Let $\alpha\in R$. Prove that $\mathbb Q(\alpha)\cong\mathbb Q(x)$ iff $\alpha$ is transcendental. Let $\alpha\in\mathbb R$. Prove that $\mathbb
    Q(\alpha)\cong\mathbb Q(x)$ if and only if $\alpha$ is
    transcendental
So is this saying that we need everything (all expressions involving x) to split a transcendental? This chapter is defeating me. I don't see what is being said here.
 A: You can easily define an isomorphism $\phi: \mathbb{Q}[x]\to \mathbb{Q}[\alpha]$ namely $\phi(p(x))=p(\alpha)$ (it's clearly surjective and the kernel is $0$ since $\alpha$ is trascendental). From this, it follows that $\mathbb{Q}(x)$ is isomorphic to $\mathbb{Q}(\alpha)$ (remember that $\mathbb{Q}(x)$ is the field of fractions of the ring $\mathbb{Q}[x]$)
A: Think about what happens if $\alpha$ is something like $\sqrt2$.  Then in $\mathbb Q(\alpha)$ you have things happening like $\alpha^2-2=0$.  But in $\mathbb Q(x)$, $x$ is a variable and so $x^2-2\not=0$, in fact no polynomial is zero in $\mathbb Q(x)$ except 0 itself.  So if $\alpha$ is transcendental it does not satisfy any polynomial so $P(\alpha)$ is not zero for any polynomial, just like in $\mathbb Q(x)$.  
A: This is essentially the same as Daniel’s answer, but a bit fleshed out with more background.

Look up localization and especially its universal property if you don’t know about this yet.
There’s a ring homomorphism $φ\colon ℚ[X] → ℚ(α),~X ↦ α$ with $ℚ[α] ⊂ \operatorname{img} φ$. This map has trivial kernel if and only if $α$ isn’t root of any polynomials with rational coefficients, i.e. if and only if $α$ is transcendental. But in this case, the situation of the universal property of localization for $S = ℚ[X]\setminus \{0\}$ is given because $ℚ(α)$ is a field and now $φ(S) ⊂ ℚ(α) \setminus \{0\}$.
The localization of $ℚ[X]$ by $S$ is $S^{-1}ℚ[X] = ℚ(X)$. You only need to show that the resulting map $ℚ(X) → ℚ(α)$ stays injective and becomes surjective.
