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In $\mathbb{N}$, we can define an equivalence relation on the Cartesian product $\mathbb{N}^2$ as $(a,b) \sim (c,d)$ if and only if $a + d = b + c$. Then, the quotient set $\mathbb{N}^2 / \sim$ is isomorphic to $\mathbb{Z}$ (this is one way $\mathbb{Z}$ is defined in some books).

Similarly, there is an equivalence relation on the set of all Cauchy sequences in $\mathbb{Q}$, denoted $\mathbb{L}$, where $p_n \sim q_n$ if and only if $\lim_{n \to \infty}(p_n - q_n) = 0$. The quotient set $\mathbb{L} / \sim$ is isomorphic to $\mathbb{R}$.

My question is: Is there an equivalence relation on the $n$-fold Cartesian product $\mathbb{Q}^n$ such that the quotient set $\mathbb{Q}^n / \sim$ is isomorphic to the algebraic numbers $\mathbb{A}$, which are the roots of polynomials with integer coefficients? (I believe $n$ can be finite because $\mathbb{A}$ is countable.)

Furthermore, can such an equivalence relation be defined by some binary operation as in the first example? If so, is this binary operation commutative and associative?

Clarification: By “isomorphic to,” I mean isomorphic as a group. My consideration is: Is there a binary operation on $\mathbb{R}$ that makes $\mathbb{A}$ the smallest field closed under this operation, similar to how $\mathbb{Z}$ is closed under subtraction and $\mathbb{R}$ is the closure under limits, so that I can naturally extend $\mathbb{Q}$ into $\mathbb{A}$?

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  • $\begingroup$ There's an issue here about what you mean by "isomorphic to". In both of your examples, the quotient sets are isomorphic to the structures after defining the operations of $+$ and $\times$ in some way. e.g. there is a natural bijection $\mathbb{N}^2/{\sim}\to \mathbb{Z}$, and there is a natural way to define operations on $\mathbb{N}^2/{\sim}$ in such a way that this bijection turns into an isomorphism of rings. So I think what you really want to ask is whether we can define operations on $\mathbb{Q}^n/{\sim}$ in such a way that it is isomorphic to $\mathbb{A}$. $\endgroup$
    – Alex Kruckman
    Commented Sep 26 at 15:36
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    $\begingroup$ "isomorphic" in what sense? As sets, you can even take $n=1$ and $\sim$ to be trivial; this is indeed just a statement of countability. If you mean as a group, then this is not possible, since $\mathbb A$ is infinite-dimensional over $\mathbb Q$. $\endgroup$
    – Wojowu
    Commented Sep 26 at 15:37
  • $\begingroup$ But now to make the question non-trivial, you need to put some restriction on how you want the operations to be defined. After all, any countable set $X$ can be put in bijection with $\mathbb{A}$ (since $\mathbb{A}$ is countable), and this bijection can be turned into an isomorphism of rings by transferring the ring operations from $\mathbb{A}$ to $X$ along the isomorphism. $\endgroup$
    – Alex Kruckman
    Commented Sep 26 at 15:37
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    $\begingroup$ One way to make the question non-trivial is to ask whether the ring $\mathbb{A}$ is interpretable in the ring $\mathbb{Q}$ in the sense of first-order logic. But this might not be what you really intended to ask, since the construction of $\mathbb{R}$ from $\mathbb{Q}$ by Cauchy sequences is not a first-order interpretation. $\endgroup$
    – Alex Kruckman
    Commented Sep 26 at 15:39
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    $\begingroup$ As a group, $\mathbb A$ is the same as $\mathbb Q^{(\omega)}$ (the direct sum of countably many copies of $\mathbb Q$). Thus, polynomials are irrelevant anyway. $\endgroup$
    – Emil Jeřábek
    Commented Sep 27 at 6:43

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