Same minimal polynomial gives isomorphism

Let L and P be two field extensions of K. For $a \in L$ and $b\in P$ algebraic over K with the same minimal polynomial $f_a=f_b$ then there exists an isomorphism $w: K(a) \rightarrow K(b)$ with $w(a)=b$.

Proof attempt: Since $f$ is irreducible and has $a$ as root then $K(a)$ is isomorphic to $K[x]/(f)$ with the isomorphism $X+f\rightarrow a$. And we can do the same for $b$ and we get $K(b)$ is isomorphic to $K[x]/(f)$. Therefore $K(a)$ is isomorphic with $K(b)$.

My Questions:

1.What do I do now?

2."$f$ is irreducible and has $a$ as root then $K(a)$ is isomorphic to $K[x]/(f)$" we took it for granted in the lecture but why does this hold?

They might be trivial questions but I would really like to get an understanding of it.

Thanks in advance for the help.

For the first question, we just need to check the last condition $w(a) = b$. But just looking at the definition of the two isomorphisms in the composite $$ w : K(a) \to K[x]/(f) \to K(b), $$ the first maps $a$ to $x + (f)$, and the second maps $x + (f)$ to $b$. So $w(a) = b$, as we desire.

For the second question, we start by considering the evaluation $K$-algebra homomorphism $\phi : K[x] \to K(a)$ which maps $x$ to $a$. Then since $f(a) = 0$, every $g \in (f)$ is mapped to zero by $\phi$. (Just note that $\phi(g) = g(a)$.) Thus $\phi$ descends to a $K$-algebra homomorphism $\widetilde{\phi} : K[x] / (f) \to K(a)$. To show that $K(a) \cong K[x]/(f)$, it just remains to show that $\widetilde{\phi} : K[x] / (f) \to K(a)$ is an isomorphism. Indeed, being a map from a field, $\widetilde{\phi}$ is automatically injective.

Finally, to check that $\widetilde{\phi}$ is surjective, we just need a characterization of an arbitrary element of $K(a)$. We'll use that $K(a) = K[a]$, which is proved in many places (including I'm sure on this site). Given this there is essentially nothing to do: any element $v$ of $K[a]$ is just a polynomial in $a$ with cofficients in $K$, say $v = 3 a^2 + 42 a^3 + 6$, and replacing each occurrence of $a$ with $x$ gives a polynomial (e.g. $3 x^2 + 42 x^3 + 6$) which is mapped to $v$ by $\phi$. This completes the proof.

Remark. You might ask: "Hang on, this seems to make sense but how did we use anywhere that $f$ was a minimal polynomial?". Indeed, so long as $f$ was any polynomial over $K$ with the root $a$ we could still construct the map $\widetilde{\phi} : K[x]/(f) \to K(a)$ and see that this map is surjective. The problem is that we lose injectivity: $f$ will in general of course not be irreducible (equivalently, up to a scalar the minimal polynomial of $a$), so $K[x]/(f)$ won't be a field.

To expand on lhf's comments: the polynomial ring $K[X]$ has the property that for any commutative ring extension of the form $K[\alpha]$, there is a unique homomorphism $K[X]\to K[\alpha]$ fixing $K$ and sending $X\mapsto\alpha$. This is just a general property of polynomial rings. Now, if $\alpha$ is algebraic over $K$, then the kernel of this homomorphism is the ideal $(f_\alpha)$ generated by its minimal polynomial, and by the fundamental theorem on homomorphisms we then have $K[\alpha]\cong K[X]/(f_\alpha)$, and the isomorphism fixes $K$ and sends $\alpha\mapsto \overline X$.

Now if $a,b$ are algebraic with the same minimal polynomial $f$, the corresponding ring extensions $K[a]$ and $K[b]$ are isomorphic to $K[X]/(f)$, both via an isomorphism sending $a\mapsto\overline X$ or $b\mapsto\overline X$ and fixing $K$. Now just compose one of the isomorphisms with the inverse of the other one to get an isomorphism between $K[a]$ and $K[b]$ which fixes $K$ and maps $a\mapsto b$.

Now finally note that $K[\alpha]=K(\alpha)$ for algebraic $\alpha$, so the isomorphism also applies to the field extensions in question.