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Consider the polynômial $x^2-3$ over $\mathbb{Q}$. A splitting field would be $\mathbb{Q}(\sqrt{3})$. I also know, via some theorem that if I have another splitting field $S$ over $\mathbb{Q}$), it should be isomorphic to $\mathbb{Q}(\sqrt{3})$ and the isomorphisme $\phi$ is identity on the elements of $\mathbb{Q}$. I understand it should be at least isomorphic, but can S be different of $\mathbb{Q(\sqrt{3})}$ (in the set equality sense, not the isomorphic sense)? I thought of this because it is not specified that the isomorphism is the identity on $S \setminus \mathbb{Q}$.

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  • $\begingroup$ Take a splitting field $K$, where all elements have the same color as the rational numbers, which (we know) are pale green; then paint red the elements of $K\setminus\mathbb{Q}$. You have a very simple example of “different” splitting fields. $\endgroup$ – egreg Sep 28 '18 at 13:58
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$\mathbb{Q}(\sqrt{3}) \subseteq \mathbb R$ and $\mathbb{Q}[x]/(x^2-3)$ are different sets.

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  • $\begingroup$ Yes, but I was wondering, for subsets $S$ and $S'$ that contain $\mathbb{Q}$ and not just imbed it. $\endgroup$ – roi_saumon Sep 28 '18 at 14:09

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