I need to prove wether the following statement is true or false, but i have no idea on how to solve it:

Being $E/K$ an algebraic field extension and $\alpha, \beta \in E$ algebraics elements over $K$. If there's a field isomorfism $\phi:K(\alpha)\to K(\beta)$ so that $\phi(k)=k \enspace \forall k\in K$ $\Rightarrow \enspace \exists p(x)\in K[x]$ irreducible so that $p(\alpha)=p(\beta)=0$.

Any ideas? Thanks a lot!

  • 1
    $\begingroup$ You need $\phi|_K = Id$ and $\phi(\alpha) = \beta$. With those conditions you can apply $\phi$ and $\phi^{-1}$ to $p(\alpha)=0,q(\beta)=0$ where $p,q$ are the minimal polynomials. $\endgroup$ – reuns Oct 14 at 21:47

No. Take $K = \mathbb{Q}$ and $\alpha = 0, \beta =1$, $\phi= id_{\mathbb{Q}}$. Any polynomial $p$ over $\mathbb{Q}$ that vanishes in $0$ and $1$ contains the factors $X, X-1$. I.e.

$$p= X(X-1) A(X)$$

and clearly $p$ is not irreducible.


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