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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 $L[x]$.

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Suppose $L$ is the rationals, $\alpha=2$, $\beta=3$, so $q(t)=t-2$, $p(t)=t-3$. Is there an automorphism $\psi$ of ${\bf Q}[x]$ such that $\psi(t-2)=t-3$? –  Gerry Myerson Feb 14 '13 at 5:29
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@GerryMyerson Sure $t\mapsto t-1$. –  JSchlather Feb 14 '13 at 5:31
    
OK, never mind. –  Gerry Myerson Feb 14 '13 at 6:16
    
I assume you meant $L[t]$, not $L[x]$. –  Hurkyl Feb 14 '13 at 6:19
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3 Answers 3

up vote 2 down vote accepted

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 $b$.

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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.

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It still has to be the case that the automorphisms preserve degree, so it wouldn't matter how the base field was permuted. I'm wondering if we strengthen the condition what happens. So if I required that $L(\alpha)=L(\beta)$ in the splitting field of $p(t)$ and $q(t)$ for instance or that $L(\alpha)$ and $L(\beta)$ were isomorphic as field extensions. –  JSchlather Feb 14 '13 at 6:31
    
Regarding your final sentence (and the essence of this example) it would be natural to interpret the requirement $L(\alpha)\cong L(\beta)$ of the question as an isomorphism of $L$-algebras, i.e., the isomorphism should commute with the obvious embedding of $L$; then this example is no longer one. –  Marc van Leeuwen Feb 14 '13 at 12:36
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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 $a,b\in\mathbf R$.

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I thought about (something much like) this and rejected it, on the grounds that if $\psi(X)=(1/2)X$ then $\psi(X^2+1)=(1/4)(X^2+4)$, which of course has the same roots as $X^2+4$. But it's not literally equal to $X^2+4$, so it's a perfectly good example. –  Gerry Myerson Feb 14 '13 at 23:00
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