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In Algebraic Extensions of Fields by Paul J. McCarthy (p. 68), the following definition is given:

A field $k$ is quasi-finite if it is perfect and if it has, in any algebraic closure, exactly one extension of degree $n$ for each integer $n > 0$.

So here it means that we don't even allow having two cubic extensions $F,F' \subset \overline k$ of $k$ such that $F \cong F'$ as $k$-algebras, but $F \neq F'$ ? (For instance $\Bbb Q(\zeta_3\sqrt[3]2) \cong \Bbb Q(\sqrt[3]2)$ as fields [hence as $\Bbb Q$-algebras, since the characteristic is $0$] even if these fields are not equal).

Or is it meant that we only require having only one extension of degree $n$, up to isomorphism? If so, it is up to "field isomorphism", or up to "$k$-algebra isomorphism"?

Even if we fix an algebraic closure of $k$, this is not clear to me what the precise condition should be.


If the definition is really "exactly one" (without even considering isomorphisms), then:

Is there a perfect field which is not quasi-finite but has exactly one extension of degree $n$, up to field isomorphisms (for any $n \geq 1$)?

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    I think if it says one extension, it means one extension. That's the way finite fields work, and it's that fact about finite fields that is being generalized and called quasi-finite.2017-02-02
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    The next question would be "is there a perfect field which is not quasi-finite but has exactly one extension of degree $n$, up to field isomorphism (for any $n \geq 1$)?"2017-02-02
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    A weaker question is: find a field $F$ and an integer $n \geq 2$ such that $F$ has infinitely many distinct field extensions of degree $n$, but all of them are (pairwise) isomorphic as fields (or even as $F$-algebras).2017-02-02
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    I agree with Gerry Myerson's comment above: "exactly one extension" means "exactly one", i.e. we don't identify isomorphic extensions. Doing this would amount to saying "exactly one _Galois_ extension".2018-06-27
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    The important point being to mention that all the extensions lie "in any (fixed) algebraic closure".2018-08-27

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