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Let $F$ be a field such that for every $x \in F$ there exists a $k >0$ such that $x^k=1$. Does this imply that the characteristic of $F$ is strictly greater than zero?

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Yes. If the characteristic is zero then the prime subfield is isomorphic to $\mathbb{Q}$, and this contains elements - e.g. 2 - whose nonzero powers are never equal to 1.

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Every field $F$ of characteristic $0$ contains (up to isomorphism) the field of rational numbers. If you take the rational number $\frac{1}{2}$, say, then there is no integer $k>0$ such that $\left(\frac{1}{2}\right)^k = 1$.

For a finite field with $q$ (prime power) elements, one has $x^{q-1}=1$ for each element $x\ne 0$ and $x^q=x$ for each element $x$.

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