What are the left and right ideals of matrix ring? How about the two sided ideals?

Martin Sleziak
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    Hint: as a left module over itself, the matrix ring decomposes into a direct sum of copies of the module of column vectors. ā€“ Rasmus Mar 13 '11 at 16:47
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    Concerning twosided ideals: try taking a non-zero matrix and multiply from the left and from the right with varius matrices. See what you can get this way. ā€“ Rasmus Mar 13 '11 at 16:48

2 Answers2


The left ideals of the matrix ring $S = M_n(R)$ over a (unital, associative) ring $R$ are in a $1ā€“1$ correspondence with the left submodules of a free left $R$-module $R^n$ of rank $n$.

Given any left $R$-submodule $K\le R^n$ define the ideal $I(K)$ as those matrices whose rows are elements of $K$. If $M$ is such a matrix and $A$ is an arbitrary $n\times n$ matrix over $R$, then $AM$ is a matrix whose rows are $R$-linear combinations of the rows of $M$, and so also contained in $K$. Similarly, by multiplying by elementary matrices (matrix units even), you get that every left ideal $I$ determines a submodule $K(I)$ consisting of the first rows of elements of the ideal. You then check that $I(K(I))=I$ and $K(I(K))=K$.

This is called Morita equivalence. The left ideal $P = I(R\oplus0\oplus\dots\oplus 0) = I(R\oplus 0^{nāˆ’1})$ is a direct summand of the left $S$-module $S$, and so is called a projective module. It is faithful, since $S \simeq P^n$ as left $S$-modules. In other words, it is a progenerator. $I(K) = _SP\otimes_R K$ and $K(I) = _R\operatorname{Hom}_S(P,I)$.

The short version is just that modules over $S$ are just big versions of modules over $R$. In particular, left ideals of $S$ are just left submodules of $R^n$.

Over some rings $R$, the submodules of $R^n$ can be significantly more complicated. Even for fields, the submodule lattice of $R=\mathbb R$ is very simple: just $0 \le R$. For $R^n$ one has the Grassmannian, which to my mind is quite a bit more complex than $R$. Other rings can be even worse, so one should be careful of leaning too heavily on the intuition for fields when dealing with matrix rings, though any Morita property will transfer between $R$ and $S$ without trouble using the "I()" and "K()" constructions.

Jack Schmidt
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For two-sided ideals, see the previous question which deals with fields. In the comments I state the description of ideals over a commutative ring with $1$: they are precisely the subrings that consist of matrices with coefficients in $\mathcal{J}$, where $\mathcal{J}$ is an ideal of $R$. The result for not-necessarily commutative rings is the obvious one. For rings without an identity, it gets a bit more complicated, but essentially the same kind of argument works.

For one-sided ideals, think about some of the argument given in the 2-sided case. You can do some of the stuff described, though perhaps not all. For instance, for left ideals, you can isolate rows. So perhaps you can modify the answer yoyo gives for the division ring case to a general case, along the same lines as the result for two-sided ideals generalizes from fields to the general case.

Arturo Magidin
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