Lifting isomorphisms of fields to automorphisms of polynomial rings Let $L$ be a field and $\alpha, \beta$ algebraic over $L$ such that $L(\alpha)\cong L(\beta)$. If $q(t)$ and $p(t)$ are the minimum polynomials of $\alpha$ and $\beta$, respectively, does it follow that there exists an automorphism $\psi$ of $L[x]$ such that $\psi(q(t))=p(t)$. 
The converse to this question is immediate by pushing $\psi$ down to $L[x]/\langle q(t)\rangle$. I don't see a way though to lift the isomorphism between $L[x]/\langle q(t) \rangle$ and $L[x]/\langle p(t) \rangle$ up to $L[x]$. 
 A: Does this work? $L$ is the rationals, $\alpha=\root3\of2$, $\beta=\root3\of4$. So $L(\alpha)=L(\beta)$ (as is requested in the comment on the answer by @Hurkyl), $q(t)=t^3-2$, $p(t)=t^3-4$. Any automorphism of $L[t]$ has to take $1$ to $1$ and $t$ to $at+b$ for some rational $a$ and $b$, so we'd need $(at+b)^3-2=t^3-4$ as polynomials, and that won't happen with rational $a$ and $b$. 
A: A counterexample is to take $L = k(x^4)$, $\alpha = x^2$, and $\beta = x$.
Then $\alpha$ is algebraic over $L$ with minimal polynomial $f(t) = t^2 - x^4$.
Also $\beta$ is algebraic over $L$ with minimal polynomial $g(t) = t^4 - x^4$.
We have $k(\beta) \cong k(\alpha) \cong k(x^4)$
However, automorphisms of $L[t]$ that fix $L$ are all of the form $t \mapsto a + bt$, and so there is no automorphism of $L[t]$ that sends $f \to g$ or $g \to f$.
(I don't think anything relevantly weird happens if we allow automorphisms that don't fix $L$)
That said, while the fields $k(x)$ and $k(x^2)$ are isomorphic, the field extensions $k(x) / L$ and $k(x^2) / L$ are not isomorphic.
A: Not really much different than the answer by Gerry Myerson, but even simpler. Take $L=\mathbf R$, $\alpha=\mathbf i$, $\beta=2\mathbf i$, so that $q(X)=X^2+1$ and $p(X)=X^2+4$. But $(aX+b)^2+1=a^2X^2+2abX+b^2+1\neq X^2+4$ for all $a,b\in\mathbf R$.
