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I was thinking about the matrix ring $M_n(\mathbb{F})$ (set of all $n\times n$ matrices over the field $\mathbb{F}$) and I determined the fact that every matrix in this ring is either a unit or a zero divisor. However, I think in general this would not imply that the only ideals of $M_n(\mathbb{F})$ are $M_n(\mathbb{F})$ and $\{0\}$. Does knowing that the fact that I proved help determine what the ideals of $M_n(\mathbb{F})$ are?

I of course know other facts which are that the only ideals of the field $\mathbb{F}$ are $0$ and $\mathbb{F}$ and that if $I$ is an ideal in $\mathbb{F}$ then $M_n(I)$ is an ideal of $M_n(\mathbb{F})$.

Any help is appreciated, thank you.

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Sarah
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  • This question might help. – Alex Vong Mar 09 '17 at 13:03
  • I'm not sure if knowing that fact helps. The proof of what you want to prove that I know of doesn't use that fact. It is simple in itself though. It goes like this: a non-zero matrix has a non-zero element; we can multiple that matrix with some other (2, to be exact) matrices to get a matrix with exactly one nonzero entry; we can then multiply that matrix with exactly one nonzero entry with some other 2 matrices to land that element in any position; hence the set of linear combinations of all matrices we could generate from one nonzero matrix covers the whole ring. – Tunococ Mar 09 '17 at 13:09
  • It may be easier for you if you think of $M_n(\mathbb{F})$ as the ring of all $\mathbb{F}-$linear operators on $\mathbb{F}^n$ (or any other $n$-dimensional vector space over $\mathbb{F}$). – Mark Mar 09 '17 at 13:34
  • The Question would be better phrased as the "only two-sided ideals" are the trivial ones. Since $M_n(\mathbb{F})$ is a non-commutative ring, the difference in phrasing is crucial. – hardmath Mar 10 '17 at 00:57

2 Answers2

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There are rings in which every element is either a unit or a zero divisor and in which there exist nonzero proper ideals —k[x]/(x^2)$ is an example— so there is no way to prove what you want using only that property of the matrix ring.

Mariano Suárez-Álvarez
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There are many conditions sufficient to make a ring the union of its units and zero divisors. One easy one is that this is true in a right Artinian ring.

But there are lots of Artinian rings with very different ideal structures, and it does not say anything about the simplicity of such a ring.

rschwieb
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