# Abstract Nonsense

## Matrix Rings (Pt. II)

Point of Post: This is a continuation of this post.

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Ideals of $\text{Mat}_n(R)$

As always whenever we define a new construct our first question after basic considerations is “what is the ideal structure of this new construct?” Of course, rings of matrices are no different. Ideally what we would like to do is come up with some kind of classification of these ideals, something of the form “Every ideal of $\text{Mat}_n(R)$ is of the form ______, and everything of the form ______ is an ideal of $\text{Mat}_n(R)$.” We had a similar goal when we were trying to classify the ideals of product rings where we saw, at least for the finite product of unital rings, that the ideals of $R_1\times\cdots\times R_n$ were precisely the sets of the form $\mathfrak{a}_1\times\cdots\times\mathfrak{a}_n$ where $\mathfrak{a}_k$ is an ideal of $R_k$. Taking this as a template we see that our classification would be best if the _______ were $\text{Mat}_n(\mathfrak{a})$. What we shall see is that, very similar to the case for product rings, is that we will get half our classification (sets of the form $\text{Mat}_n(\mathfrak{a})$ are ideals) for general rings and the full classification only for unital rings. So ,let’s get on with it!

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Before we prove that every set of the form $\text{Mat}_n(\mathfrak{a})$ is an ideal, we make note about a certain ‘functorial’ property about the matrix ring construction. Namely:

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Theorem: Let $R$ and $S$ be rings, then every morphism $f:R\to S$ induces a morphism $F(f):\text{Mat}_n(R)\to\text{Mat}_n(S)$ given by $(r_{i,j})\mapsto (f(r_{i,j}))$ with the property that $\text{id}_R\leadsto\text{id}_{\text{Mat}_n(R)}$ and $g\circ f\leadsto F(g)\circ F(f)$.

Proof: Evidently $F(f)$ is additive and

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\displaystyle \begin{aligned}F(f)\left((r_{i,j})(s_{i,j})\right) &=F(f)\left(\left(\sum_{t=1}^{n}r_{i,t}s_{t,j}\right)\right)\\ &=\left(f\left(\sum_{t=1}^{n}r_{i,t}s_{t,j}\right)\right)\\ &=\left(\sum_{t=1}^{n}f(r_{i,t})f(s_{t,j})\right)\\ &= \left(f(r_{i,j})\right)\left(f(s_{i,j}\right)\\ &= F(f)((r_{i,j}))F(f)((s_{i,j}))\end{aligned}

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Evidently $F(\text{id}_R)=\text{id}_{\text{Mat}_n(R)}$ and $F(g\circ f)(r_{i,j})=(g(f(r_{i,j}))=F(g)((f(r_{i,j})))=F(f)\left(F(g)((r_{i,j}))\right)$ and since $(r_{i,j})$ was arbitrary we may conclude that $F(g\circ f)=F(g)\circ F(f)$. The conclusion follows. $\blacksquare$

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Remark: For those who know the lingo the above theorem says that $R\mapsto \text{Mat}_n(R)$ and $f\mapsto F(f)$ is an endofunctor on $\bold{Rng}$. Moreover, it’s evident that $F$ reduces to an endofunctor on $\bold{Ring}$.

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From basic category theory we know that functors take isomorphisms to isomorphisms (for those who haven’t had exposure to basic category theory merely note that if $f:R\to S$ is an isomorphism then $F(f)\circ F(f^{-1)}=F(f\circ f^{-1})=F(\text{id}_S)=\text{id}_{\text{Mat}_n(S)}$ and similarly $F(f^{-1})\circ F(f)=\text{id}_{\text{Mat}_n(R)}$ and so $F(f)$ is an isomorphism with $F(f)^{-1}=F(f^{-1})$). Thus, we get as a simple corollary:

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Theorem: If $R\cong S$ then $\text{Mat}_n(R)\cong\text{Mat}_n(S)$

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More specifically, we see that $F$ takes monomorphisms to monomorphism and epimorphisms to epimorphisms as is evident from $F$‘s specific nature.

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With this theorem we have much less messy work to do with the following theorem:

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Theorem: Let $R$ be a ring, then for every $\mathfrak{a}\in\mathfrak{L}(R)$ we see that $\text{Mat}_n(\mathfrak{a})$ is an ideal of $\text{Mat}_n(R)$  and

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$\displaystyle \frac{\text{Mat}_n(R)}{\text{Mat}_n(\mathfrak{a})}\cong\text{Mat}_n\left(R/\mathfrak{a}\right)$

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Proof: Since the canonical epimorphism (projection) $\pi:R\twoheadrightarrow R/\mathfrak{a}$ is an epimorphism we have by prior comment that the map $f:\text{Mat}_n(R)\to\text{Mat}_n(R/\mathfrak{a}):(r_{i,j})\mapsto (r_{i,j}+\mathfrak{a})$ is an epimorphism since $f=F(\pi)$. Since evidently $\ker f=\text{Mat}_n(\mathfrak{a})$ we have, by common theorem, that $\text{Mat}_n(\mathfrak{a})$ is an ideal of $\text{Mat}_n(R)$ and the rest follows from thefirst isomorphism theorem. $\blacksquare$

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We now show the converse for unital rings as claimed, in particular:

Theorem: Let $R$ be a unital ring, then every ideal of $\text{Mat}_n(R)$ is of the form $\text{Mat}_n(\mathfrak{a})$ for some ideal $\mathfrak{a}\in\mathfrak{L}(R)$.

Proof: Let $\mathcal{I}$ be the ideal in $\text{Mat}_n(R)$ and let $J$ be the set of all elements of $R$ which appear as an entry in some matrix in $\mathcal{I}$. We claim that $J$ is an ideal of $\text{Mat}_n(R)$. Indeed, let $r\in J$ be arbitary, by assumption there is some matrix $M\in\mathcal{I}$ for which $r$ is an entry, say entry $(i,j)$. We note by assumption that $\mathcal{I}$ is an ideal that $E_{1,i}ME_{j,1}\in\mathcal{I}$. That said, a quick check shows that $E_{1,i}ME_{1,j}$ is equal to $\text{diag}(r,0,\cdots,0)$, and so $\text{diag}(r,0,\cdots,0)\in\mathcal{I}$. So, if $s,t\in R$ are arbitrary then by assumption we have that

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$\text{diag}(s,0,\cdots,0)\,\text{diag}(r,0,\cdots,0)\,\text{diag}(t,0,\cdots,0)=\text{diag}(srt,0,\cdots,0)\in\mathcal{I}$

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and so by definition $srt\in J$. Since $s,r,t$ were arbitrary we may conclude that $J$ is an ideal of $R$ as claimed.

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It now suffices to that $\mathcal{I}=\text{Mat}_n(J)$. Evidently the $\subseteq$ inclusion is obvious and so it suffices to show the reverse inclusion. To prove this we let $(r_{i,j})$ be any matrix with elements from $J$  and note that by previous comment $\text{diag}(r_{i,j},0,\cdots,0)\in\mathcal{I}$ and so $E_{i,1}\text{diag}(r_{i,j},0,\cdots,0)E_{1,j}\in\mathcal{I}$, noting then that $\displaystyle (r_{i,j})=\sum_{i,j=1}^{n}E_{i,1}\text{diag}(r_{i,j},0,\cdots,0)E_{1,j}$ finishes the argument. $\blacksquare$

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This really is quite the comforting phenomenon. There is no reason (and I challenge anyone to contest this) that there is no reason to guess a priori that the ideals would be this simple–they could be some kind of crazy, hard to write down, heap of mess. Moreover, their sensitivity to the amount of ideals the underlying ring has means that homomorphisms for matrix rings over division rings, or fields are very nice. In particular, we have the following:

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Theorem: Let $D$ be a division ring and $S$ an arbitrary ring, then any non-zero ring morphism $f:\text{Mat}_n(D)\to S$ is surjective.

Proof:We know that it suffices to prove $\ker f$ is trivial. To do this we merely note that $\ker f$ is an ideal of $\text{Mat}_n(D)$ and by the previous theorem that this implies $\ker f=\text{Mat}_n(\mathfrak{a})$ for some ideal $\mathfrak{a}$ of $D$. But, by previous theorem that the only ideals of $D$ are trivial, and since $\ker f\ne\text{Mat}_n(D)$ by assumption we may conclude that $\ker f=\text{Mat}_n(\{0\})=\{0\}$ and so the conclusion follows by previous remark. $\blacksquare$

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.