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.

  • $\begingroup$ Sorry for posting an elementary question, but I am really stuck. $\endgroup$
    – awllower
    Feb 18, 2011 at 5:28
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    $\begingroup$ 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. $\endgroup$ Feb 18, 2011 at 5:54
  • $\begingroup$ @Pete: And any book on noncommutative rings, since it provides the standard example of an ideal that is prime but not completely prime. $\endgroup$ Feb 18, 2011 at 6:09
  • $\begingroup$ @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". $\endgroup$ Jun 13, 2011 at 9:37
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    $\begingroup$ 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). $\endgroup$ Jan 17, 2012 at 16:31

2 Answers 2


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 $\mathfrak{I}$?


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

  • $\begingroup$ 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. $\endgroup$
    – awllower
    Feb 18, 2011 at 5:47
  • $\begingroup$ 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. $\endgroup$
    – awllower
    Feb 18, 2011 at 5:55
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    $\begingroup$ @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 $R$. $\endgroup$ Feb 18, 2011 at 5:58
  • $\begingroup$ It reminds me that the basic number theory by Andre Weil has contained this, doesn't it? BTW I do feel better now ^^, thanks. $\endgroup$
    – awllower
    Feb 18, 2011 at 6:02
  • $\begingroup$ @awllower: I'm not in my office, so I can't check. Sorry. $\endgroup$ Feb 18, 2011 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:

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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. $\blacksquare$

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    $\begingroup$ So if $R$ is a simple ring, then so too is $M_n(R)$, right? $\endgroup$ Sep 16, 2015 at 12:57
  • $\begingroup$ @goblin That's right. But for more specific results about simple rings, see Wedderburn's theorem and Thm.11.3.8, p.368 in Grillet. $\endgroup$
    – Leo
    Sep 20, 2015 at 10:43

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