Abstract Algebra - The Equality of Numbers - 0 in the Reals equals 0 in Zero Ring? I've been learning a bit about Abstract Algebra and it has me thinking a lot about numbers. One thing that has occurred to me is that if we group numbers into sets based on legal operations (the definitions of groups, rings, and fields for example define the properties of the numbers in their respective sets). Are numbers themselves unique? Or are they only defined within their own classes? Can we say for example that the number $0$ in the Zero-Ring is equal to (or even identical to) the number $0$ in say the real numbers: $$0 \in \left\{0\right\} = 0 \in \mathbb{R}$$
Are they the same number? Is there an equality that can be defined through some kind of morphism? Or are these numbers different from each other because they belong to different classes, and so are defined differently?
 A: Every ring has an additive identity - that's part of the definition of a ring.
We (almost) always call that element "$0$" because it's convenient. It helps us remember how it behaves in that ring's arithmetic.
But those $0$'s aren't the same thing at all (except when one ring is a subring of another). They are different things that happen to have the same name. Context disambiguates, so you know what each $0$ "really is".
A: The answer (to the more general question of whether the zeroes of different structures are equal) is going to unavoidably depend on your formalism, and tell you essentially nothing, regardless of what it is.
For example, if we define $\mathbb{Q}$ as the set of all equivalence classes of pairs $(a,b)$ where $a$ and $b$ are integers with $b$ non-zero under the equivalence relation generated by $(a,b) \sim (c,d)$ if $bc = ad$, and use ZFC as our formalism (and also define ordered pairs by $(a,b) = \{\{a\},\{a,b\}\}$ as is standard), then we cannot have $0_\mathbb{Q} = 0_\mathbb{Z}$, as we have $0_\mathbb{Z} \in \{0_\mathbb{Z}\} \in \{\{0_{\mathbb{Z}}\},\{0_\mathbb{Z},1_\mathbb{Z}\}\} = (0,1) \in [(0,1)] = 0_\mathbb{Q}$, so $0_\mathbb{Q} = 0_\mathbb{Z}$ would give a loop of set membership, violating the axiom of constructibility.
However, if we instead define $\mathbb{Q}$ first, then define $\mathbb{Z}$ as the subring generated by $1_\mathbb{Q}$, then we obviously have $0_\mathbb{Z} = 0_\mathbb{Q}$ in the most literal of senses.
