Let $F$ be an algebraically closed field. This certainly implies that every non-zero element $x$ of $F$ has at least one $n$-th root, for each positive integer $n$. Does it in fact imply the condition that every non-zero element has $n$ distinct $n$-th roots?
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https://math.stackexchange.com/questions/2917312/proof-that-there-are-exactly-n-distinct-nth-roots-of-unity-in-fields-of-char This seems to have some relevant info – Ryan Shesler Jun 14 '20 at 01:39
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Hint: Recall that if $F$ is a field of characteristic $p,$ then $(a + b)^p = a^p + b^p.$ How many distinct $p$th roots of unity does this imply are there in $F$?
Hint: What equation must a $p$th root of unity satisfy? Once you have this, apply the first hint.
Stahl
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@user107952 that's correct; $x^p - 1 = x^p - 1^p = (x - 1)^p,$ so there is only one distinct $p$th root of unity in any field of characteristic $p.$ – Stahl Jun 14 '20 at 02:44
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What about the case of characteristic of zero? I think in that case, there would be n distinct n-th roots, because AFAIK all algebraically closed fields of characteristic zero are elementarily equivalent to the complex number field. – user107952 Jun 14 '20 at 10:12
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If the field is algebraically closed of characteristic 0, then there are indeed $n$ distinct $n$th roots of any nonzero element. See the link in Ryan Shesler's comment under your question for details. However, not all algebraically closed fields of characteristic 0 are equivalent to $\Bbb C$: you must assume the field has the same cardinality as $\Bbb C$ for this to be true. – Stahl Jun 15 '20 at 14:18
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The link discusses roots of unity in particular, but the ideas involving formal derivatives in the comments work more generally. – Stahl Jun 15 '20 at 14:20