If $F(\alpha)=F(\beta)$, must $\alpha$ and $\beta$ have the same minimal polynomial?

Let's consider a field $F$ and $\alpha,\beta\in\overline{F}-F$ (where $\overline{F}$ is an algebraic closure). If $F(\alpha)=F(\beta)$, is it true that $\alpha$, $\beta$ have the same minimal polynomial over $F$?

For reference, I explain my way. (From here forward, all field isomorphisms are the identity on $F$.)

There exist polynomials $f,g\in F[x]$ such that $f(\beta)=\alpha$, $g(\alpha)=\beta$. Let $p$ and $q$ be the minimal polynomials of $\alpha$ and $\beta$ over $F$. There are field isomorphisms $$\sigma:F[x]/(p)\to F(\alpha),\qquad \tau:F[x]/(q)\to F(\beta)$$ (these two isomorphisms are constructed in the proof of the fundamental theorem of field theory), and the trivial isomorphism $i:F(\alpha)\to F(\beta)$ (I hope that there is isomorphism $j$ with $j(\alpha)=\beta$, but I can't prove it. How do you think about this?) Then $$\rho:=(\tau^{-1}\circ i\circ\sigma)$$ is also an isomorphism with $$\rho(x+(p))=f(x)+(q),\qquad \rho^{-1}(x+(q))=g(x)+(p).$$ From this, I got $q\mid f\circ p$ but not $q\mid p$. Is this way right? If it is, please proceed this way.

• This is hopelessly, almost universally false. For example, consider $\alpha=\sqrt{2}$, $\beta=3+\sqrt{2}$, $F={\mathbb Q}$ ... Commented Jul 5, 2015 at 9:47
• @EwanDelanoy: This is hopelessly, almost universally false. Well put, I was going to write almost the same thing. Commented Jul 5, 2015 at 9:53

Let $a$ and $b$ be any two distinct elements of $F$. Then we have $F(a)=F(b)=F$ but the minimal polynomial of $a$ over $F$ is $x-a$, whereas the minimal polynomial for $b$ over $F$ is $x-b$, which are different.
This is just a trivial example of the general behavior. Let $F$ be any infinite field, and let $a$ be anything that's algebraic over $F$ (regardless of whether it's an element of $F$ or not). Then there are infinitely many $b$'s that are algebraic over $F$ such that $F(a)=F(b)$ — consider $b=a+c$ as $c$ ranges over all elements of $F$ — but only finitely many roots of the minimal polynomial of $a$ (since a polynomial of degree $n$ over a field has at most $n$ distinct roots). Thus, all but finitely many of the $b$'s such that $F(a)=F(b)$ do not share a minimal polynomial with $a$.
If you really want a concrete example, observe that $\mathbb{Q}(\sqrt{2})=\mathbb{Q}(1+\sqrt{2})$ but $$\text{min poly of }\sqrt{2} = x^2-2,\qquad \text{min poly of }1+\sqrt{2}=x^2-2x-1$$
• I missed trivial case. Then, how do you think about nontrivial case with condition $a,b\notin F$? Commented Jul 5, 2015 at 9:43
Let $$\alpha$$ be an algebraic number (over the rationals) whose degree is a prime number $$p>2$$. Let $$f(x)$$ be a polynomial with integer/rational coefficients of degree less than $$p$$.
Then $$\mathbf{Q}[\alpha] = \mathbf{Q}[ f(\alpha)]$$. There are infinitely many polynomials $$f(x)$$ satisfying above condition, all their values at $$\alpha$$ are distinct, and so all these $$f(\alpha)$$ will have different minimal polynomial than $$\alpha$$, excepting for the finite number (which is $$p$$) of conjugates of $$\alpha$$.