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Let $\mathcal{F}$ be the class of free fields of characteristic zero and let $X\neq \emptyset$ be a set. How would one show that there are no free fields in $\mathcal{F}$ over $X$?

Also, how would one identify the free field of characteristic zero that is generated by the empty set?

user1729
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Alvey
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    What is a free field? What does $X$ have to do with the question? – Gerry Myerson Aug 18 '13 at 00:28
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    Yes, Googling "free field" turns up a physics term and a probability term. In neither case is "characteristic zero" meaningful. Perhaps a translation issue? – Thomas Andrews Aug 18 '13 at 00:48
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    I don't understand why this question was closed. I remember once asking the same question. There are free groups, free semigroups, and lots of other free objects, however there are no free fields (over a non-empty set). The OP is asking why is this so? – user1729 Aug 18 '13 at 10:39
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    The question asks if there is a field $F(X)$ of characteristic $0$ with the property that $\hom_{\mathrm{Fld}}(F(X),K) \cong \hom_{\mathrm{Set}}(X,|K|)$ for every field $K$ of characteristic $0$. I hope now everybody gets it? – Martin Brandenburg Aug 18 '13 at 13:18
  • I don't know how 1729 knew to edit in the phrase, "over $X$", and I don't know how Martin got from the question as written to the question in the comment, but I guess I'm out of my depth here. – Gerry Myerson Aug 19 '13 at 12:46

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I assume that "free" is meant in the sense of universal algebra. In the class $\scr F_0$ of fields of characteristic $0$, the field $\mathbb Q$ of rational numbers is free on $\varnothing$ (usually expressed by calling it an initial object), i.e., it admits a unique homomorphism to each field in $\scr F_0$. On the other hand, as soon as $X$ has an element $x$, there cannot be a field in $\scr F_0$ free on $X$. To prove it, suppose $F$ were such a field. It has $x$ as en element (strictly speaking, the element I mean is the image of $x\in X$ under the canonical map from $X$ into $F$, but I'll omit that notational complication). If $x$ is transcendental, then there cannot be a homomorphism from $F$ to another field $K\in\scr F_0$ taking $x$ to an algebraic element of $K$. If, on the other hand, $x$ is algebraic in $F$, then there cannot be a homomorphism from $F$ to another field $K\in\scr F_0$ taking $x$ to a transcendental element of $K$ (or to an algebraic element with a different minimal polynomial). Either way, the universal property that defines "free" is violated.

Andreas Blass
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    I thought universal algebras could not describe fields. Maybe I'm misremembering... – Thomas Andrews Aug 18 '13 at 02:02
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    From Wikipedia on Universal Algebras: A more fundamental restriction is that universal algebra cannot study the class of fields, because there is no type in which all field laws can be written as equations (inverses of elements are defined for all non-zero elements in a field, so inversion cannot simply be added to the type). – Thomas Andrews Aug 18 '13 at 02:04
  • You can define free in this sense categorically, by an adjoint functor to the forgetful functor from the the category of fields of characteristic zero and the category of sets. – Thomas Andrews Aug 18 '13 at 02:13
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    @ThomasAndrews The statement you quoted from Wikipedia strikes me as excessively restrictive. It seems to say that universal algebra must limit itself to varieties ("laws written as equations"), whereas I would surely permit quasi-varieties as a legitimate study in universal algebra. Fields (whether or not of a fixed characteristic) don't even form a quasi-variety, and I would not recommend using universal algebra to study them, because almost none of universal algebra's theorems apply. But ... [continued in next comment] – Andreas Blass Aug 18 '13 at 02:16
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    But if someone (like the OP here) says "free" in connection with fields (or other algebraic structures), then I'd assume, absent evidence to the contrary, that the universal-algebraic meaning of "free" is intended. In the present case, that assumption is reinforced by (my best guess as to) what is to be proved --- existence of a free structure on $\varnothing$ and nonexistence on other sets. – Andreas Blass Aug 18 '13 at 02:18
  • Well, I remembered the fact about fields and universal algebra from a course I took in Universal Algebra at MIT in 1988 or 1989. I vaguely recall lots of the most useful UA results are not possible for general cases. – Thomas Andrews Aug 18 '13 at 02:20
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    @ThomasAndrews You're quite right that most of universal algebra disappears when you try to apply it to the class of fields of characteristic $0$. As this question showed, free structures don't exist (except for the initial $\mathbb Q$). Products don't exist, not even the empty product. The only congruence on a field is equality (whereas in universal algebra, there is at least the "big" congruence that identifies everything). All these properties of varieties fail for fields but hold for quasi-varieties. – Andreas Blass Aug 18 '13 at 02:25
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    Left-adjoint to forgetful is a good way to define "free" when free things on arbitrary sets exist, but in the present situation all we can say about this adjoint functor is that it doesn't exist. Reformulating the notion of "free" in terms of a universal property allows one to express more detail: The free thing on $X$ exists iff $X=\varnothing$. – Andreas Blass Aug 18 '13 at 02:29