If two elements a and b in a field extension L/K are algebraic in K and have the same minimal polynomial, prove that $K(a)=K(b)$

Question

I have a field extension $L/K$ and two elements $a,b \in K$, which are algebraic over $K$ and both have the same minimal polynomial, and need to prove that then $K(a)=K(b)$. I can see how this can be true but can't formally prove it. Should I use the fact that both minimal polynomials are irreducible in $K$?

Answer 1

They are isomorphic, but not necessarily equal.

If $p(x)$ is the common minimal polynomial, then $$ K(a) \cong K[x]/(p(x)) \cong K(b) $$

However, it may happen that $K(a) \neq K(b)$ : For instance if $p(x) =x^3 - 2 \in \mathbb{Q}[x]$, taking $L = \overline{\mathbb{Q}}$, you can check that $$ \mathbb{Q}(\sqrt[3]{2}) \neq \mathbb{Q}(\alpha) $$ where $\alpha$ is a complex root of $p(x)$

To require that $K(a) = K(b)$ for all roots $b$ of $p(x)$ is to say that $K(a)$ is the splitting field of $p(x)$.

Answer 2

The assertion is not true: let $K:=\mathbb{Q}$ and consider the polynomial $x^3-2$. It is irreducible over $\mathbb{Q}$ and is thus the minimal polynomial of each of its roots in $L:=\mathbb{C}$. However it has a real root $a$, thus $\mathbb{Q}(a)\subset\mathbb{R}$, while its other roots are non-real.