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

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?

Solution 1:

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.