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
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!