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Let $E$ and $F$ be two regular Galois extensions of $\mathbb{Q}(t)$ with group $G$, and let $E_{t_0}$ resp. $F_{t_0}$ be the residue fields corresponding to specializing $t\mapsto t_0\in \mathbb{Q}$. If $\{E_{t_0}\mid t_0\in \mathbb{Q}\}=\{F_{t_0}\mid t_0\in \mathbb{Q}\}$, can we conclude that $E$ and $F$ are the same up to some automorphism of $\mathbb{Q}(t)$ (extended to $E$)? (One certainly has to allow automorphisms here, as e.g.\ $\mathbb{Q}(\sqrt{t})$ and $\mathbb{Q}(\sqrt{t-1})$ of course yield the same set of specializations.)

In other words, does the set of specializations already fix the regular extension $E$ up to $PGL_2$-equivalence? If so, how far can the condition $\{E_{t_0}\mid t_0\in \mathbb{Q}\}=\{F_{t_0}\mid t_0\in \mathbb{Q}\}$ be relaxed? (I'm thinking something like "equality up to finitely many exceptions", or "up to a thin set")

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  • $\begingroup$ I don't understand what it means to specialize a Galois extension of $\mathbb{Q}(t)$ at some $t_0 \in \mathbb{Q}$. For example, what is the specialization of $\mathbb{Q}(t)[x]/(x^2 - \frac{1}{t})$ at $t = 0$? $\endgroup$ Dec 1, 2015 at 20:20
  • $\begingroup$ If the answer is that I'm supposed to rewrite this extension as $\mathbb{Q}(t, \sqrt{t})$ and specialize it to just $\mathbb{Q}$, then how do I know that the field I get doesn't depend on the choice of rewrite? $\endgroup$ Dec 1, 2015 at 20:27
  • $\begingroup$ @QiaochuYuan: If you want it formally unambiguous, you can define $E_{t_0}$ as the residue field of a place of $E$ lying above the place $t\mapsto t_0$ of $\mathbb Q(t)$. $\endgroup$ Dec 1, 2015 at 21:38
  • $\begingroup$ @Peter: and is it clear that this doesn't depend on the choice of place of $E$? $\endgroup$ Dec 1, 2015 at 21:41
  • $\begingroup$ @QiaochuYuan: Yes. The places are conjugate, therefore the residue fields are isomorphic. But they are Galois over $\mathbb Q$, so there is only one of them. $\endgroup$ Dec 1, 2015 at 21:53

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