# Isomorphism between algebraic extensions and irreducible polynomials

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!

• 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. – 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.