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My book states the following theorem:

Suppose $K$ and $L$ are fields and $i:K\to L$ an isomorphism between them. Let $K(\alpha)/K$ and $L(\beta)/L$ be simple algebraic extensions, such that $\alpha$ has minimum polynomial $m_\alpha$ over $K$, and $\beta$ has minimal polynomial $m_\beta$ over $L$. Suppose further that $i(m_\alpha(t))=m_\beta(t)$. Then there exists an isomorphism $j:K(\alpha)\to L(\beta)$ such that $j(k)=i(k)$ for all $k\in K$, and $j(\alpha)=\beta$.

Is it true, in general, that if two simple algebraic field extensions are isomorphic, then such an $I$ needn't exist? Looking at $K=L=\mathbb{Q}$, $m_\alpha=t^2-2$ and $m_\beta=t^2-4t+2$, the extensions are isomorphic, but there is no automorphism $i:\mathbb{Q}\to \mathbb{Q}$ such that $i(m_\alpha(t))=m_\beta(t)$.

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  • $\begingroup$ Yes, in general such an isomorphism need not exist, the condition on the minimal polynomials is essential. $\endgroup$ – Starfall Sep 13 '17 at 14:22
  • $\begingroup$ @Starfall Thanks! A bit of a long shot: Does that mean that we can relax the conditions and still have two isomorphic extensions $K(\alpha)/K$ and $L(\beta)/L$? $\endgroup$ – Stensrud Sep 13 '17 at 15:11
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That's correct. An isomorphism $i$ as in the theorem exists iff $K(\alpha)$ and $L(\beta)$ are not just isomorphic but are isomorphic via an isomorphism which maps $K$ to $L$ and $\alpha$ to $\beta$ (in which case $i$ is the restriction of that isomorphism to $K$). It's entirely possible $K(\alpha)$ and $L(\beta)$ could be isomorphic but an isomorphism between them cannot send $\alpha$ to $\beta$, or cannot send $K$ to $L$.

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