Describe the elements in $\mathbb{Q}(\pi)$
Attempt: $\mathbb{Q}(\pi)$ is the smallest field which contains $\mathbb{Q}$ and $\pi$
We know that $\nexists~ f(x) \in \mathbb{Q}[x]$ such that $f(\pi)=0$
Hence, $\mathbb{Q}[x]/\langle p(x) \rangle \not\approx \mathbb{Q}(\pi)~~\forall~~p(x) \in \mathbb{Q}[x]~~ ;~~ p(x)$ is irreducible in $\mathbb{Q}[x]$.
How do I move ahead? Please note that this exercise happens to be in the field extension chapter before algebraic extensions or finite fields are introduced.
EDIT: So, it means the the kernel of the mapping $f: \mathbb{Q}[x] \rightarrow \mathbb{Q}[\pi]$ is an isomorphism since kernel $f=0$. Hence, their field of fractions of the integral domain is also isomorphic. I get till this point, but how do we relate this to $\mathbb Q[\pi]$ . Does this $\implies \mathbb{Q}(\pi)= \{a_0+a_1 \pi + a_2 \pi^2 + \cdots + a_n \pi^n ~~|~~a_0,a_1,\cdots a_n \in \mathbb Q\}$
Thank you for your help.
If $\alpha$ is a complex number that is "transcendental over" $\mathbb{Q}$ then $\mathbb{Q}(\alpha)$ is isomorphic to $\mathbb{Q}(x)$ i.e. the field of rational functions over $\mathbb{Q}$.
To see why define the map, $ f: \mathbb{Q}[x] \to \mathbb{Q}[\alpha]$ where $\mathbb{Q}[\alpha]$ is the smallest ring containing $\alpha$ and $f(p(x)) = p(\alpha)$ where $p(x)\in \mathbb{Q}[x]$ is a polynomial. This mapping is a homomorphism, clearly surjective, and most importantly injective. The reason for this if there was a non-trivial polynomial $q(x)$ such that $f(q(x)) = 0$ then it would mean $q(\alpha) = 0$, which is impossible as we assumed that $\alpha$ was transcendental. Thus, we conclude that $\mathbb{Q}[x]$ and $\mathbb{Q}[\alpha]$ are isomorphic as rings. Since they both are integral domains we can form their field of fractions and get the claimed result.