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Let $D$ be an integral domain and $F$ be its field of quotients. Let $[(x, y)]$ denote the equivalence class of $(x, y)$. If $D$ is finite, then $D \simeq F$, so $|D| = |F|$. If $D$ is infinite, then the function mapping from $D$ to $F$ given by $d \mapsto [(d, 1)]$ is an injection. But what function is an injection from $F$ to $D$? I know there is an injection from $F$ to $D \times D$ given by $[(p, q)] \mapsto (p, q)$, but is there an injection from $D \times D$ to $D$? Is it even true that $|F| = |D|$ in this case?

Is it true that $|S \times S| = |S|$ for any infinite set $S$? If any of these are true, where could I find a proof?

Mason
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1 Answers1

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Let us write

$$\pi_1: \begin{cases}D\times D\to F\\(a,b)\mapsto [(a,b)]\end{cases}\\ \\ \pi_2:\begin{cases}F\to D\\ a/b\mapsto a\end{cases}$$

We have

$$D\times D\overset{\pi_1}{\twoheadrightarrow}F\overset{\pi_2}{\twoheadrightarrow} D$$ (Here the double headed arrows are used to represent surjections) Switching to cardinalities:

$$|D\times D|\ge |F|\ge |D|$$

As it has been proved by Tarski (see here for a proof and an historical digression), AC is equivalent to

$$|D\times D|=|D|$$

Thus we obtain

$$|D|\ge |F|\ge |D|$$

which implies, thanks to Schröder-Bernstein theorem

$$|D|=|F|$$