$F(\alpha)$ is isomorphic to the field $F(x)$ of rational functions over $F$ in the indeterminate $x$ where $\alpha$ is transcendental over $F$ 
Q.$(a)$ If $\alpha$ is transcendental over a field $F$, then show that
$(i)$ the map $\mu : F[x] → F[\alpha],f (x) \mapsto f (\alpha)$ is an
isomorphism; $(ii)$ $F(\alpha)$ is isomorphic to the field $F(x)$ of
rational functions over $F$ in the indeterminate $x$. $(b)$ Show that
the field $\mathbb{Q}(\pi)$ is isomorphic to the field $\mathbb{Q}(x)$
of rational functions over $\mathbb{Q}$.

First of allWhat is the meaning of indeterminate $x$?
By the way hers what I have tried
$(i)$ Now clearly the map $\mu : F[x] → F[\alpha],f (x) \mapsto f (\alpha)$ is an epimorphism. If two polynomial expressions $f_1(\alpha)$ and $f_2(\alpha)$ are equal in $F(\alpha)$, their coefficients must be equal term by term, otherwise, the difference $f_1(α) − f_2(\alpha)$ will give a polynomial equation in α with coefficients, not all zero. This contradicts our assumption that $\alpha$ is transcendental over $F$. This shows that the map $\mu$ is a bijection and hence it is an isomorphism under usual operations of polynomials.
$(ii)$ Define the map $\varphi : F(\alpha) → F(x),\frac{f (\alpha)}{g(\alpha} \mapsto \frac{f (x)}{g(x)}$. Exactly the same reason as we have in $(i)$ holds for $(ii)$, so $ \varphi $ is an isomorphism
$(b)$ if we take $\alpha=\pi$ which is transcendental over $\mathbb{Q}$, from $(ii)$ we have the desired result.
It's quite absurd that $\mathbb{Q}(\pi) \cong \mathbb{Q(x)}$ because $\mathbb{Q(x)}$ is a ratio of polynomials. Or am I misunderstood something?
Is my reasoning correct? Or there is another better way to prove this.  Any help is appreciated thanks
 A: There is no mistake in your logic. The result you are calling absurd should be the reason why numbers $\alpha$ that don't satisfy a polynomial over a field $K$ are said to be transcendental. They literally transcend the comprehension of the field $K$. The field $K$ has no idea how to think about $\alpha$. Another way to interpret your isomorphism is as the field $\mathbb Q$ saying "I literally searched every polynomial expression and even rational expression that $\pi$ might satisfy.. and I failed to find anything". This is basically just one step away from the definition of transcendental numbers.
If you ask the abstract field of rational numbers $\mathbb Q$ how it thinks about the alien number $\sqrt2$, they will tell you "Well it's a number that I don't have within but when you square it, you get a number $2$ which I am very familiar with." If you ask it the same question about $\pi$, it will tell you: "I have no freaking clue. I know as much about it as I do about an indeterminate $x$. Neither of them satisfy any relationship that I am familiar with". So as far as the field $\mathbb Q$ is concerned, transcendental numbers (like $e$ and $\pi$ and $\tau$ etc ) basically behave like indeterminates $x$.
Now, why then, do we feel that $\pi$ is something very concrete and much less ethereal than something like an $x$? It's because we are so used to thinking of the field $\mathbb Q$ embedded in $\mathbb R$ which carries a natural metric structure( i.e. distance ) which lets us think things like: "Oh $\pi$ is kind of close to $3$ or even closer to $\frac{31}{10}$. But this is all additional metric structure we get by the embedding $\mathbb Q \hookrightarrow \mathbb R$. In fact, without this embedding, you cannot even define $\pi$. The object $\mathbb Q$ in isolation with only it's field structure knows none of this. As far it is concerned, transcendental numbers behave like indeterminates. In fact, if you take a look back at parts i and ii, this is what you have proven!
A: If $F(x)\rightarrow F(\alpha):f(x)\mapsto f(\alpha)$ is an isomorphism, then the kernel is $\{0\}$. Thus there is no algebraic equation $g(\alpha)=0$ for some non-zero polynomial $g(x)$. This means that $\alpha$ is not algebraic over $F$ and so can be viewed as an indeterminate over $F$.
