As far as I know, one can define a ring with or without a multiplicative identity.

My question is: what kind of properties, theorems, etc. get lost when one talks about rngs instead of rings, and where and how are rngs relevant?

For example, does anything change when we discuss torsionless/reflexive modules?


First of all, there are many rings without units that are very interesting. A prominent example is the $C^*$-algebra $C_0(X)$, that is $C_0(X)=\left\{f\in C(X)\mid \mbox{ $f$ vanishes at infinity}\right\}$. Here $X$ is a Hausdorff topological space. In fact, $C_0(X)$ has a unit if and only if $X$ is compact. These spaces are very important since any commutative $C^*$-algebra is of this form.

Now given a ring $R$ without unit, one can always embed $R$ into a unital ring of the same characteristic (which can still be defined for non-unital rings). So one can use the usual ring theory on this unitization. (A unitization is not unique, so the usefulness depends on what you try to do).

Obviously if you work in a non-unital setting, ring theory becomes more difficult. Normally if $R$ and $S$ are rings and $f:R\rightarrow S$ is a ring morphism, one requires that $f(1_R)=1_S$, in a non-unital setting we no longer have this and thus ring morphisms are more difficult.

Your question is very general, and there are many things that can go wrong. For example if $R$ is a ring with unit, then any ideal in the matrix ring $M_{n\times n}(R)$ is of the form $M_{n\times n}(A)$. This is no longer true if $R$ has no unit. Hence the theory of matrices over non-unital rings is quite different.

  • $\begingroup$ Yes, I was trying to find a way to make it a bit more specific but couldn't come up with more than the last question posed, about the reflexive/torsionless modules. Do you have any comments on that? It came up when I write something like "Let $R$ be a ring", and then started to worry whether I should then specifiy whether it has a unit or not. Of course I can just say, well, my ring is going to have unity so I can just say that, but then there is this desire to be always be as general as possible (which is not always a good thing I think). $\endgroup$ – B. Pasternak May 14 '16 at 13:36
  • $\begingroup$ Anyways thanks for your answer, as I understand it, the unit property really makes the ring behave more nicely, in a broad way. $\endgroup$ – B. Pasternak May 14 '16 at 13:38
  • $\begingroup$ Hmm, haven't given it much thought, but this page is somewhat interesting : math.stackexchange.com/questions/1596880/…. But I haven't dealt with unitless rings too much except in $C^*$-algebra theory. $\endgroup$ – Mathematician 42 May 14 '16 at 13:50

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