This is exercise 6.8 of Falko, Algebra I. I think I solved most part of the problem but I still have some saddle points to handle.
Let $f,g$ be irreducible polynomials over $K$ field both without repeated roots in algebraic closure of $K$. $L=K(a),E=K(b)$ be the extension with $f(a)=g(b)=0$. Show if $f=\prod f_i,g=\prod g_i$ are prime factorizations over $K(b),K(a)$ respectively, we have $r=s$, and after reordering we have $E[x]/f_i\cong L[x]/g_i$. In particular $[K(b):K]deg(f_i)=[K(a):K]deg(g_i)$.
The first thing to notice is $K(a)[x]/(g)=\frac{K[y][x]}{(f,g)}=K(b)[x]/(f)$. The result is basically based on chinese remainder theorem and identification of minimal polynomial of $a$ over $K(b)$ and $b$ over $K(a)$ respectively. Since $K(a,b)=L(b)=E(a)$, I can remove the $K(a,b)$ part only from the product of $K-$algebras which are $E[x]/(f)\cong \prod_iE[x]/(f_i)$. Then induction on $r$ of $\prod_{ i\leq r}E[x]/(f_i)\cong\prod_{j\leq s} L[x]/(g_i)$ where I assumed that minimal polynomials of $a,b$ in $K(b)[x],K(a)[x]$ are $f_i,g_i$ respectively.
Is there any straightforward way to do this problem?
What I am doing here is too Ad Hoc? Any particular reason why I need to perform each step? Say I am telling a story and my story is quite unpleasant.
What is this problem telling me here? What is the context or what is $E[x]/(f)$?
The non-repeated roots here removed the multiplicity issue here. Will I still somehow maintain the isomorphism/ structure if I remove no-repeated roots condition?