abstract algebra - What is the field $mathbb{Q}(pi)$?

I'm having a hard time understanding section 29,30,31 of Fraleigh.

In 29.16 example, what is the field $\mathbb{Q}(\pi)$? and why is it isomorphic to the field $\mathbb{Q}(x)$ of rational functions over $\mathbb{Q}$?

(According to the definition, the field $\mathbb{Q}(\pi)$ is the smallest subfield of $E$ (extension field of $\mathbb{Q}$) containing $\mathbb{Q}$ and $\pi$.)

Thank you!

Answer

Define a ring homomorphism $f\colon \mathbb{Q}[x]\to\mathbb{R}$ by
$$f(p)=p(\pi)$$
so that (for example) $$f(\tfrac{1}{3})=\tfrac{1}{3},\quad f(x)=\pi,\quad f(2x^2+5)=2\pi^2+5,$$
and so on. The image of a ring homomorphism is a subring of the codomain; in the case of this particular ring homomorphism $f$, the image is given the name $\mathbb{Q}[\pi]$. It is the "smallest" subring of $\mathbb{R}$ that contains $\mathbb{Q}$ and $\pi$.

What is the kernel of this homomorphism $f$? That is, what polynomials $p\in\mathbb{Q}[x]$ have $\pi$ as a root? The answer is none (other than the obvious $p=0$). This is what it means for $\pi$ to be transcendental (in 1882, Lindemann proved that $\pi$ is transcendental). The first isomorphism theorem for rings now tells us that
$$\mathbb{Q}[x]/(\ker f)\cong \mathbb{Q}[\pi]$$

but since the kernel of $f$ is trivial, this statement just says that $\mathbb{Q}[x]\cong \mathbb{Q}[\pi]$. In other words, we see that $f$ is a ring isomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}[\pi]$.

Now see if you can prove the following general fact: if two integral domains $D_1$ and $D_2$ are isomorphic, then their respective fields of fractions $\mathrm{Frac}(D_1)$ and $\mathrm{Frac}(D_2)$ are also isomorphic.