Dtyjlui

In ring theory, is every real number divisible by any other nonzero real number, precisely because nonzero division is closed in $mathbb{R}$?

And so in general this idea of "closure under division" can be used to describe divisibility in a ring? E.g. one might say:

$mathbb{R}^*$ is closed under division, hence each one of its elements is divisible by any other one. (Which for example, in $mathbb{Z}$, remains false.)

The answer to the first question is unequivocally yes. This is why it's important that divisibility (and hence primality) is only meaningful relative to a base ring.

Mainly answered to say that since the previous answers weren't being sufficiently definite about it. That said:

As to the second question, one would normally just say that $mathbb{R}$ is a field. An element that every element is divisible by is called a unit, so you equivalently say that every nonzero element is a unit. The reverse condition, being divisible by every nonzero element, doesn't have a name I'm aware of, but is a nontrivial condition in certain rings.

People are objecting to the term "closure" on the grounds that it doesn't make sense except in the context of an operation defined on a larger set. This is a valid objection which also applies to a common way you'll see group theory taught: it's fine to say "a subgroup is a subset closed under the group operations," but it's not okay to say that "a group is a set closed under multiplication such that..." because what would it even mean for it not to be closed?

I don't think it is wise to talk about the division operator being closed in the ring minus the $0$ element, but rather that every nonzero element has a inverse element such that $xcdot x^{-1}=1$. Every real number "is divisible" by every nonzero real number because every nonzero real number has a multiplicative inverse.

I want to make this seemingly pedantic distinction because in a more general ring it is not clear what "division" should be for elements that don't already have a defined inverse. The best we can do is build the field of fractions, but again it is all defined in terms of the multiplication operation and the idea of what a multiplicative inverse is.

Well, yes, it is true that if $dinmathbb R setminus {0}$ and $ain mathbb R$, then $d|a$, since this means
$$exists kin mathbb R, ;a=dk,$$
(and is enough to take $k=frac ad$). But this is equivalent to have a multiplicative inverse for every non-zero element of the ring, which implies that this is in fact a field (or a division ring, if it were not commutative).

I would not say that $mathbb R^*$ is "closed under division", since for a ring $R$ the set $R^*$ is by definition the set of elements of $R$ with an inverse, so this is kind of the same. For instance, $$mathbb Z^*={-1,1},$$ and so $mathbb Z^*$ is also "closed under division", since
$$1|1, quad 1|-1, quad -1|1 quad text{and}quad -1|-1.$$