1
$\begingroup$

I know that the ring $\mathbb{Z}$ has the binary operation under addition, and when we are trying to identify whether a given ring is a subring of $\mathbb{Z}$, the subring must contain: the identity, inverse, and be closed under addition. My question is that does the subring also have to be closed under multiplication because multiplication is not the binary operation on $\mathbb{Z}$?

$\endgroup$
  • 1
    $\begingroup$ possible duplicate of How to check a set of ring is a subring? $\endgroup$ – Dietrich Burde Dec 18 '14 at 21:34
  • 1
    $\begingroup$ @DietrichBurde No this isn't a duplicate (of that question anyway). The subring test for this particular ring can be weakened. $\endgroup$ – Bill Cook Dec 18 '14 at 22:08
3
$\begingroup$

A subring has to be a ring, and a ring has two operations, not just one. Yes, it needs to be closed under multiplication.

$\endgroup$
3
$\begingroup$

In general a subring must be closed under multiplication.

However, in your specific case. No. You don't need to check.

Why? Because multiplication in $\mathbb{Z}$ is merely repeated addition/subtraction. Since you already have closure under addition and additive inverses, closure under multiplication follows for free.

The ring of integers is kind of weird in this way. It turns out that (1) subgroups, (2) cyclic subgroups, (3) normal subgroups, (4) subrings, (5) ideals, and (6) principal ideals all coincide for this ring.

$\endgroup$
  • $\begingroup$ Actually $\mathbb{Z}$ has no non-trivial subrings (in the usual terminology, where rings have a multiplicative unit). As you say, any additive subgroup of $\mathbb{Z}$ is an ideal and hence a subrng (where rngs don't necessarily have the multiplicative unit). $\endgroup$ – Rob Arthan Dec 18 '14 at 22:40
  • $\begingroup$ Ok then. "subrngs". Not everybody assumes their rings have multiplicative identities. Many universities use Joe Gallian's text. It's an example of a text which allows for rings without (multiplicative) identity. $\endgroup$ – Bill Cook Dec 19 '14 at 5:02

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.