The above question was given to me as an assignment but I'm a bit stuck and I think the way I've been going about it is a dead end, or if not a dead end, I can't figure out a way to finish it off
So in a previous question we had to prove that $\mathbb{Q}/(x^{2}-5) \simeq\mathbb{Q}[\sqrt{5}]$, and I did this with the isomorphism $\varphi:\mathbb{Q}/(x^{2}-5)\to\mathbb{Q}[\sqrt{5}],\;ax+b\mapsto a\sqrt{5}+b$
I then thought that I could do the same with this, question, isomorphically mapping $\varphi:K(a) \to K[x]/(f)$ where $f$ is the minimal polynomial of $a$, and then mapping $\psi:K[x]/(f) \to K(b)$ using something of a similar form to my above example. What I came up with was
$$\varphi: K(a) \to K[x]/(f),\; \sum_{i=0}^{n-1}c_{i}a^{i} \mapsto \sum_{i=0}^{n-1}c_{i}x^{i}$$
where $n = \deg(f)$, but I don't know how to prove this is an isomorphism or if I was wrong and it actually isn't one. The additive property is easy and works out fine, but I don't know how to prove the multiplicative property because of when the power of $a$ goes above $n-1$. I looked around for answers to this before posting and I know that $K(a) \simeq K[x]/(f) \simeq K(b)$ is true, but I couldn't find a proof anywhere. Is what I'm doing a dead end or am I barking up the completely wrong tree, and if so, how should I go about proving this?
I agree with the comments. This is just an overview of the formal proof.
Define, $\phi:K[x]\to K(a)$ by $\phi(q(x))=q(a)$ for all $q(x)\in K[x]$ where $K(a)$ is the subfield of $L$ that contains both $K$ and $\{a\}$. Now it is a homomorphism (check that!) and for any $r\in K(a)$ we set the constant polynomial $q(x)=r,\forall x$ we see $\phi(q(x))=q(a)=r$ so $\phi$ is onto-homomorphism with kernel $Ker(\phi)=\{q(x)\in K[x]:q(a)=0\}=\{q(x):a\text{ is a root of }q(x)\}=(p(x))$, ideal generated by $p(x)$, $p(x)$ being the minimal polynomial having zero at $a$.
Now by, First isomorphism theorem, $$K[x]/(p(x))\cong K(a)$$ Now $a,b$ both have same minimal polynomial so, $$K[x]/(p(x)\cong K(b))$$ and thus the results follows.