there?" (And I didn't answer the question!) It is, BTW, an extremely standard non commutative algebra question. Any basic text which treats central simple algebras over a field should cover this. –  Pete L. Clark Feb 18 '11 at 5:54
    
@Pete: And any book on noncommutative rings, since it provides the standard example of an ideal that is prime but not completely prime. –  Arturo Magidin Feb 18 '11 at 6:09
    
@Pete Do you really need to look in a book on central simple algebras to find this fact? Actually, this is one of the first theorems I learnt after the definition of an "ideal". –  Amitesh Datta Jun 13 '11 at 9:37
1  
A very belated comment: as mentioned above, the classification of two-sided ideals of $M_n(R)$ appears as Theorem 22 in $\S$ 1.10 of my noncommutative algebra notes: math.uga.edu/~pete/noncommutativealgebra.pdf. The statement and proof are virtually identical to the excerpt from Grillet's book in Leon Lampret's answer (notwithstanding the fact that I don't own Grillet's book, i.e., this is indeed a very standard result). –  Pete L. Clark Jan 17 '12 at 16:31

2 Answers 2

up vote 12 down vote accepted

Suppose that you have an ideal $\mathfrak{I}$ which contains a matrix with a nonzero entry $a_{ij}$. Multiplying by the matrix that has $0$'s everywhere except a $1$ in entry $(i,i)$, kill all rows except the $i$th row; multiplying by a suitable matrices on the right, kill all columns except the $j$th column; now you have a matrix, necessarily in $\mathfrak{I}$, which contains exactly one nonzero entry, namely $a_{ij}$ in position $(i,j)$.

Now show that $\mathfrak{I}$ must contain all matrices in $M_{n\times n}(k)$. This will show that a $2$-sided ideal consists either of only the $0$ matrix, or must be equal to the entire ring.

Added. Now that you have a matrix that has a single nonzero entry, can you get a matrix that has a single nonzero entry on whatever coordinate you specify, and such that this nonzero entry is whatever element of $k$ you want, by multiplying this matrix (on either left, or right, or both) by suitable elementary matrices? Will they all be in $\mathfrak{I}$?

And...

$$\left(\begin{array}{cc} a&b\\ c&d \end{array}\right) = \left(\begin{array}{cc} a & 0\\ 0 & 0 \end{array}\right) + \cdots$$

share|improve this answer
    
Well, @Arturo Magidin, the last paragraph is why I cannot get it, i.e. why does this imply that $\mathfrak{I}$ must contain all matrices in $M_{n*n}$? Can you be more specific, please, thanks in any case. –  awllower Feb 18 '11 at 5:47
    
I got it!! Thanks very much, it clarified all the things such that I feel I was a dumb! In any case, thanks very much. –  awllower Feb 18 '11 at 5:55
3  
@awllower: Well, to make you feel better, prove the following generalization: if $R$ is a commutative ring with identity, then the ideals of $M_{n\times n}(R)$ are exactly the subrings of the form $M_{n\times n}(\mathfrak{I})$, where $\mathfrak{I}$ is an ideal of $R$. –  Arturo Magidin Feb 18 '11 at 5:58
    
It reminds me that the basic number theory by Andre Weil has contained this, doesn't it? BTW I do feel better now ^^, thanks. –  awllower Feb 18 '11 at 6:02
    
@awllower: I'm not in my office, so I can't check. Sorry. –  Arturo Magidin Feb 18 '11 at 6:06

A faster, and more general result, which Arturo hinted at, is obtained via following proposition from Grillet's Abstract Algebra, section "Semisimple Rings and Modules", page 360:

enter image description here

Consequence: if $R:=D$ is a division ring, then $M_n(D)$ is simple.

Proof: Suppose there existed an ideal of $M_n(D)$. By the proposition, it'd be of the form $M_n(I)$, for $I\unlhd D$, but division rings do not have any ideals (other than $0$ and $D$), so this is a contradiction. $\blacksquare$

share|improve this answer
    
Thanks for the good answer. –  awllower Feb 15 '13 at 6:30

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.