I have a set $R$ of $n\times n$ matrices where $n=1,2,3,\ldots$, and two binary operations $\oplus$ and $\otimes$. I can prove that $(R,\oplus,\otimes)$ is a "ring" if I am permitted to allow all the additive inverses $J_n$, the set of all $n\times n$ zero matrices, and allow all the multiplicative inverses $I_n$, the set of all $n\times n$ identity matrices (1's on the main diagonal, zeros elsewhere), i.e. all $I_n$ and $J_n$ where $n=1,2,3,\ldots$.

Can I actually claim this to be a ring, or does this fall outside the definition of ring. If so, what object is it?

I'm pretty sure it's not exactly a ring because the additive and multiplicative identities must be unique.

Update1 I think this could be related to the idea of multi-ring.

Update2 Perhaps a better reference of multiring is to be found here. Apparently, they were introuced in 1956, 1983 and then independently in 2006 (see [10,11,15] in reference list in ibid.) I think they're called hyperring and multiring synonymously, but not totally sure on that just yet.

  • $\begingroup$ It's not a ring. I don't know a name for this type of object, but I'm not an algebraist. $\endgroup$ – John Hughes Nov 4 '16 at 10:02
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    $\begingroup$ This almost certainly is not a ring, but it is not because the unit and zero element are not unique. This is not actually a part of the definition of a ring, but a consequence. $\endgroup$ – user259242 Nov 4 '16 at 10:07
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    $\begingroup$ In other words, if indeed $(R,\oplus,\otimes)$ were a ring, one could show that the unit and the zero element are unique. $\endgroup$ – user259242 Nov 4 '16 at 10:07
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    $\begingroup$ @Algorist I have never heard of "multi-ring spaces" but yes, maybe they work too. I wouldn't be surprised if it was actually the same thing I'm describing though, just in more mainstream terms. $\endgroup$ – rschwieb Nov 4 '16 at 12:22
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    $\begingroup$ @rschwieb not sure about the credibility of that link for "multi-rings" now... $\endgroup$ – Pixel Nov 5 '16 at 23:52

Of course this all depends on the operations which you have been a little sketchy about:

(from the comments)

By way of ⊗ and ⊕, which are defined to extend the sizes of the matrices so that addition and multiplication is possible in the usual way.

That could be done in many ways.

If you are, for example, just padding the smaller square matrix with zeros and then adding/multiplying as normal, then I think you are just looking at the direct limit of these rings in the category of rings without identity.

If you padded with zeros except for $1$'s along the diagonal, I think that makes it a direct limit in the category of rings with unity.

(I've heard of another construction that takes the LCM of the sizes of the matrices, and then uses block-diagonal matrices build from the original two so that you have two matrices with the LCM as the size which you can multiply and add. This seems more complicated than you probably intended.)

The thing is that with all these approaches, not all of the things you included ("all of the $n\times n$ identity matrices"/"all of the $n\times n$ zero matrices") stay distinct. If one matrix "extends" to a second, then those two things should be considered equal. So there is still only one additive identity, the common limit of all the finite additive identities. If you padded with $1$'s on the diagonal, there is still only one multiplicative identity, the common limit of all the finite identities.

  • $\begingroup$ Thanks so much for your answer, it gives me a few leads. $\endgroup$ – Pixel Nov 4 '16 at 12:17
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    $\begingroup$ @Algorist If you know(soon know) the definition of direct limits, then I hope the candidate homomorphisms for the direct system ('cone') are obvious. $\endgroup$ – rschwieb Nov 4 '16 at 12:24
  • $\begingroup$ It is also interesting to note that not all these ways of embedding give the same limit. There are uncountably many such limits in fact (at least if we take traceless matrices and consider the limit of the Lie algebras, so probably also for this case). $\endgroup$ – Tobias Kildetoft Nov 5 '16 at 10:09
  • $\begingroup$ @rschwieb your last paragraph got me thinking. Are you saying that If I define my objects so as to identify different sized similar objects, then i do have a ring because there would be one additive and one multiplicative identity which is just chosen depending on the sizes? So long as I can define a set of varying sized objects to reach a limiting object, so that everything is well defined... $\endgroup$ – Pixel Nov 6 '16 at 10:37
  • $\begingroup$ Either of the first two constructions might suit you then. $\endgroup$ – rschwieb Nov 6 '16 at 11:35

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