I was stuck on a problem for a while, and I'd just like to clarify something.
Let $K$ be a field and $S$ a subset of $K$. $S$ is not necessarily a field.
Let $K'$ be the intersection of all subfields of $K$ which contain $S$. This is equivalent to $K'$ being the smallest subfield of $K$ containing $S$.
The thing that caused me trouble for a while was proving that $K'$ is the set of all elements of $K$ that can be obtained from elements of $S$ by a finite sequence of field operations.
My understanding is that $+$ and $\times$ are the operations of a field, so all I can do is perform repeated combinations of addition and multiplication.
The books says that $\{1\}$ generates $\mathbb Q$.
As far as I could tell: if $+$ and $\times$ are the only field operations then $\{1\}$ can only generate $\mathbb N = \{1,2,3,\ldots\}$, which isn't even a field!
Am I right to assume that $-$ and $\div$ are field operations? If we allow $-$ and $\div$ then $\{1\}$ generates $\mathbb Q$.
My understanding was that subtraction was addition of the additive inverse: $x-y = x+(-y)$ and that division was multiplication by the multiplicative inverse: $x \div y = x \times y^{-1}$.
My worry was that $S$ was a subset, and not necessarily a field, so for any $x \in S$, it might not be true that $(-x) \in S$ or that $x^{-1} \in S$, and so do $x-y$ and $x\div y$ make sense?
Edit
I understand that $K'$ is supposed to the field of rational expressions $K(S)$, i.e. quotients of polynomials in elements of $S$. But I can't build that without division and subtraction.
In this context, the author of your book must have intended $-$ and $\div$ to be considered as field operations. Otherwise, the result that "$K'$ is the set of all elements that can be obtained from $K$ using a finite number of operations" would be false.
$x-y$ and $x\div y$ make sense in $K$, and $S$ is a subset of $K$, so they make sense in $S$. The result of the operation will not be in $S$, but that is not a problem; $-$ and $\div$ are not operations on $S$, they are operations on $K$.