Integer rings and UFDs in transcendental field extensions of $\mathbb{Q}$ I recently began to study algebraic field extensions of $\mathbb{Q}$ aka number fields and especially the definition of algebraic integers in these fields. Some rings of algebraic integers are unique factorization domains (UFDs) e.g. $\mathbb{Z}[i]$ and $\mathbb{Z} \left[\frac{1 + \sqrt{-19}}{2} \right]$ whereas others are not e.g. $\mathbb{Z}[\sqrt{-5}]$.
In example 2.14 of Frazer Jarvis' book Algebraic number theory it is stated that:



*$\mathbb{Q(π)}$ is not a number field; $\pi$ does not satisfy any polynomial equation over $\mathbb{Q}$ (as it is transcendental); therefore $[\mathbb{Q}(π) : \mathbb{Q}]$ is infinite.


An integer in a number field of degree $n$ is defined as a root of some polynomial $X^n + a_1 X^{n-1} + \dots + a_n$ where $a_i \in \mathbb{Z}$. But as the degree of a transcendental extension is infinite, it seems to me that this definition does not seem viable for these field.
This makes me wonder whether there is actually a way to define transcendental integers in the field $\mathbb{Q(π)}$ and other transcendental extensions such that this set forms a ring under the usual arithmetic operations. If so, does the ring of integers have an integral basis and can it be shown to be a UFD in some cases?
Also can you recommend any book to dive deeper into the theory of transcendental field extensions? Thanks in advance.
 A: 
An integer in a number field of degree $n$ is defined as a root of some
polynomial $X^n+a_1X^{n-1}+\ldots+a_n$ where $a_i \in \mathbb{Z}$.
But as the degree of a
transcendental extension is infinite, it seems to me that this
definition does not seem viable for these field.

This is imprecise. An algebraic integer can be defined without any reference to a particular number field: an algebraic integer is simply a root of a polynomial with integer coefficients and leading coefficient $1$, of any degree. You can view the set $\mathcal{O}$ of algebraic integers inside the complex numbers $\mathbb{C}$ or a fixed algebraic closure $\overline{\mathbb{Q}}/\mathbb{Q}$. It turns out to be a ring: the sum and product of algebraic integers is an algebraic integer.
Given a field $K/\mathbb{Q}$, the ring of integers of $K$ is simply the intersection $\mathcal{O}_K = \mathcal{O} \cap K$. Once again, there's nothing in this definition that compels $K/\mathbb{Q}$ to be finite, although in Algebraic Number Theory we naturally tend to focus on number fields. For example, if $K = \mathbb{Q}(\pi)$ then $\mathcal{O} \cap K = \mathbb{Z}$, simply because $\mathcal{O} \subset \overline{\mathbb{Q}}$ while $\mathbb{Q}(\pi) \cap \overline{\mathbb{Q}} = \mathbb{Q}$. You may have something else in mind when you speak of "transcendental integers", but the direct application of the usual definition of algebraic integers to the case of purely transcendental extensions does not yield anything new.
Transcendental field extensions are studied from many different perspectives: function fields play a central role throughout algebraic geometry, for example. Specifically for transcendental numbers inside $\mathbb{C}$, you may be interested in the field of Transcendental Number Theory, for example in Alan Baker's classic monograph.
