Let \(n\ge3\). For the splitting field of \(x^n-2\) over \(\QQ\), prove the affine embedding \[ \Gal\hookrightarrow (\ZZ/n\ZZ)\rtimes(\ZZ/n\ZZ)^\times \] and prove that the Galois group is dihedral of order \(2n\) exactly for \(n=3,4,6\).
Compute all intermediate fields of \(\QQ(\sqrt2,\sqrt3)\), prove \[ \QQ(\sqrt2,\sqrt3)=\QQ(\sqrt2+\sqrt3), \] and find the minimal polynomial of \(\sqrt2+\sqrt3\).
Compute all intermediate fields of \(\QQ(2^{1/4},\zeta_8)\).
Show that \(\QQ(2^{1/3})\) and \(\QQ(\zeta_3 2^{1/3})\) are isomorphic fields but are not identical as subfields of \(\CC\).
Let \(L/K\) be a finite separable extension. Show that \(L/K\) is normal if and only if \(L\) is the splitting field over \(K\) of some polynomial in \(K[x]\).
Is \(\QQ(2^{1/3})/\QQ\) normal?
Show that \(\GF(p^n)\) is the splitting field of \(x^{p^n}-x\) over \(\FF_p\).
Show that \[ \GF(p^d)\subseteq\GF(p^n) \iff d\mid n. \]
Compute the Galois group of \(x^n-1\) over \(\QQ\).
For an odd prime \(p\), identify all elements of the Galois group of \(x^p-2\) over \(\QQ\).
Compute the Galois group of \(x^{15}+2\) over \(\QQ\) and identify its Sylow \(2\)-complement structure.
Show that the Galois group of \(x^3+4x+2\) over \(\QQ\) is \(S_3\).
Solution
1. The polynomial \(x^n-2\).Let \(\alpha=2^{1/n}\) and let \(\zeta_n\) be primitive. The roots are \(\zeta_n^j\alpha\), so the splitting field is\[ L=\QQ(\alpha,\zeta_n). \]Every automorphism has the form\[ \sigma(\alpha)=\zeta_n^a\alpha, \qquad \sigma(\zeta_n)=\zeta_n^b, \]with \(a\in\ZZ/n\ZZ\) and \(b\in(\ZZ/n\ZZ)^\times\), giving an injection into the affine group with law\[ (a,b)(c,d)=(a+bc,bd). \]
For \(n=3,4,6\), the cyclotomic field has degree \(2\) and intersects the real pure field \(\QQ(\alpha)\) trivially, so \([L:\QQ]=2n\). The translation \(\alpha\mapsto\zeta_n\alpha\) has order \(n\), complex conjugation has order \(2\), and conjugation inverts the translation. Hence the group is \(D_{2n}\).
Conversely, if the group were \(D_{2n}\), then its cyclotomic quotient \((\ZZ/n\ZZ)^\times\) would be an abelian quotient of a dihedral group, hence would have order at most \(4\) and exponent at most \(2\). This reduces to \(n=3,4,5,6,8,10,12\). The cases \(5,10\) are excluded because the unit group is \(C_4\). The case \(8\) has order \(16\) but generators satisfying \(srs=r^3\) rather than \(r^{-1}\), and the case \(12\) would force \(\sqrt2\in\QQ(\zeta_{12})\), impossible. Thus exactly \(n=3,4,6\) are dihedral.
2. The biquadratic field.Let\[ L=\QQ(\sqrt2,\sqrt3). \]Then\[ \Gal(L/\QQ)\cong C_2\times C_2, \]so the proper nontrivial intermediate fields are\[ \QQ(\sqrt2),\qquad \QQ(\sqrt3),\qquad \QQ(\sqrt6). \]If \(\alpha=\sqrt2+\sqrt3\), then\[ \alpha^{-1}=\sqrt3-\sqrt2, \]so both \(\sqrt2\) and \(\sqrt3\) lie in \(\QQ(\alpha)\). Hence \(L=\QQ(\alpha)\). Moreover\[ (\alpha^2-5)^2=24, \]so the minimal polynomial is\[ \boxed{x^4-10x^2+1}. \]
3. The field \(\QQ(2^{1/4},\zeta_8)\).Put \(\alpha=2^{1/4}\). Since \(\zeta_8=(1+i)/\sqrt2\),\[ L=\QQ(\alpha,i) \]is the splitting field of \(x^4-2\), with Galois group \(D_8\).The three quadratic intermediate fields are\[ \QQ(i),\qquad \QQ(\sqrt2),\qquad \QQ(\sqrt{-2}). \]The five quartic intermediate fields are\[ \QQ(\zeta_8),\quad \QQ(\alpha),\quad \QQ(i\alpha),\quad \QQ((1+i)\alpha),\quad \QQ((1-i)\alpha). \]Together with \(\QQ\) and \(L\), these exhaust the subgroup lattice of \(D_8\).
4. Two isomorphic but unequal cubic fields.Let\[ \alpha=2^{1/3},\qquad \beta=\zeta_3\alpha. \]Both have irreducible minimal polynomial \(x^3-2\), so\[ \QQ(\alpha)\cong\QQ[x]/(x^3-2)\cong\QQ(\beta). \]But \(\QQ(\alpha)\subset\RR\) while \(\beta\) is nonreal, so the two subfields of \(\CC\) are not equal.
5. Finite separable normal extensions are exactly finite splitting fields.If \(L/K\) is finite, separable, and normal, choose generators\[ L=K(\alpha_1,\ldots,\alpha_r). \]Let \(m_i\) be the minimal polynomial of \(\alpha_i\) over \(K\). Normality implies each \(m_i\) splits in \(L\), so \(L\) is the splitting field of\[ f=\prod_i m_i. \]Conversely, a splitting field is normal, and separability of \(L/K\) ensures the relevant irreducible factors are separable. Hence the equivalence.
6. The real cubic field is not normal.The minimal polynomial of \(2^{1/3}\) is \(x^3-2\), whose other roots are\[ \zeta_3 2^{1/3},\qquad \zeta_3^2 2^{1/3}. \]They are nonreal, while \(\QQ(2^{1/3})\subset\RR\). Thus the polynomial does not split there, so the extension is not normal.
7. Finite fields as splitting fields.The roots of\[ x^{p^n}-x \]in an algebraic closure are exactly the elements fixed by the \(n\)th iterate of Frobenius. They form the unique field with \(p^n\) elements, namely \(\GF(p^n)\). Since the derivative is \(-1\), the polynomial is squarefree, so \(\GF(p^n)\) is its splitting field.
8. Subfields of finite fields.The extension\[ \GF(p^n)/\FF_p \]is cyclic Galois of degree \(n\), generated by Frobenius. Its subgroups are in bijection with divisors of \(n\). Therefore\[ \GF(p^d)\subseteq\GF(p^n) \iff d\mid n. \]
9. Cyclotomic Galois groups.The splitting field of \(x^n-1\) is \(\QQ(\zeta_n)\), and\[ \Gal(\QQ(\zeta_n)/\QQ) \cong (\ZZ/n\ZZ)^\times \]via\[ a\longmapsto(\zeta_n\mapsto\zeta_n^a). \]
10. The Galois group of \(x^p-2\) for odd prime \(p\).Let \(\alpha=2^{1/p}\). Since\[ [\QQ(\alpha):\QQ]=p, \qquad [\QQ(\zeta_p):\QQ]=p-1, \]their intersection is \(\QQ\). Hence the splitting field has degree\[ p(p-1). \]Every affine pair occurs, so\[ \Gal(x^p-2/\QQ) \cong C_p\rtimes(\ZZ/p\ZZ)^\times \cong C_p\rtimes C_{p-1}. \]Its elements are precisely\[ \sigma_{a,b}(\alpha)=\zeta_p^a\alpha, \qquad \sigma_{a,b}(\zeta_p)=\zeta_p^b, \]for \(a\in\ZZ/p\ZZ\) and \(b\in(\ZZ/p\ZZ)^\times\).
11. The Galois group of \(x^{15}+2\).If \(\alpha=2^{1/15}\), then \(-\alpha\) is a root of \(x^{15}+2\), so the splitting field is\[ L=\QQ(\alpha,\zeta_{15}). \]The pure field has degree \(15\) and the cyclotomic field has degree \(\varphi(15)=8\); these degrees are coprime, so the intersection is \(\QQ\). Hence\[ [L:\QQ]=120 \]and every affine pair occurs:\[ \Gal(L/\QQ) \cong C_{15}\rtimes(\ZZ/15\ZZ)^\times. \]Since\[ (\ZZ/15\ZZ)^\times\cong C_4\times C_2, \]this unit group is a Sylow \(2\)-subgroup \(S_2\), and\[ \boxed{\Gal(L/\QQ)\cong C_{15}\rtimes S_2}. \]
12. The cubic \(x^3+4x+2\).The polynomial has no rational root, so it is irreducible over \(\QQ\). Its discriminant is\[ \Delta=-4(4)^3-27(2)^2=-364, \]which is not a square in \(\QQ\). An irreducible cubic has transitive Galois group, hence either \(A_3\) or \(S_3\), and it lies in \(A_3\) exactly when the discriminant is a square. Therefore\[ \Gal(x^3+4x+2/\QQ)\cong S_3. \]