Appendix B Hints and Answers to Selected Exercises
1 Preliminaries
1.4 Exercises
1.4.2.
1.4.6.
1.4.10.
1.4.14.
1.4.17.
1.4.18.
1.4.20.
1.4.22.
Hint.
(a) Let \(x, y \in A\text{.}\) Then \(g(f(x)) = (g \circ f)(x) = (g \circ f)(y) = g(f(y))\text{.}\) Thus, \(f(x) = f(y)\) and \(x = y\text{,}\) so \(g \circ f\) is one-to-one. (b) Let \(c \in C\text{,}\) then \(c = (g \circ f)(x) = g(f(x))\) for some \(x \in A\text{.}\) Since \(f(x) \in B\text{,}\) \(g\) is onto.
1.4.23.
1.4.24.
Hint.
(a) Let \(y \in f(A_1 \cup A_2)\text{.}\) Then there exists an \(x \in A_1 \cup A_2\) such that \(f(x) = y\text{.}\) Hence, \(y \in f(A_1)\) or \(f(A_2) \text{.}\) Therefore, \(y \in f(A_1) \cup f(A_2)\text{.}\) Consequently, \(f(A_1 \cup A_2) \subset f(A_1) \cup f(A_2)\text{.}\) Conversely, if \(y \in f(A_1) \cup f(A_2)\text{,}\) then \(y \in f(A_1)\) or \(f(A_2)\text{.}\) Hence, there exists an \(x\) in \(A_1\) or \(A_2\) such that \(f(x) = y\text{.}\) Thus, there exists an \(x \in A_1 \cup A_2\) such that \(f(x) = y\text{.}\) Therefore, \(f(A_1) \cup f(A_2) \subset f(A_1 \cup A_2)\text{,}\) and \(f(A_1 \cup A_2) = f(A_1) \cup f(A_2)\text{.}\)
1.4.25.
1.4.28.
2 The Integers
2.4 Exercises
2.4.1.
Hint.
The base case, \(S(1): [1(1 + 1)(2(1) + 1)]/6 = 1 = 1^2\) is true. Assume that \(S(k): 1^2 + 2^2 + \cdots + k^2 = [k(k + 1)(2k + 1)]/6\) is true. Then
\begin{align*}
1^2 + 2^2 + \cdots + k^2 + (k + 1)^2 & = [k(k + 1)(2k + 1)]/6 + (k + 1)^2\\
& = [(k + 1)((k + 1) + 1)(2(k + 1) + 1)]/6\text{,}
\end{align*}
and so \(S(k + 1)\) is true. Thus, \(S(n)\) is true for all positive integers \(n\text{.}\)
2.4.3.
2.4.8.
Hint.
Follow the proof in Example 2.4.
2.4.11.
Hint.
The base case, \(S(0): (1 + x)^0 - 1 = 0 \geq 0 = 0 \cdot x\) is true. Assume \(S(k): (1 + x)^k -1 \geq kx\) is true. Then
\begin{align*}
(1 + x)^{k + 1} - 1 & = (1 + x)(1 + x)^k -1\\
& = (1 + x)^k + x(1 + x)^k - 1\\
& \geq kx + x(1 + x)^k\\
& \geq kx + x\\
& = (k + 1)x\text{,}
\end{align*}
so \(S(k + 1)\) is true. Therefore, \(S(n)\) is true for all positive integers \(n\text{.}\)
2.4.17. Fibonacci Numbers.
Hint.
For (a) and (b) use mathematical induction. (c) Show that \(f_1 = 1\text{,}\) \(f_2 = 1\text{,}\) and \(f_{n + 2} = f_{n + 1} + f_n\text{.}\) (e) Use part (b) and Exercise 2.4.16.
2.4.19.
2.4.23.
2.4.27.
2.4.29.
3 Groups
3.5 Exercises
3.5.1.
3.5.2.
3.5.6.
3.5.8.
3.5.15.
3.5.16.
3.5.17.
3.5.18.
Hint.
Let
\begin{equation*}
\sigma =
\begin{pmatrix}
1 & 2 & \cdots & n \\ a_1 & a_2 & \cdots & a_n
\end{pmatrix}
\end{equation*}
be in \(S_n\text{.}\) All of the \(a_i\)s must be distinct. There are \(n\) ways to choose \(a_1\text{,}\) \(n - 1\) ways to choose \(a_2, \ldots\text{,}\) 2 ways to choose \(a_{n - 1}\text{,}\) and only one way to choose \(a_n\text{.}\) Therefore, we can form \(\sigma\) in \(n(n - 1) \cdots 2 \cdot 1 = n!\) ways.
3.5.25.
3.5.31.
3.5.35.
3.5.41.
3.5.46.
3.5.49.
4 Cyclic Groups
4.5 Exercises
4.5.1.
4.5.2.
4.5.3.
Hint.
(a) \(7 {\mathbb Z} = \{ \ldots, -7, 0, 7, 14, \ldots \}\text{;}\) (b) \(\{ 0, 3, 6, 9, 12, 15, 18, 21 \}\text{;}\) (c) \(\{ 0 \}\text{,}\) \(\{ 0, 6 \}\text{,}\) \(\{ 0, 4, 8 \}\text{,}\) \(\{ 0, 3, 6, 9 \}\text{,}\) \(\{ 0, 2, 4, 6, 8, 10 \}\text{;}\) (g) \(\{ 1, 3, 7, 9 \}\text{;}\) (j) \(\{ 1, -1, i, -i \}\text{.}\)
4.5.4.
Hint.
(a)
\begin{equation*}
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},
\begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix},
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}\text{.}
\end{equation*}
(c)
\begin{equation*}
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},
\begin{pmatrix}
1 & -1 \\
1 & 0
\end{pmatrix},
\begin{pmatrix}
-1 & 1 \\
-1 & 0
\end{pmatrix}, \\
\begin{pmatrix}
0 & 1 \\
-1 & 1
\end{pmatrix},
\begin{pmatrix}
0 & -1 \\
1 & -1
\end{pmatrix},
\begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix}\text{.}
\end{equation*}
4.5.10.
4.5.11.
4.5.15.
4.5.16.
4.5.17.
4.5.18.
4.5.22.
4.5.27.
4.5.31.
4.5.37.
5 Permutation Groups
5.4 Exercises
5.4.1.
5.4.2.
5.4.3.
5.4.4.
5.4.5.
5.4.8.
5.4.11.
5.4.17.
5.4.25.
5.4.29.
5.4.30.
6 Cosets and Lagrange’s Theorem
6.5 Exercises
6.5.1.
6.5.2.
6.5.3.
6.5.4.
6.5.5.
Hint.
(a) \(\langle 8 \rangle\text{,}\) \(1 + \langle 8 \rangle\text{,}\) \(2 + \langle 8 \rangle\text{,}\) \(3 + \langle 8 \rangle\text{,}\) \(4 + \langle 8 \rangle\text{,}\) \(5 + \langle 8 \rangle\text{,}\) \(6 + \langle 8 \rangle\text{,}\) and \(7 + \langle 8 \rangle\text{;}\) (c) \(3 {\mathbb Z}\text{,}\) \(1 + 3 {\mathbb Z}\text{,}\) and \(2 + 3 {\mathbb Z}\text{.}\)
6.5.7.
6.5.12.
6.5.19.
6.5.22.
Hint.
7 Introduction to Cryptography
7.4 Exercises
7.4.1.
7.4.3.
7.4.4.
7.4.7.
7.4.9.
7.4.10.
8 Algebraic Coding Theory
8.6 Exercises
8.6.2.
8.6.3.
8.6.4.
8.6.6.
8.6.7.
Hint.
-
\((00000), (00101), (10011), (10110)\)\begin{equation*} G = \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{pmatrix} \end{equation*}
-
\((000000), (010111), (101101), (111010)\)\begin{equation*} G = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \\ 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{pmatrix} \end{equation*}
8.6.9.
8.6.11.
Hint.
(a) A canonical parity-check matrix with standard generator matrix
\begin{equation*}
G =
\begin{pmatrix}
1 \\ 1 \\ 0 \\ 0 \\ 1
\end{pmatrix}\text{.}
\end{equation*}
(c) A canonical parity-check matrix with standard generator matrix
\begin{equation*}
G =
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
1 & 1 \\
1 & 0
\end{pmatrix}\text{.}
\end{equation*}
8.6.12.
8.6.15.
8.6.19.
8.6.23.
9 Isomorphisms
9.4 Exercises
9.4.1.
Hint.
9.4.2.
9.4.3.
9.4.6.
9.4.8.
9.4.11.
9.4.16.
9.4.19.
9.4.20.
9.4.25.
9.4.27.
9.4.38.
9.4.45.
10 Normal Subgroups and Factor Groups
10.4 Exercises
10.4.1.
10.4.8.
10.4.11.
10.4.12.
10.4.14.
Hint.
(a) Let \(g \in G\) and \(h \in G'\text{.}\) If \(h = aba^{-1}b^{-1}\text{,}\) then
\begin{align*}
ghg^{-1} & = gaba^{-1}b^{-1}g^{-1}\\
& = (gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1})\\
& = (gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1}\text{.}
\end{align*}
We also need to show that if \(h = h_1 \cdots h_n\) with \(h_i = a_i b_i a_i^{-1} b_i^{-1}\text{,}\) then \(ghg^{-1}\) is a product of elements of the same type. However, \(ghg^{-1} = g h_1 \cdots h_n g^{-1} = (gh_1g^{-1})(gh_2g^{-1}) \cdots (gh_ng^{-1})\text{.}\)
11 Homomorphisms
11.4 Exercises
11.4.2.
11.4.4.
11.4.5.
11.4.9.
11.4.17.
12 Matrix Groups and Symmetry
12.4 Exercises
12.4.1.
Hint.
\begin{align*}
\frac{1}{2} \left[ \|{\mathbf x} + {\mathbf y}\|^2 + \|{\mathbf x}\|^2 - \| {\mathbf y}\|^2 \right] & = \frac{1}{2} \left[ \langle x + y, x + y \rangle - \|{\mathbf x}\|^2 - \| {\mathbf y}\|^2 \right]\\
& = \frac{1}{2} \left[ \| {\mathbf x}\|^2 + 2 \langle x, y \rangle + \| {\mathbf y}\|^2 - \|{\mathbf x}\|^2 - \| {\mathbf y}\|^2 \right]\\
& = \langle {\mathbf x}, {\mathbf y} \rangle\text{.}
\end{align*}
12.4.3.
12.4.5.
12.4.7.
12.4.10.
12.4.13.
12.4.17.
13 The Structure of Groups
13.4 Exercises
13.4.1.
13.4.4.
Hint.
(a) \(\{ 0 \} \subset \langle 6 \rangle \subset \langle 3 \rangle \subset {\mathbb Z}_{12}\text{;}\) (e) \(\{ (1) \} \times \{ 0 \} \subset \{ (1), (1 \, 2 \, 3), (1 \, 3 \, 2) \} \times \{ 0 \} \subset S_3 \times \{ 0 \} \subset S_3 \times \langle 2 \rangle\subset S_3 \times {\mathbb Z}_4\text{.}\)
13.4.7.
13.4.12.
13.4.16.
13.4.21.
14 Group Actions
14.5 Exercises
14.5.1.
Hint.
Example 14.1: \(0\text{,}\) \({\mathbb R}^2 \setminus \{ 0 \}\text{.}\) Example 14.2: \(X = \{ 1, 2, 3, 4 \}\text{.}\)
14.5.2.
Hint.
(a) \(X_{(1)} = \{1, 2, 3 \}\text{,}\) \(X_{(1 \, 2)} = \{3 \}\text{,}\) \(X_{(1 \, 3)} = \{ 2 \}\text{,}\) \(X_{(2 \, 3)} = \{1 \}\text{,}\) \(X_{(1 \, 2 \, 3)} = X_{(1 \, 3 \, 2)} = \emptyset\text{.}\) \(G_1 = \{ (1), (2 \, 3) \}\text{,}\) \(G_2 = \{(1), (1 \, 3) \}\text{,}\) \(G_3 = \{ (1), (1 \, 2)\}\text{.}\)
14.5.3.
14.5.6.
Hint.
The conjugacy classes for \(S_4\) are
\begin{gather*}
{\mathcal O}_{(1)} = \{ (1) \},\\
{\mathcal O}_{(12)} = \{ (1 \, 2), (1 \, 3), (1 \, 4), (2 \, 3), (2 \, 4), (3 \, 4) \},\\
{\mathcal O}_{(1 \, 2)(3 \, 4)} = \{ (1 \, 2)(3 \, 4), (1 \, 3)(2 \, 4), (1 \, 4)(2 \, 3) \},\\
{\mathcal O}_{(123)} = \{ (1 \, 2 \, 3), (1 \, 3 \, 2), (1 \, 2 \, 4), (1 \, 4 \, 2), (1 \, 3 \, 4), (1 \, 4 \, 3), (2 \, 3 \, 4), (2 \, 4 \, 3) \},\\
{\mathcal O}_{(1234)} = \{ (1 \, 2 \, 3 \, 4), (1 \, 2 \, 4 \, 3), (1 \, 3 \, 2 \, 4), (1 \, 3 \, 4 \, 2), (1 \, 4 \, 2 \, 3), (1 \, 4 \, 3 \, 2) \}\text{.}
\end{gather*}
The class equation is \(1 + 3 + 6 + 6 + 8 = 24\text{.}\)
14.5.8.
14.5.11.
Hint.
The group of rigid motions of the cube can be described by the allowable permutations of the six faces and is isomorphic to \(S_4\text{.}\) There are the identity cycle, 6 permutations with the structure \((abcd)\) that correspond to the quarter turns, 3 permutations with the structure \((ab)(cd)\) that correspond to the half turns, 6 permutations with the structure \((ab)(cd)(ef)\) that correspond to rotating the cube about the centers of opposite edges, and 8 permutations with the structure \((abc)(def)\) that correspond to rotating the cube about opposite vertices.
14.5.15.
14.5.17.
14.5.22.
15 The Sylow Theorems
15.4 Exercises
15.4.1.
15.4.2.
15.4.5.
Hint.
Since \(|G| = 96 = 2^5 \cdot 3\text{,}\) \(G\) has either one or three Sylow \(2\)-subgroups by the Third Sylow Theorem. If there is only one subgroup, we are done. If there are three Sylow \(2\)-subgroups, let \(H\) and \(K\) be two of them. Therefore, \(|H \cap K| \geq 16\text{;}\) otherwise, \(HK\) would have \((32 \cdot 32)/8 = 128\) elements, which is impossible. Thus, \(H \cap K\) is normal in both \(H\) and \(K\) since it has index \(2\) in both groups.
15.4.8.
15.4.10.
15.4.17.
Hint.
If \(G\) is abelian, then \(G\) is cyclic, since \(|G| = 3 \cdot 5 \cdot 17\text{.}\) Now look at Example 15.14.
15.4.23.
15.4.26.
16 Rings
16.7 Exercises
16.7.1.
16.7.3.
Hint.
(a) \(\{1, 3, 7, 9 \}\text{;}\) (c) \(\{ 1, 2, 3, 4, 5, 6 \}\text{;}\) (e)
\begin{equation*}
\left\{
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix},
\begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix},
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix},
\begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix},
\begin{pmatrix}
0 & 1 \\
1 & 1
\end{pmatrix},
\right\}\text{.}
\end{equation*}
16.7.4.
16.7.7.
16.7.8.
16.7.13.
16.7.16.
16.7.18.
16.7.26.
16.7.28.
16.7.33.
16.7.37.
17 Polynomials
17.5 Exercises
17.5.2.
17.5.3.
17.5.5.
17.5.7.
17.5.8.
17.5.10.
17.5.13.
17.5.14.
17.5.16.
17.5.20. Cyclotomic Polynomials.
17.5.26.
18 Integral Domains
18.4 Exercises
18.4.1.
18.4.2.
18.4.4.
18.4.9.
18.4.15.
18.4.16.
19 Lattices and Boolean Algebras
19.5 Exercises
19.5.2.
Hint.
19.5.4.
19.5.5.
19.5.6.
19.5.8.
19.5.10.
19.5.14.
Hint.
Let \(I, J\) be ideals in \(R\text{.}\) We need to show that \(I + J = \{ r + s : r \in I \text{ and } s \in J \}\) is the smallest ideal in \(R\) containing both \(I\) and \(J\text{.}\) If \(r_1, r_2 \in I\) and \(s_1, s_2 \in J\text{,}\) then \((r_1 + s_1) + (r_2 + s_2) = (r_1 + r_2) +(s_1 + s_2)\) is in \(I + J\text{.}\) For \(a \in R\text{,}\) \(a(r_1 + s_1) = ar_1 + as_1 \in I + J\text{;}\) hence, \(I + J\) is an ideal in \(R\text{.}\)
19.5.18.
19.5.20.
Hint.
\(( \Rightarrow)\text{.}\) \(a = b \Rightarrow (a \wedge b') \vee (a' \wedge b) = (a \wedge a') \vee (a' \wedge a) = O \vee O = O\text{.}\) \(( \Leftarrow)\text{.}\) \(( a \wedge b') \vee (a' \wedge b) = O \Rightarrow a \vee b = (a \vee a) \vee b = a \vee (a \vee b) = a \vee [I \wedge (a \vee b)] = a \vee [(a \vee a') \wedge (a \vee b)] = [a \vee (a \wedge b')] \vee [a \vee (a' \wedge b)] = a \vee [(a \wedge b') \vee (a' \wedge b)] = a \vee 0 = a\text{.}\) A symmetric argument shows that \(a \vee b = b\text{.}\)
20 Vector Spaces
20.5 Exercises
20.5.3.
20.5.5.
20.5.7.
20.5.10.
20.5.12.
20.5.15. Linear Transformations.
Hint.
(a) Let \(u, v \in \ker(T)\) and \(\alpha \in F\text{.}\) Then
\begin{gather*}
T(u +v) = T(u) + T(v) = 0\\
T(\alpha v) = \alpha T(v) = \alpha 0 = 0\text{.}
\end{gather*}
Hence, \(u + v, \alpha v \in \ker(T)\text{,}\) and \(\ker(T)\) is a subspace of \(V\text{.}\)
(c) The statement that \(T(u) = T(v)\) is equivalent to \(T(u-v) = T(u) - T(v) = 0\text{,}\) which is true if and only if \(u-v = 0\) or \(u = v\text{.}\)
20.5.17. Direct Sums.
21 Fields
21.5 Exercises
21.5.1.
21.5.2.
21.5.3.
21.5.5.
Hint.
21.5.8.
21.5.14.
Hint.
Suppose that \(E\) is algebraic over \(F\) and \(K\) is algebraic over \(E\text{.}\) Let \(\alpha \in K\text{.}\) It suffices to show that \(\alpha\) is algebraic over some finite extension of \(F\text{.}\) Since \(\alpha\) is algebraic over \(E\text{,}\) it must be the zero of some polynomial \(p(x) = \beta_0 + \beta_1 x + \cdots + \beta_n x^n\) in \(E[x]\text{.}\) Hence \(\alpha\) is algebraic over \(F(\beta_0, \ldots, \beta_n)\text{.}\)
21.5.22.
Hint.
Since \(\{ 1, \sqrt{3}, \sqrt{7}, \sqrt{21}\, \}\) is a basis for \({\mathbb Q}( \sqrt{3}, \sqrt{7}\, )\) over \({\mathbb Q}\text{,}\) \({\mathbb Q}( \sqrt{3}, \sqrt{7}\, ) \supset {\mathbb Q}( \sqrt{3} +\sqrt{7}\, )\text{.}\) Since \([{\mathbb Q}( \sqrt{3}, \sqrt{7}\, ) : {\mathbb Q}] = 4\text{,}\) \([{\mathbb Q}( \sqrt{3} + \sqrt{7}\, ) : {\mathbb Q}] = 2\) or 4. Since the degree of the minimal polynomial of \(\sqrt{3} +\sqrt{7}\) is 4, \({\mathbb Q}( \sqrt{3}, \sqrt{7}\, ) = {\mathbb Q}( \sqrt{3} +\sqrt{7}\, )\text{.}\)
21.5.27.
Hint.
Let \(\beta \in F(\alpha)\) not in \(F\text{.}\) Then \(\beta = p(\alpha)/q(\alpha)\text{,}\) where \(p\) and \(q\) are polynomials in \(\alpha\) with \(q(\alpha) \neq 0\) and coefficients in \(F\text{.}\) If \(\beta\) is algebraic over \(F\text{,}\) then there exists a polynomial \(f(x) \in F[x]\) such that \(f(\beta) = 0\text{.}\) Let \(f(x) = a_0 + a_1 x + \cdots + a_n x^n\text{.}\) Then
\begin{equation*}
0 = f(\beta) = f\left( \frac{p(\alpha)}{q(\alpha)} \right) = a_0 + a_1 \left( \frac{p(\alpha)}{q(\alpha)} \right) + \cdots + a_n \left( \frac{p(\alpha)}{q(\alpha)} \right)^n\text{.}
\end{equation*}
Now multiply both sides by \(q(\alpha)^n\) to show that there is a polynomial in \(F[x]\) that has \(\alpha\) as a zero.
21.5.28.
Hint.
See the comments following Theorem 21.13.
22 Finite Fields
22.4 Exercises
22.4.1.
22.4.4.
22.4.5.
22.4.7.
22.4.8.
22.4.11.
22.4.12.
22.4.17.
22.4.18.
Hint.
Since \(\alpha\) is algebraic over \(F\) of degree \(n\text{,}\) we can write any element \(\beta \in F(\alpha)\) uniquely as \(\beta = a_0 + a_1 \alpha + \cdots + a_{n - 1} \alpha^{n - 1}\) with \(a_i \in F\text{.}\) There are \(q^n\) possible \(n\)-tuples \((a_0, a_1, \ldots,
a_{n - 1})\text{.}\)
22.4.24. Wilson’s Theorem.
23 Galois Theory
23.5 Exercises
23.5.1.
23.5.2.
23.5.3.
Hint.
If
\begin{equation*}
[\gf(729): \gf(9)] = [\gf(729): \gf(3)] /[\gf(9): \gf(3)] = 6/2 = 3\text{,}
\end{equation*}
then \(G(\gf(729)/ \gf(9)) \cong {\mathbb Z}_3\text{.}\) A generator for \(G(\gf(729)/ \gf(9))\) is \(\sigma\text{,}\) where \(\sigma_{3^6}( \alpha) = \alpha^{3^6} = \alpha^{729}\) for \(\alpha \in \gf(729)\text{.}\)
23.5.4.
Hint.
23.5.5.
23.5.7.
Hint.
Let \(E\) be the splitting field of a cubic polynomial in \(F[x]\text{.}\) Show that \([E:F]\) is less than or equal to \(6\) and is divisible by \(3\text{.}\) Since \(G(E/F)\) is a subgroup of \(S_3\) whose order is divisible by \(3\text{,}\) conclude that this group must be isomorphic to \({\mathbb Z}_3\) or \(S_3\text{.}\)
23.5.9.
23.5.16.
23.5.20.
Hint.
-
Clearly \(\omega, \omega^2, \ldots, \omega^{p - 1}\) are distinct since \(\omega \neq 1\) or 0. To show that \(\omega^i\) is a zero of \(\Phi_p\text{,}\) calculate \(\Phi_p( \omega^i)\text{.}\)
-
The conjugates of \(\omega\) are \(\omega, \omega^2, \ldots, \omega^{p - 1}\text{.}\) Define a map \(\phi_i: {\mathbb Q}(\omega) \rightarrow {\mathbb Q}(\omega^i)\) by\begin{equation*} \phi_i(a_0 + a_1 \omega + \cdots + a_{p - 2} \omega^{p - 2}) = a_0 + a_1 \omega^i + \cdots + c_{p - 2} (\omega^i)^{p - 2}\text{,} \end{equation*}where \(a_i \in {\mathbb Q}\text{.}\) Prove that \(\phi_i\) is an isomorphism of fields. Show that \(\phi_2\) generates \(G({\mathbb Q}(\omega)/{\mathbb Q})\text{.}\)
-
Show that \(\{ \omega, \omega^2, \ldots, \omega^{p - 1} \}\) is a basis for \({\mathbb Q}( \omega )\) over \({\mathbb Q}\text{,}\) and consider which linear combinations of \(\omega, \omega^2, \ldots, \omega^{p - 1}\) are left fixed by all elements of \(G( {\mathbb Q}( \omega ) / {\mathbb Q})\text{.}\)

