Reading Atiyah-MacDonald: Introduction to Commutative Algebra, I found the following definition of subring:

A subset $S$ of a ring $A$ is a subring of $A$ if $S$ is closed under addition and multiplication and contains the identity element of $A$. The identity mapping of S into A is then a ring homomorphism.

I know this definition is "wrong", as on the question I linked below is said:

Concept of a subring in Atiyah-Macdonald's book

But my question is, what if we change our "classic" definition of subring by this other? For me, it seems that everything remains equal and, at least, in the context of rings, there is not any contradiction.

For example, the following property is still true:

If $f: A \rightarrow B$ is a ring homomorphism, then $\operatorname{im}(f) \subset B$ is a subring.

I can't find any problem with this redefinition of subring. It will be welcome any correction or comment. Thanks everyone!

  • $\begingroup$ Sure, changing definitions won't give you a contradiction. It's just that a "subring" of a ring now isn't necessarily a ring, which is ridiculous (for example, $\mathbb{N}$ is a "subring" of $\mathbb{Z}$). $\endgroup$ – user3482749 Nov 15 '18 at 11:21
  • $\begingroup$ I understand what you say @user3482749. But if we don't name these structures "subrings" and see them as another possible substructure of rings, what are we obtaining? For example, we don't require ideals to be again rings. $\endgroup$ – DrinkingDonuts Nov 15 '18 at 11:30
  • $\begingroup$ A sub-semiring. $\endgroup$ – user3482749 Nov 15 '18 at 11:33
  • $\begingroup$ Is this fact suggesting that our theorems could be weakened because all of them are still valid? @user3482749 $\endgroup$ – DrinkingDonuts Nov 15 '18 at 11:50
  • 1
    $\begingroup$ No. A great many statements about rings are not true about semirings. $\endgroup$ – user3482749 Nov 15 '18 at 11:59

For clarity, let's use the following definitions (#-subring is not a standard term, just using it to distinguish the two possible definitions of subring).

A subset $S$ of a ring $A$ is a #-subring of $A$ if $S$ is closed under addition and multiplication and contains the identity element of $A$.

A subset $S$ of a ring $A$ is a subring of $A$ if $S$ is an additive subgroup of $A$, is closed under multiplication and contains the identity element of $A$.

Then any subring is a #-subring, so the property you've given clearly still holds. As will anything that proves some subset is a subring.

The problem is that a #-subring is not necessarily a ring: $\Bbb N$ is a #-subring of $\Bbb Z$, yet $\Bbb N$ is not a ring. Hopefully it is clear to you why a subring has to be a ring for it to be a sensible choice of definition!

A #-subring is, however, a semiring (a semiring has the same definition as a ring, but without the requirement for additive inverses - so in particular, any ring is a semiring), and in fact a #-subring of a ring $R$ is precisely a subsemiring of the ring (and thus semiring) $R$.

| cite | improve this answer | |
  • $\begingroup$ As I said in the response of the comment above, I understand the incoherence of call something a subring that is not itself a ring, but my question is, What structure has defined Atiyah and why all the theorems could be enunciated in these "new" terms? $\endgroup$ – DrinkingDonuts Nov 15 '18 at 11:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.