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Course Notes

Section The splitting field of \(x^3-2\) over \(\Q\)

We consider the following example as an illustration to the ‘Fundamental Theorem of Galois Theory’.
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Proof.

  1. The field \(K\) is the splitting field of an irreducible separable polynomial \(x^3-2\) over \(\Q\) (see Example 30). Therefore, \(K/\Q\) Galois.
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  2. This follows from Example 30 and the observation that \(\alpha^2\beta=\omega\text{.}\)
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Proof.

We consider \(\{1,\alpha,\alpha^2,\omega,\omega\alpha,\omega\alpha^2\}\) as a basis of \(K/\Q\text{.}\) Define the following automorphisms.
\begin{equation*} \sigma\colon\ \begin{array}{1}\alpha\mapsto\omega\alpha\\\omega\mapsto\omega\end{array} \quad\text{and}\quad \tau\colon\ \begin{array}{1}\alpha\mapsto\alpha\\\omega\mapsto\omega^2\end{array} \end{equation*}
One checks that \(\sigma^3=\unit_K=\tau^2\) and that \(\tau\circ\sigma\circ\tau=\sigma^{-1}=\sigma^2\text{.}\) Suppose that \(\phi\colon K\to K\) is an automorphism fixing \(\Q\text{.}\) Then, \(\phi\) is determined by its action on \(\alpha\) and \(\omega\text{.}\) We have \(\phi(\alpha)\) to be one of the roots of \(x^3-2\) in \(K\text{,}\) i.e., \(\phi(\alpha)\in\{\alpha,\omega\alpha,\omega^2\alpha\}\text{.}\) Similar reasoning implies that \(\phi(\omega)\in\{\omega,\omega^2\}\text{.}\) There are six possible combinations and one can check that \(\phi\in\{e,\sigma,\sigma^2,\tau,\tau\sigma,\tau\sigma^2\}\text{.}\)
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Of course one may directly use one of the equivalent conditions of Galois extensions, viz., \([K:\Q]=6=\lvert\Gal{K}{\Q}\rvert\text{,}\) to deduce that \(\Gal{K}{\Q}\simeq S_3\simeq\{e,\sigma,\sigma^2,\tau,\tau\sigma,\tau\sigma^2\}\text{.}\)
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Proof.

By using the fundamental theorem of Galois theory, the subfields of \(K/\Q\) corresponds precisely to the fixed fields of subgroups of \(\Gal{K}{\Q}\text{.}\) Note that following are all the subgroups of \(\Gal{K}{\Q}\text{.}\)
\begin{equation*} H_0=\{e\}, H_1=\{e,\sigma,\sigma^2\}, H_2=\{e,\tau\}, H_3=\{e,\tau\sigma\}, H_4=\{e,\tau\sigma^2\}, \text{ and }, H_5=S_3 \end{equation*}
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We calculate the fixed of of \(H_1\text{.}\) It is enough to find \(x\in K\) such that \(\sigma x=x\text{,}\) i.e., \((\sigma-\unit_K)(x)=x\text{.}\) We first find the matrix of \(\sigma\) with respect to a basis \(\mathfrak{B}=\{1,\alpha,\alpha^2,\omega,\omega\alpha,\omega\alpha^2\}\text{.}\) We get the following using \(\omega^2+\omega+1=0\in K\text{.}\)
\begin{align*} [\sigma]_{\mathfrak{B}}-I_6\amp =\begin{pmatrix}1\amp 0\amp 0\amp 0\amp 0\amp 0\\0\amp 0\amp 0\amp 0\amp -1\amp 0\\0\amp 0\amp -1\amp 0\amp 0\amp 1\\0\amp 0\amp 0\amp 1\amp 0\amp 0\\0\amp 1\amp 0\amp 0\amp -1\amp 0\\0\amp 0\amp -1\amp 0\amp 0\amp 0\end{pmatrix}-\begin{pmatrix}1\amp 0\amp 0\amp 0\amp 0\amp 0\\0\amp 1\amp 0\amp 0\amp 0\amp 0\\0\amp 0\amp 1\amp 0\amp 0\amp 0\\0\amp 0\amp 0\amp 1\amp 0\amp 0\\0\amp 0\amp 0\amp 0\amp 1\amp 0\\0\amp 0\amp 0\amp 0\amp 0\amp 1\end{pmatrix}\\ \amp =\begin{pmatrix}0\amp 0\amp 0\amp 0\amp 0\amp 0\\0\amp -1\amp 0\amp 0\amp -1\amp 0\\0\amp 0\amp -2\amp 0\amp 0\amp 1\\0\amp 0\amp 0\amp 0\amp 0\amp 0\\0\amp 1\amp 0\amp 0\amp -2\amp 0\\0\amp 0\amp -1\amp 0\amp 0\amp -1\end{pmatrix} \end{align*}
The rank of the matrix \([\sigma]_{\mathfrak{B}}-I_6\) is \(4\text{,}\) and hence the nullity is \(2\text{.}\) Note that \(\{1,\omega\}\) is a basis of its kernel. In particular, \(\left[K^{H_1}:\Q\right]=2=\lvert S_3:H_1\rvert\) and hence, \(K^{H_1}=\Q(\omega)\text{.}\) Furthermore, \(H_1\) is the normal subgroup of \(S_3\text{.}\) The fundamental theorem of Galois theory asserts that \(K^{H_1}=\Q(\omega)\) is Galois. Indeed, \(\Q(\omega)\) is the splitting field of an irreducible separable polynomial \(x^2+x+1\) over \(\Q\text{.}\)
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The matrix \([\tau]_{\mathfrak{B}}-I_6\) with respect to \(\mathfrak{B}\) is
\begin{equation*} \begin{pmatrix}I_3\amp -I_3\\0_3\amp -2\cdot I_3\end{pmatrix}, \end{equation*}
where, \(0_3\) and \(I_3\) are \(3\times 3\) zero matrix and the identity matrix, respectively. The kernel of the associated linear transformation has dimension \(3\) and \(\{1,\alpha,\alpha^2\}\) may be considered as a basis. Thus, we get that \(K^{H_2}=\Q(\alpha)\text{.}\) Note that \(\left[K^{H_2}:\Q\right]=\lvert S_3:H_2\rvert\text{.}\)
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Now we can find \(K^{H_5}\text{.}\) For any \(x\in K^{H_5}\) we must have \(x\in K^{H_1}\cap K^{H_2}\text{.}\) Therefore, there exists some \(a_1,a_2\in\Q\) and \(b_1,b_2,b_3\in\Q\) such that
\begin{equation*} x=a_1\cdot 1+a_2\cdot\omega=b_1\cdot 1+b_2\cdot\alpha+b_3\cdot\alpha^2 \end{equation*}
Since \(1,\alpha,\alpha^2,\omega\) are linearly independent over \(\Q\) we must have \(x\in\Q\text{.}\) Thus, \(K^{H_1}\cap K^{H_2}=\Q\text{.}\) In particular, \(K^{H_5}=\Q\text{.}\)
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The matrix of \([\tau\sigma]_{\mathfrak{B}}-I_6\) with respect to \(\mathfrak{B}\) is
\begin{equation*} \begin{pmatrix}0\amp 0\amp 0\amp -1\amp 0\amp 0\\0\amp -2\amp 0\amp 0\amp 0\amp 0\\0\amp 0\amp -1\amp 0\amp 0\amp 1\\0\amp 0\amp 0\amp -2\amp 0\amp 0\\0\amp -1\amp 0\amp 0\amp 0\amp 0\\0\amp 0\amp 1\amp 0\amp 0\amp -1\end{pmatrix}. \end{equation*}
The rank of the above matrix is \(3\text{.}\) Hence, the kernel of the linear transformation associated with the above matrix has dimension \(3\text{.}\) One may take \(\{1,\omega\alpha,\alpha^2+\omega\alpha^2\}\) as a basis of the kernel. We thus have \(\left[K^{H_3}:\Q\right]=3\text{.}\) As \(\omega\alpha\in K^{H_3}\text{,}\) so is \((\omega\alpha)^2=-(\alpha^2+\omega\alpha^2)\text{.}\) Furthermore, \(\Q\left(\omega\alpha\right)\simeq\Q[x]/(x^3-2)\text{,}\) and hence, \(\left[\Q\left(\omega\alpha\right):\Q\right]=3\text{.}\) Therefore, \(K^{H_3}=\Q\left(\omega\alpha\right)\text{.}\)
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The matrix The matrix \([\tau\sigma^2]_{\mathfrak{B}}-I_6\) with respect to \(\mathfrak{B}\) is
\begin{equation*} \begin{pmatrix}0\amp 0\amp 0\amp -1\amp 0\amp 0\\0\amp -1\amp 0\amp 0\amp 1\amp 0\\0\amp 0\amp -2\amp 0\amp 0\amp 0\\0\amp 0\amp 0\amp -2\amp 0\amp 0\\0\amp 1\amp 0\amp 0\amp -1\amp 0\\0\amp 0\amp -1\amp 0\amp 0\amp 0\end{pmatrix}. \end{equation*}
The rank of the above matrix is \(3\text{.}\) Hence, the kernel of the linear transformation associated with the above matrix has dimension \(3\text{.}\) One may take \(\{1,\alpha+\omega\alpha,\omega\alpha^2\}\) as a basis of the kernel. We thus have \(\left[K^{H_4}:\Q\right]=3\text{.}\) Note that \(-(\alpha+\omega\alpha)=\omega^2\alpha\text{.}\) Thus, \(\omega^2\alpha\in K^{H_4}\text{.}\) Furthermore, \(\Q\left(\omega^2\alpha\right)\simeq\Q[x]/(x^3-2)\text{,}\) and hence, \(\left[\Q\left(\omega^2\alpha\right):\Q\right]=3\text{.}\) Therefore, \(K^{H_4}=\Q\left(\omega^2\alpha\right)\text{.}\)
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