# Extensions of $\sigma:\mathbb{Q}(\xi)\rightarrow\mathbb{Q}(\xi), \ \sigma(\xi)=\xi^{2}$ to $\mathbb{Q}(\xi)(\sqrt[3]{2})$

Let $L$ be the splitting field of $T^{3}-2$ and let $\sigma:\mathbb{Q}\left(\xi\right)\rightarrow\mathbb{Q}\left(\xi\right)$ be a morphism defined by $\sigma\left(\xi\right)=\xi^{2}$. Find all extensions

• $\hat{\sigma}:L\rightarrow L$ of $\sigma$

• $\hat{\sigma}:L\rightarrow\mathbb{Q}\left(\xi\right)$ of $\sigma$.

Could you please check, if I did this right ? Please answer quickly, since I have an exam in couple of hours, where these things are bound to come!

Proof-Sketch: We have $L=\mathbb{Q}\left(\xi\right)\left(\sqrt[3]{2}\right)$.

First case: The possible extensions $\mathbb{Q}\left(\xi\right)\left(\sqrt[3]{2}\right)\rightarrow L$ of $\sigma$ are given by the roots $\xi,\xi^{2}\sqrt[3]{2},\sqrt[3]{2}$ in $L$ of the minimal polynomial $T^{3}-2$ of $\sqrt[3]{2}$ over $\mathbb{Q}\left(\xi\right)$ [this is the minimal polynomial since it is normed and annihilates $\sqrt[3]{2}$, so its degree is $\leq3$, $\sqrt[3]{2}\not\in\mathbb{Q}\left(\xi\right)$, so its degree is $\geq2$ - but how can I show that it cannot have degree $=2$ ??]: \begin{eqnarray*} & \hat{\sigma_{1}}:\xi\mapsto\xi^{2},\ \sqrt[3]{2}\mapsto\xi\\ & \hat{\sigma_{2}}:\xi\mapsto\xi^{2},\ \sqrt[3]{2}\mapsto\xi^{2}\sqrt[3]{2}\\ & \hat{\sigma_{3}}:\xi\mapsto\xi^{2},\ \sqrt[3]{2}\mapsto\sqrt[3]{2}. \end{eqnarray*} Second case: This time the minimal polynomial has only the root $\xi$ in $\mathbb{Q}\left(\xi\right)$, so the only possible extension $L\rightarrow\mathbb{Q}\left(\xi\right)$ is \begin{eqnarray*} & \hat{\sigma_{1}}:\xi\mapsto\xi^{2},\ \sqrt[3]{2}\mapsto\xi,\ \end{eqnarray*}

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What is $\xi\;$? I presume its a primitive third root of unity? In which case there can't be a homomorphism from $L$ to $\mathbb Q(\xi)$. –  JSchlather Jan 31 '13 at 7:51

($\xi$ is a primitive third root of 1.)
There is a misprint in the first case, the three roots of $T^3 - 2$ are $\sqrt[3]{2}, \xi \sqrt[3]{2}, \xi^2 \sqrt[3]{2}$. (I believe your incorrect list has had an effect on your solution in the second case, see below.)
To prove $T^3 - 2$ is indeed the minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}$, you can show that $T^3 - 2$ is irreducible in $\mathbb{Q}[x]$. Now for polynomials of degree 2 and 3, irreducibility in $F[x]$, where $F$ is a field, is equivalent to the polynomial having no roots in $F$. (This criterion fails for degree 4 and higher, see for instance $(T^2 + 1)^2 \in \mathbb{Q}[x]$.)
In the second case, $\sqrt[3]{2}$ should be mapped onto another root of $T^3 - 2$, and there are none in $\mathbb{Q}(\xi)$, a quadratic extension of $\mathbb{Q}$, so no $\hat{\sigma}$ here.
Irreducibility over the rationals is proved, as I said above, by proving it has no rational roots. To prove irreducibility over $\mathbb {Q} (\xi)$, argue indirectly. $T^2 + T + 1$ is clearly irreducible (no roots) in $\mathbb {Q} (\sqrt[3]{2})[x]$, so the extension $\mathbb {Q} (\sqrt[3]{2},\xi)$ has degree 6 over the rationals. Now see this extension as an extension of $\mathbb {Q} (\xi)$ by adding $\sqrt[3]{2}$. –  Andreas Caranti Jan 31 '13 at 8:42
Or, more simply, since there are no roots of the polynomial $T^3 - 2$ of degree 3 in $\mathbb{Q}(\xi)$, the polynomial is irreducible in $\mathbb{Q}(\xi)[x]$. –  Andreas Caranti Jan 31 '13 at 8:58