Fields of characteristic zero 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?
 A: 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.
