Q:

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!

A:

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.

A:

It is the smallest subfield of $\mathbb{C}$ that contains both $\pi$ and $\mathbb{Q}$. It can be shown that it is precisely 
$$\left\{\frac{p(\pi)}{q(\pi)}:\frac{p(T)}{q(T)}\in\mathbb{Q}(T)\right\}$$
Now, why is it isomorphic to $\mathbb{Q}(T)$?
Well, consider the smallest subring of $\mathbb{C}$ containing $\mathbb{Q}$ and $\pi$. This is just $\mathbb{Q}[\pi]$ (polynomials in $\pi$). You obviously get a $\mathbb{Q}$-algebra map $\mathbb{Q}\to\mathbb{Q}[\pi]$ defined by $f(T)\mapsto f(\pi)$ (you can verify this directly, or use the general fact that polynomial rings are the free commutative $\mathbb{Q}$-algebras). Now, since $\pi$ is trancendental (it doesn't satisfy any polynomial over $\mathbb{Q}$), this map is injective, and so you can lift this to a map $\mathbb{Q}(T)\to\mathbb{Q}(\pi)=\text{Frac}(\mathbb{Q}[\pi])$ in the obvious way. By definition of $\mathbb{Q}(\pi)$ (the explicit one), this is a surjective $\mathbb{Q}$-algebra map. But, since $\mathbb{Q}(T)$ is a field, it must actually be an isomorphism.

A:

Denoting by $\,x\,$ any variable (or unknown or transcendental element over $\,\Bbb Q\,$) , define the map
$$\phi:\Bbb Q(x)\to\Bbb Q(\pi)\;,\;\;\phi(f(x)):=f(\pi)$$
Now prove the above is a bijective ring homomorphism...

