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If we have a ring $R$ then I can form a ring of matrices isomorphic to $R$ by setting $r \overset{\phi}{\mapsto} \left( \begin{array}{ccc} r & 0 \\ 0 & 0 \end{array} \right) $ and defining my ring of matrices to be the image of $\phi$. However, if I want to define this rigorously, I must specify which sets $\phi$ maps between, so I say $\phi:R \to M_2(R)$. However now $\phi$ is not technically a ring homomorphism as it does not map the identity of $R, \, (1)$ to the identity of $M_2(R)\,\, (=I_2)$ (though it does map to $1$ to the identity of the ring I want to form). So is it just impossible to define this ring as the image of a homomorphism without defining the ring already? I think I can define the set $$\tilde R=\left( \begin{array}{ccc} R & 0 \\ 0 & 0 \end{array} \right) = \left\{ \left( \begin{array}{ccc} r & 0 \\ 0 & 0 \end{array} \right):r \in R \right\}$$ and then give this a ring structure by defining $\phi : R \to \tilde R$, but is there no better way?

Edit I'm finding it difficult to phrase exactly what my question is, so I apologise for being unclear. My question is how I construct the specific matrix ring I give above, not just any matrix ring isomorphic to $R$. Though I'm aware that the construction @RobertLewis gives obviously works, this is a different (but isomorphic) ring. The reason I ask the question is that I want some 'neat' way of defining my ring above, but doing it via homomorphisms seems tricky. It's weird that $\phi:R \to M_2(R)$ isn't a ring homomorphism, but $\phi:R \to \tilde R$ is, even though $\tilde R \subset M_2(R)$ as sets. My question is basically to get to the heart of this, as it seems deeply unintuitive/wrong to me, even though it looks formally correct.

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The only alternative I can think of would be to consider everything as a non-unital ring, a.k.a. "rng" (Wikipedia link). Then the map $\phi:R\to\mathrm{M}_2(R)$ is a perfectly fine rng homomorphism, and since the rng $R$ happens to have an identity, so will the image $\phi(R)\subset\mathrm{M}_2(R)$.

If you want to stick with rings instead of rngs, I would have to say "yes, there is no better way than to just define $\tilde{R}$".

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  • $\begingroup$ So you first make it a rng homomorphism and then extend it to a ring homomorphism, once you note that the image has an identity? $\endgroup$ – James May 27 '15 at 22:46
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    $\begingroup$ That's right. Just to flesh it out a little more, you could proceed like this: the map $\phi:R\to\mathrm{M}_2(R)$ is a rng homomorphism, so its image $\phi(R)$ is a sub-rng of $\mathrm{M}_2(R)$. Since the rng $R$ has an unit element $1$, we know that $\phi(1)$ is an unit element $\phi(R)$, since for any $\phi(a)\in\phi(R)$, we have $$\phi(a)\cdot\phi(1)=\phi(a\cdot 1)=\phi(a)$$ Thus, in fact $\phi(R)$ is a ring, and $\phi:R\to\phi(R)$ is a ring homomorphism. $\endgroup$ – Zev Chonoles May 27 '15 at 22:49
  • $\begingroup$ Ok, thanks. This really cleared thing up :) $\endgroup$ – James May 27 '15 at 22:51
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    $\begingroup$ @ZevChonoles: A a matter of mathematical interest, suppose I ask, "Find all injective homomorphisms $\phi$ which map $R$ into $M_n(R)$". Any ideas about that one? $\endgroup$ – Robert Lewis May 27 '15 at 22:52
  • $\begingroup$ @RobertLewis: Not sure - however, given any rng homomorphism $\phi:R\to\mathrm{M}_n(R)$, the element $\phi(1)\in\mathrm{M}_n(R)$ must be idempotent, so I guess I'd start by trying to find all idempotent elements of $\mathrm{M}_n(R)$ - however, there's no guarantee that there is a $\phi:R\to \mathrm{M}_n(R)$ with $\phi(1)=A$ for every idempotent $A\in\mathrm{M}_n(R)$. $\endgroup$ – Zev Chonoles May 27 '15 at 23:07
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I'm a little unclear about the phrase "define this ring as the image of a homomorphism without defining the ring already"; but nevertheless:

What if we map $r \to rI$; that is, define a map

$\theta:R \to M_n(R); \;\; \theta(r) = rI, \tag{1}$

where $I \in M_n(R)$ is the $n \times n$ identity matrix? Then

$\theta(r + s) = (r + s)I = rI + sI = \theta(r) + \theta(s) \tag{2}$

and

$\theta(rs) = (rs)I = (rI)(sI) = \theta(r) \theta(s); \tag{3}$

$\theta$ is clearly a ring homomorphism. Also,

$\theta(1) = 1I = I; \tag{4}$

$\theta$ maps the identity of $R$ to the identity of $M_n(R)$, as requested. Also, if

$\theta(r) = 0, \tag{5}$

then

$rI = 0, \tag{6}$

which clearly implies $r = 0$; $\theta$ is in fact in injective homomorphism into a subring of $M_n(R)$; $R$ is, via $\theta$, isomorphic to the subring $RI$ of $M_n(R)$.

You can define a subring of $M_n(R)$ to be the image $R$ under $\theta$.

How's that?

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  • $\begingroup$ That's doesn't seem to be the subring of $\mathrm{M}_2(R)$ that the OP is talking about. You have $$\left\{\begin{pmatrix} r& 0\\ 0 & r\end{pmatrix}:r\in R\right\}$$ whereas they want $$\left\{\begin{pmatrix} r& 0\\ 0 & 0\end{pmatrix}:r\in R\right\}$$ $\endgroup$ – Zev Chonoles May 27 '15 at 22:40
  • $\begingroup$ @RobertLewis See my edit $\endgroup$ – James May 27 '15 at 22:44
  • $\begingroup$ @ZevChonoles: Well, truth to tell, as I said, I found a couple of things a little unclear. But it seems to me that a major concern of the OP lies in finding a homomorphic image of $R$ in $M_n(R)$ which maps $1_R$ to the identity of $M_n(R)$; thus I offered my suggestion. What are your thoughts on the matter? $\endgroup$ – Robert Lewis May 27 '15 at 22:44

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