Field Extensions III

English Solution

Claim: \[ L_1L_2 =F(\alpha_1,\ldots,\alpha_m,\beta_1,\ldots,\beta_n). \]

Let \[ M=F(\alpha_1,\ldots,\alpha_m,\beta_1,\ldots,\beta_n). \]

Since \(M\) contains \(L_1\) and \(L_2\), \[ L_1L_2\subseteq M. \]

Conversely, \(L_1L_2\) contains \(F\), all \(\alpha_i\), and all \(\beta_j\). Hence \[ M\subseteq L_1L_2. \]

Therefore, \[ \boxed{L_1L_2 =F(\alpha_1,\ldots,\alpha_m,\beta_1,\ldots,\beta_n).} \] \(\blacksquare\)

English Solution

Claim 1: If \(L_1/F\) and \(L_2/F\) are finite, then \[ [L_1L_2:F]\le [L_1:F][L_2:F]. \]

By the tower law, \[ [L_1L_2:F] =[L_1L_2:L_1][L_1:F]. \] Also, \[ [L_1L_2:L_1]\le[L_2:F]. \] Hence \[ \boxed{[L_1L_2:F]\le[L_1:F][L_2:F].} \]

Example: Take \[ L_1=L_2=\mathbb Q(\sqrt2). \] Then \[ [L_1L_2:\mathbb Q]=2<4 =[L_1:\mathbb Q][L_2:\mathbb Q]. \]

Claim 2: If \[ \gcd([L_1:F],[L_2:F])=1, \] then \[ \boxed{[L_1L_2:F]=[L_1:F][L_2:F].} \]

Indeed, both \([L_1:F]\) and \([L_2:F]\) divide \([L_1L_2:F]\). Since they are coprime, their product divides \([L_1L_2:F]\). Together with the above inequality, equality follows.

English Solution

Claim: \[ \operatorname{Gal}(L/K) \] is a group under composition.

Proof:

Closure: If \(\sigma,\tau\in\operatorname{Gal}(L/K)\), then \[ \sigma\circ\tau \] is again a \(K\)-automorphism.

Associativity: Composition of functions is associative.

Identity: The identity map \[ \operatorname{id}_L \] belongs to \(\operatorname{Gal}(L/K)\).

Inverse: If \(\sigma\in\operatorname{Gal}(L/K)\), then \(\sigma^{-1}\) is also a field automorphism fixing \(K\) pointwise.

Hence all group axioms hold. Therefore, \[ \boxed{\operatorname{Gal}(L/K)\text{ is a group}.} \] \(\blacksquare\)

English Solution

(a) The extension \(\mathbb Q(\sqrt2)/\mathbb Q\) is quadratic, so \[ \operatorname{Gal}(\mathbb Q(\sqrt2)/\mathbb Q)\cong C_2, \] with the nontrivial automorphism sending \(\sqrt2\mapsto -\sqrt2\).

(b) In \(\mathbb Q(\sqrt2,\sqrt3)\), each square root can independently change sign, so \[ \operatorname{Gal}(\mathbb Q(\sqrt2,\sqrt3)/\mathbb Q)\cong C_2\times C_2. \]

(c) The field \(\mathbb Q(\sqrt2,\sqrt[3]3)\) contains the real cube root of \(3\), but the other complex cube roots do not lie in the field. Hence only the sign of \(\sqrt2\) can vary, and \[ \operatorname{Gal}(\mathbb Q(\sqrt2,\sqrt[3]3)/\mathbb Q)\cong C_2. \]

(d) In \(\mathbb Q(\sqrt2,\sqrt3,\sqrt5)\), each square root can independently change sign, so \[ \operatorname{Gal}(\mathbb Q(\sqrt2,\sqrt3,\sqrt5)/\mathbb Q)\cong C_2^3. \]

(e) The field \(\mathbb Q(\sqrt[3]3)\) is generated by a real root of an irreducible cubic, and the other two roots are nonreal. Thus the only automorphism is the identity: \[ \operatorname{Gal}(\mathbb Q(\sqrt[3]3)/\mathbb Q)=\{\mathrm{id}\}. \]

(f) For \(\mathbb Q(i)/\mathbb Q\), the nontrivial automorphism sends \(i\mapsto -i\), so \[ \operatorname{Gal}(\mathbb Q(i)/\mathbb Q)\cong C_2. \]

(g) The extension \(\mathbb C/\mathbb R\) has nontrivial automorphism given by complex conjugation, hence \[ \operatorname{Gal}(\mathbb C/\mathbb R)\cong C_2. \]

(h) For \(k(t)/k(t^2)\): if \(\operatorname{char}k\neq2\), then the automorphism group is \(\{\mathrm{id}, t\mapsto -t\}\), so it is isomorphic to \(C_2\); if \(\operatorname{char}k=2\), the two values coincide and the group is trivial.

English Solution

(1) \(k[t]/(t^2+1)\): This quotient is a field exactly when \(t^2+1\) is irreducible over \(k\), i.e. when \(-1\) is not a square in \(k\). If \(\alpha\) denotes the class of \(t\), then \(\alpha^2=-1\). The nontrivial automorphism sends \(\alpha\mapsto-\alpha\), so when \(\operatorname{char}k\neq2\), \[ \boxed{\operatorname{Gal}(K/k)\cong C_2.} \] If \(\operatorname{char}k=2\), then \(t^2+1=(t+1)^2\), so the extension is not separable and the automorphism group is trivial.

(2) \(k[t]/(t^2+t+1)\): This quotient is a field exactly when \(t^2+t+1\) is irreducible over \(k\). If \(\beta\) is one root, then the other root is \(-1-\beta\), and the nontrivial automorphism exchanges the roots. Therefore \[ \boxed{\operatorname{Gal}(K/k)\cong C_2.} \]

General rule: For a quadratic extension, the automorphism group is either trivial or cyclic of order \(2\); in the nontrivial case the nonidentity element exchanges the two roots.

English Solution

Claim 1: If \(H\le\operatorname{Gal}(K/F)\), then \[ K^H \] is an intermediate field.

If \(x,y\in K^H\), then for every \(\sigma\in H\), \[ \sigma(x+y)=x+y,\qquad \sigma(xy)=xy. \] Also, \[ \sigma(-x)=-x,\qquad \sigma(x^{-1})=x^{-1}\quad(x\neq0). \] Hence \(K^H\) is a subfield and \[ F\subseteq K^H\subseteq K. \]

Claim 2: \[ K^S=K^{\langle S\rangle}. \]

An element fixed by every element of \(S\) is also fixed by every element generated by \(S\). Therefore, \[ \boxed{K^S=K^{\langle S\rangle}.} \] \(\blacksquare\)

English Solution

Let \[ K=\mathbb Q(\sqrt2,\sqrt[3]3). \] By Exercise 1.34, \[ [K:\mathbb Q]=6. \]

The cube root \(\sqrt[3]3\) has two non-real conjugates, so they are not in \(K\subset\mathbb R\). Hence every automorphism fixes \(\sqrt[3]3\).

The element \(\sqrt2\) may be sent to \(\pm\sqrt2\). Therefore, \[ \boxed{\operatorname{Gal}(K/\mathbb Q)\cong C_2.} \]

The two subgroups are \[ \{\mathrm{id}\},\qquad C_2. \] Their fixed fields are \[ K,\qquad \mathbb Q(\sqrt[3]3). \]

However, the intermediate fields also include \[ \mathbb Q,\qquad \mathbb Q(\sqrt2). \]

Thus there are more intermediate fields than subgroups. This happens because \(K/\mathbb Q\) is not Galois.

English Solution

Example: Let \[ K=k(t),\qquad F=k(t^p), \] where \(\operatorname{char}k=p>0\).

We have \[ X^p-t^p=(X-t)^p. \] Thus \(t\) is the only root of its minimal polynomial.

Therefore any \(F\)-automorphism must satisfy \[ \sigma(t)=t. \] Hence \[ \boxed{\operatorname{Gal}(K/F)=\{\mathrm{id}\}.} \]

But \[ [K:F]=p>1. \] Thus the extension is not Galois. This shows that the Galois correspondence can fail in positive characteristic for purely inseparable extensions.

English Solution

Let \[ K=\mathbb Q(\sqrt2,\sqrt3). \] Then \[ [K:\mathbb Q]=4. \]

The automorphisms are obtained by independently changing the signs of \(\sqrt2\) and \(\sqrt3\). Hence \[ \operatorname{Gal}(K/\mathbb Q) \cong C_2\times C_2. \]

This group has five subgroups: \[ \{1\},\quad \langle\sigma\rangle,\quad \langle\tau\rangle,\quad \langle\sigma\tau\rangle,\quad C_2\times C_2. \]

The corresponding fixed fields are \[ K,\quad \mathbb Q(\sqrt3),\quad \mathbb Q(\sqrt2),\quad \mathbb Q(\sqrt6),\quad \mathbb Q. \]

Hence the Galois correspondence is exact in this case.

\[ \boxed{K/\mathbb Q\text{ is Galois and the correspondence is one-to-one.}} \] \(\blacksquare\)