Field extension with transcendental number

Question

Is it something meaningful to write: $\Bbb Q(t)$. Where $t$ is not algebraic. If so for which values it is meaningful and what it means.

Answer 1

There are two distinct field constructions happening here. One is to recognize one field as a subfield of another $k\subseteq K,$ to find an element of the larger field $\alpha\in K$, and consider the smallest subfield of $K$ containing both the subfield $k$ and the element $\alpha.$ This field is denoted $k(\alpha).$

The other is to adjoin an indeterminate $t$ to an existing field $k$, giving the field of rational functions $k(t)$.

Both $k(t)$ and $k(\alpha)$ are extension fields of $k$. What they have in common and the reason for the common notation for them, is that $k(a)$ is the set of rational expressions in $a$ with coeffcients in $k$, for any symbol $a$.

The two operations yield isomorphic fields $k(t)\cong k(\alpha)$ if and only if $\alpha$ transcendental. This is essentially the definition of transcendental.

Conversely however, if $\alpha$ is not transcendental, $k(t)\not\cong k(\alpha).$ There is a relationship between them, which is that $k(t)$ is the field of fractions of the polynomial ring $k[t]$, and there is a ring homomorphism $k[t]\to k[\alpha]$, which is given by evaluation at $t=\alpha.$ But this does not extend to a field homomorphism $k(t)\to k(\alpha).$

So for example, $\mathbb{Q}(t)\cong\mathbb{Q}(\pi),$ as both are transcendental field extensions (extensions of infinite degree) of transcendence degree one. But $\mathbb{Q}(t)\not\cong\mathbb{Q}(\sqrt{2}),$ as the latter is an extension of degree two; an algebraic extension.

Answer 2

The ring $k[X]$ of polynomials over an integral domain $k$ is an integral domain. So you can built the quotient field $k(t):=Quot(k[t])$. (Observe that $\mathbb{Q}$ is an integral domain, so you can choose $k=\mathbb{Q}$). If $t$ is algebraic $k(t)=k[t]$ holds.