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I am reading an introduction to abstract algebra by Keith Nicholson. There is a theorem that goes like this :

Let $\sigma : F \rightarrow{\overline{F}}$ be an isomorphism of fields, let $f \in F[x]$ be a nonconstant polynomial, and let $f^{\sigma}$ be its image by $\sigma$. If $E \supset F$ is a splitting field for $f$ and $\bar{E} \supset \bar{F}$ is a splitting field for $f^{\sigma}$, there is an isomorphism $E \rightarrow \bar{E}$ that extends $\sigma$

But if I consider the extension $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}(\sqrt{2})$, where we have that $\mathbb{Q}(\sqrt[4]{2})$ is the splitting field of $x^2-\sqrt{2}$ over $\mathbb{Q}(\sqrt{2})$. Then consider the isomorphism $\sigma : \mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2})$ that sends $\sqrt{2} \mapsto -\sqrt{2}$. So the theorem tells us that $\sigma$ extends to an automorphism $\hat{\sigma}$ of $\mathbb{Q}(\sqrt[4]{2})$. But we we have to have $\hat{\sigma} : \sqrt[4]{2} \mapsto \pm \sqrt[4]{2}$ but then $\sigma(\sqrt{2})=\sqrt{2}$ (so $\hat{\sigma}$ doesn't actually extend $\sigma$, no?). What am I doing wrong here?

roi_saumon
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1 Answers1

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Your mistake is expecting $\sigma$ to extend to an automorphism of $\mathbb{Q}(\sqrt[4]{2})$. The theorem doesn't say that $\sigma$ extends to an automorphism of $E$, it says that it extends to an isomorphism $E\to \overline{E}$. And $E$ and $\overline{E}$ are splitting fields of different polynomials, namely $f$ and $f^\sigma$.

We have $F = \overline{F} = \mathbb{Q}(\sqrt{2})$, $\sigma$ the automorphism of $F$ determined by $\sqrt{2}\to -\sqrt{2}$, and $f$ the polynomial $x^2-\sqrt{2}$. Then $f^\sigma$ is $x^2 + \sqrt{2}$. The splitting field $E$ of $f$ over $F$ is $\mathbb{Q}(\sqrt[4]{2})$, and the splitting field $\overline{E}$ of $f^\sigma$ over $\overline{F}$ is $\mathbb{Q}(\sqrt[4]{2}i)$.

And indeed, $E$ is isomorphic to $\overline{E}$ by a map $\widehat{\sigma}$ determined by $\sqrt[4]{2}\mapsto \sqrt[4]{2}i$ (note that $\sqrt[4]{2}\mapsto -\sqrt[4]{2}i$ would also work). As expected, $\widehat{\sigma}$ extends $\sigma$, since $$\widehat{\sigma}(\sqrt{2}) = \widehat{\sigma}((\sqrt[4]{2})^2) = (\widehat{\sigma}(\sqrt[4]{2}))^2 = (\sqrt[4]{2}i)^2 = -\sqrt{2}.$$

Alex Kruckman
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