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In essence, $f(\Bbb Z) \subset \Bbb Q$, where $f$ is an isomorphism. $f(a)=\frac{a}{1}$. It maps $\Bbb Z$ to its fraction field $Q$. The elements of $Q$ are equivalence classes. I wonder why accepting $\Bbb Z \subset \Bbb Q$ does not cause contradictions since it violates set theory. What exactly is the function of isomorphism? Do I need more tools to fully understand it?

@Ned clarified my thoughts, I am curious about the relation between subset and embedding. Embedding can be viewed as subset sometimes. Why it does not cause contradictions?

I added an example to make my question clearer. Many books prove $R[X]$ over a UFD is a UFD using a lemma (see http://people.math.binghamton.edu/mazur/teach/gausslemma.pdf --Theorem 3.). But $f \in R[X]$ means it can never be an element of $R$ or $K[X]$(the polynomial ring over the fraction field of $R$).

Andrew_Ren
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    Where do you see a violation of set theory? – lulu Feb 03 '23 at 15:55
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    What you have written is very unclear. Perhaps you are asking something like "how is it possible that a set can be in $1:1$ correspondence with a proper subset of itself?" But that is a characteristic of infinite sets. The natural numbers are clearly in $1:1$ correspondence with the even natural numbers, for example. – lulu Feb 03 '23 at 15:58
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    It doesn't violate set theory because when we work in $\mathbb{Q}$ we actually work with the set $f(\mathbb{Z})$, not with $\mathbb{Z}$ itself. So no contradictions. It causes no confusion because all the algebraic operations, ordering and other properties in $f(\mathbb{Z})$ are absolutely the same as in $\mathbb{Z}$. You basically ask why do we write $n$ instead of $\frac{n}{1}$. Just easier notation, nothing more. – Mark Feb 03 '23 at 16:05
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    The OP isn't asking about countability -- it's about "subset" vs. "embedding" i.e. the integer $5$ vs. "equivalence class of the pair $(5,1)$". – Ned Feb 03 '23 at 16:08
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    This abuse of notation is ubiquitous because it (greatly) simplifies notation (that may otherwise be so complex that it obfuscates more essential matters). It is used not only for (natural) embeddings but also for any (natural) maps, e.g. we often (ab)use common integer notation in any ring with $1,$ e.g. $,2,$ denotes $1+1$ in a polynomial ring $ R[x]$ (vs. $,2x^0 = x^0 + x^0)$ and in quotient rings such as $\Bbb Z/m\Bbb Z,$ (vs. $,2!+!m\Bbb Z = 1!+!m\Bbb Z + 1!+!m\Bbb Z).\ \ $ – Bill Dubuque Feb 03 '23 at 18:24
  • Further (but less common) $,1/2,$ denotes $,2^{-1},$ in any ring with odd characteristic (e.g. modular fractions in $\Bbb Z/m\Bbb Z$ for odd $m),,$ using the universal map arising from localization (partial rings of fractions). In computer algebra systems such "type coercions" are automated using ideas from category or type theory. $\ \ $ – Bill Dubuque Feb 03 '23 at 18:24
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    We can make it true set-theoretically by an obvious set-theoretic surgery in $\Bbb Q$ that replaces the subset $f(\Bbb Z)$ by $\Bbb Z.,$ But in algebra there is no need for such Frankenstein sets because we are generally not interested in internal (set-theoretic) structure (representations) of the elements of an algebraic structure but rather we are only interested in the structure up to isomorphism (see here), i.e. structures with the same operation tables (up to order) are considered the same - no matter how the elements are represented. – Bill Dubuque Feb 03 '23 at 18:41

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Your question isn't entirely clear, but this confusion probably comes from an "abuse of notation" involving the integers. Sometimes we use the symbol $\mathbb Z$ to refer to the integers as defined by pairs of natural numbers $(a,b)$ modulo the equivalence relation $(a_1,b_1)\sim (a_2,b_2)\iff a_1+b_2 = a_2+b_1$, but sometimes we use $\mathbb Z$ to refer to the subring of $\mathbb Q$ which is the image of the embedding $f$ that you describe in the question. You're absolutely right that these are different sets - in particular, the former is a set of equivalence classes whose elements are from $\mathbb N^2$, whereas the latter is a set of equivalence classes of pairs of equivalence classes (pairs of integers, that is) whose elements are from $\mathbb N^2$.

Perhaps this is overkill, but you might be more satisfied by the way this is handled in type theory. As opposed to sets, distinct types are always disjoint, and there's no notion of "subtype" analogous to the notion of a "subset" in set theory. In type theory, we only work with embeddings, so the only way in which we would refer to the integers $\mathbb Z$ as being "contained in" the rational numbers $\mathbb Q$ would be with respect to a particular embedding $f:\mathbb Z\hookrightarrow\mathbb Q$.

Franklin Pezzuti Dyer
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Embedding can be viewed as subset sometimes. Why it does not cause contradictions?

You understand that an embedding is not the same as a subset. The answer to your question lies in the "can be viewed as".

There is no confusion (or contradiction) because how you think about the relationship between the integers and the rationals depends on the kind of mathematics you are doing.

When you are interested in number theory, you want the integers to be a subset of the rationals. To do otherwise would clutter the ideas you care about with clumsy irrelevant notation.

When you are interested in how the rationals might be constructed from the integers you want to establish the embedding as part of your construction.

By analogy, when you want to drive your car somewhere you just turn the key and go. You don't think about what goes on under the hood. When it's time for a new car you think about whether to spring for an all electric. That there are two modes never causes confusion.

Ethan Bolker
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