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It is known, that there are no free fields (a free field would have a morphism to every other field even though there are no morphisms between fields of different characteristic). What is the situation, when we restrict or attention to fields of a given characteristic? Let $\mathsf{Fld}_k$ be the category of fields of characteristic $k$:

For what $k$ does the forgetful functor $\mathsf{Fld}_k \to \mathsf{Set}$ have a left adjoint?

This question deals with the case $k=0$, but I'm interested, if anything is known about the general case or at least other cases.

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Stefan Perko
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    Doesn't my answer to the $k=0$ question work for other characteristics also? The prime field $\mathbb Z/p$ is initial (i.e., free on $\varnothing$), but there is no free field-of-characteristic-$p$ on any nonempty set $X$ because the elements of $X$ wold have to map to transcendental elements and also to various algebraic elements. – Andreas Blass Nov 20 '15 at 20:53
  • @AndreasBlass I don't know, if you say so, I believe you (I will check it eventually). Currently, I'm just collecting examples for adjunctions. I don't actually know anything about field theory. – Stefan Perko Nov 20 '15 at 21:02

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It never has a left adjoint. The argument in this answer will work with essentially no modifications.

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Qiaochu Yuan
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  • So I guess I will be trying to learn what operads are tomorrow. For now I upvote your answer. – Stefan Perko Nov 20 '15 at 21:14
  • @Stefan: you don't need to know what operads are. Read starting from "claim." – Qiaochu Yuan Nov 20 '15 at 21:17
  • So I'm confused (and don't know much about fields). What is a field 'over' $\mathbb{Q}$? – Stefan Perko Nov 20 '15 at 21:22
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    @StefanPerko It's just a field of characteristic zero -- such a field contains $\mathbb{Q}$ as a prime subfield, and if $K \subset L$ is an embedding of fields then $L$ is a vector space over $K$. I guess it's the source of the terminology. – Najib Idrissi Nov 20 '15 at 22:31
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    I finally understand the argument, especially, that it has nothing to do with the characteristic being $0$, rather that field morphisms are injective and hence respect both $0$ and invertible elements. Thank you all. – Stefan Perko Nov 21 '15 at 11:02
  • @StefanPerko Oh by the way, operads are very interesting objects -- you can learn what they are regardless! It won't be a waste of time. – Najib Idrissi Nov 21 '15 at 12:27
  • @NajibIdrissi Your comment may actually motivate me to do so soon^^. – Stefan Perko Nov 21 '15 at 12:34