$K[T]/(f(T))\simeq F[T]/(f(T))\Rightarrow K\simeq F$

Question

I'm trying to prove/disprove the following claim:

Let $F/K$ be a finite field extension. Suppose $f(T)\in K[T]$ is irreducible in both $K[T]$ and $F[T]$. Then the fields $K[T]/(f(T))$ and $F[T]/(f(T))$ are isomorphic $\Leftrightarrow K$ and $F$ are isomorphic.

This looks very intuitive to me, since the "constants" of $K':=K[T]/(f(T))$ are all in $K$ and the "constants" of $F':=F[T]/(f(T))$ are all in $F$, so $K'=F'\Rightarrow K=F$.

How can I formalize this?

Answer 1

Let $k=\Bbb{C}(x_1,x_2,\ldots)$ the field of rational functions in infinitely many undeterminates.

The splitting field of the cubic $T^3+x_n T+1\in k[T]$ is $L_n=k(\alpha_n,\Delta_n^{1/2})$ where $\alpha_n$ is a root and $\Delta_n=-4x_n^3-27$ is the discriminant, let $E_n=k(\alpha_n)$ and $Q_n= k(\Delta_n^{1/2})$ the cubic and quadratic subfield,

Let

$$K = \prod_{n=1}^\infty L_{2n}E_{2n+1},\qquad F =K(\Delta_1^{1/2})=Q_1\prod_{n=1}^\infty L_{2n}E_{2n+1}$$ where $\prod$ means the compositum of all those algebraic extensions of $k$.

Then we have the two cubic extensions $$K(\alpha_1)=E_1\prod_{n=1}^\infty L_{2n}E_{2n+1} \cong L_1\prod_{n=1}^\infty L_{2n}E_{2n+1}= F(\alpha_1)$$

but (since every copy of $Q_1$ in $K$ is contained in a copy of $L_1$)

$K $ is not isomorphic to $F$.

Answer 2

The ring homomorphism $K[T]\to F[T]$ provides a field homomorphism $K[T]/(f)\to F[T](f)$. From the dimension formula for field extensions you get the desired result.