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Is there an infinite field such that every non-zero element has finite multiplicative order and the set of all orders is bounded?

In Is there an infinite field such that every non-zero element has finite multiplicative order? , Clément Guérin remarks that the closure of $\mathbb{F}_p$ has the property that every element has finite order, and $\overline{\mathbb{F}_p}$ is infinite. The set of orders in this example is not bounded above.

Is there an example in which the set of orders is not bounded above, or is this impossible?

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  • $\begingroup$ @nobel-cat Your example is not a field. $\endgroup$
    – J.-E. Pin
    Commented Dec 20, 2021 at 17:00

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No. Suppose that $n$ is an upper bound on the orders. Then every nonzero $x$ is a root of $X^{n!} - 1$. Then there are at most $n!$ nonzero elements of $\mathbb{F}$. But this contradicts the fact that $\mathbb{F}$ is infinite.

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