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Appendix D Hints and Answers to Selected Exercises

1 Rings
1.6 Exercises

1.6.1.

Hint.
(a) \(7 {\mathbb Z}\) is a ring but not a field; (c) \({\mathbb Q}(\sqrt{2}\, )\) is a field; (f) \(R\) is not a ring.

1.6.3.

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*}

1.6.4.

Hint.
(a) \(\{0 \}\text{,}\) \(\{0, 9 \}\text{,}\) \(\{0, 6, 12 \}\text{,}\) \(\{0, 3, 6, 9, 12, 15 \}\text{,}\) \(\{0, 2, 4, 6, 8, 10, 12, 14, 16 \}\text{;}\) (c) there are no nontrivial ideals.

1.6.7.

Hint.
Assume there is an isomorphism \(\phi: {\mathbb C} \rightarrow {\mathbb R}\) with \(\phi(i) = a\text{.}\)

1.6.8.

Hint.
False. Assume there is an isomorphism \(\phi: {\mathbb Q}(\sqrt{2}\, ) \rightarrow {\mathbb Q}(\sqrt{3}\, )\) such that \(\phi(\sqrt{2}\, ) = a\text{.}\)

1.6.13.

Hint.
(a) \(x \equiv 17 \pmod{55}\text{;}\) (c) \(x \equiv 214 \pmod{2772}\text{.}\)

1.6.16.

Hint.
If \(I \neq \{ 0 \}\text{,}\) show that \(1 \in I\text{.}\)

1.6.18.

Hint.
(a) \(\phi(a) \phi(b) = \phi(ab) = \phi(ba) = \phi(b) \phi(a)\text{.}\)

1.6.26.

Hint.
Let \(a \in R\) with \(a \neq 0\text{.}\) Then the principal ideal generated by \(a\) is \(R\text{.}\) Thus, there exists a \(b \in R\) such that \(ab =1\text{.}\)

1.6.33.

Hint.
Let \(a/b, c/d \in {\mathbb Z}_{(p)}\text{.}\) Then \(a/b + c/d = (ad + bc)/bd\) and \((a/b) \cdot (c/d) = (ac)/(bd)\) are both in \({\mathbb Z}_{(p)}\text{,}\) since \(\gcd(bd,p) = 1\text{.}\)

1.6.37.

Hint.
Suppose that \(x^2 = x\) and \(x \neq 0\text{.}\) Since \(R\) is an integral domain, \(x = 1\text{.}\) To find a nontrivial idempotent, look in \({\mathbb M}_2({\mathbb R})\text{.}\)

2 Polynomials
2.4 Exercises

2.4.2.

Hint.
(a) \(9x^2 + 2x + 5\text{;}\) (b) \(8x^4 + 7x^3 + 2x^2 + 7x\text{.}\)

2.4.3.

Hint.
(a) \(5 x^3 + 6 x^2 - 3 x + 4 = (5 x^2 + 2x + 1)(x -2) + 6\text{;}\) (c) \(4x^5 - x^3 + x^2 + 4 = (4x^2 + 4)(x^3 + 3) + 4x^2 + 2\text{.}\)

2.4.5.

Hint.
(a) No zeros in \({\mathbb Z}_{12}\text{;}\) (c) \(3\text{,}\) \(4\text{.}\)

2.4.10.

Hint.
One factorization is \(x^2 + x + 8 = (x + 2)(x + 9)\text{.}\)

2.4.16.

Hint.
Let \(\phi : R \rightarrow S\) be an isomorphism. Define \(\overline{\phi} : R[x] \rightarrow S[x]\) by \(\overline{\phi}(a_0 + a_1 x + \cdots + a_n x^n) = \phi(a_0) + \phi(a_1) x + \cdots + \phi(a_n) x^n\text{.}\)

2.4.20. Cyclotomic Polynomials.

Hint.
The polynomial
\begin{equation*} \Phi_n(x) = \frac{x^n - 1}{x - 1} = x^{n - 1} + x^{n - 2} + \cdots + x + 1 \end{equation*}
is called the cyclotomic polynomial. Show that \(\Phi_p(x)\) is irreducible over \({\mathbb Q}\) for any prime \(p\text{.}\)

3 Integral Domains
3.3 Exercises

3.3.1.

Hint.
Note that \(z^{-1} = 1/(a + b\sqrt{3}\, i) = (a -b \sqrt{3}\, i)/(a^2 + 3b^2)\) is in \({\mathbb Z}[\sqrt{3}\, i]\) if and only if \(a^2 + 3 b^2 = 1\text{.}\) The only integer solutions to the equation are \(a = \pm 1, b = 0\text{.}\)

3.3.2.

Hint.
(a) \(5 = -i(1 + 2i)(2 + i)\text{;}\) (c) \(6 + 8i = -i(1 + i)^2(2 + i)^2\text{.}\)

3.3.9.

Hint.
Let \(z = a + bi\) and \(w = c + di \neq 0\) be in \({\mathbb Z}[i]\text{.}\) Prove that \(z/w \in {\mathbb Q}(i)\text{.}\)

3.3.15.

Hint.
Let \(a = ub\) with \(u\) a unit. Then \(\nu(b) \leq \nu(ub) \leq \nu(a)\text{.}\) Similarly, \(\nu(a) \leq \nu(b)\text{.}\)

3.3.16.

Hint.
Show that \(21\) can be factored in two different ways.

4 Vector Spaces
4.4 Exercises

4.4.3.

Hint.
\({\mathbb Q}(\sqrt{2}, \sqrt{3}\, )\) has basis \(\{ 1, \sqrt{2}, \sqrt{3}, \sqrt{6}\, \}\) over \({\mathbb Q}\text{.}\)

4.4.5.

Hint.
The set \(\{ 1, x, x^2, \ldots, x^{n-1} \}\) is a basis for \(P_n\text{.}\)

4.4.7.

Hint.
(a) Subspace of dimension \(2\) with basis \(\{(1, 0, -3), (0, 1, 2) \}\text{;}\) (d) not a subspace

4.4.10.

Hint.
Since \(0 = \alpha 0 = \alpha(-v + v) = \alpha(-v) + \alpha v\text{,}\) it follows that \(- \alpha v = \alpha(-v)\text{.}\)

4.4.12.

Hint.
Let \(v_0 = 0, v_1, \ldots, v_n \in V\) and \(\alpha_0 \neq 0, \alpha_1, \ldots, \alpha_n \in F\text{.}\) Then \(\alpha_0 v_0 + \cdots + \alpha_n v_n = 0\text{.}\)

4.4.15. 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 \(u = v\text{.}\)

4.4.17. Direct Sums.

Hint.
(a) Let \(u, u' \in U\) and \(v, v' \in V\text{.}\) Then
\begin{align*} (u + v) + (u' + v') & = (u + u') + (v + v') \in U + V\\ \alpha(u + v) & = \alpha u + \alpha v \in U + V\text{.} \end{align*}

5 Fields
5.4 Exercises

5.4.1.

Hint.
(a) \(x^4 - (2/3) x^2 - 62/9\text{;}\) (c) \(x^4 - 2 x^2 + 25\text{.}\)

5.4.2.

Hint.
(a) \(\{ 1, \sqrt{2}, \sqrt{3}, \sqrt{6}\, \}\text{;}\) (c) \(\{ 1, i, \sqrt{2}, \sqrt{2}\, i \}\text{;}\) (e) \(\{1, 2^{1/6}, 2^{1/3}, 2^{1/2}, 2^{2/3}, 2^{5/6} \}\text{.}\)

5.4.3.

Hint.
(a) \({\mathbb Q}(\sqrt{3}, \sqrt{7}\, )\text{.}\)

5.4.5.

Hint.
Use the fact that the elements of \({\mathbb Z}_2[x]/ \langle x^3 + x + 1 \rangle\) are 0, 1, \(\alpha\text{,}\) \(1 + \alpha\text{,}\) \(\alpha^2\text{,}\) \(1 + \alpha^2\text{,}\) \(\alpha + \alpha^2\text{,}\) \(1 + \alpha + \alpha^2\) and the fact that \(\alpha^3 + \alpha + 1 = 0\text{.}\)

5.4.14.

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 \(F(\beta_0, \ldots, \beta_n)\text{.}\)

5.4.22.

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, \({\mathbb Q}( \sqrt{3}, \sqrt{7}\, ) = {\mathbb Q}( \sqrt{3} +\sqrt{7}\, )\text{.}\)

5.4.27.

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.

6 Finite Fields
6.2 Exercises

6.2.4.

Hint.
There are eight elements in \({\mathbb Z}_2(\alpha)\text{.}\) Exhibit two more zeros of \(x^3 + x^2 + 1\) other than \(\alpha\) in these eight elements.

6.2.5.

Hint.
Find an irreducible polynomial \(p(x)\) in \({\mathbb Z}_3[x]\) of degree \(3\) and show that \({\mathbb Z}_3[x]/ \langle p(x) \rangle\) has \(27\) elements.

6.2.7.

Hint.
(a) \(x^5 -1 = (x+1)(x^4+x^3 + x^2 + x+ 1)\text{;}\) (c) \(x^9 -1 = (x+1)( x^2 + x+ 1)(x^6+x^3+1)\text{.}\)

6.2.17.

Hint.
If \(p(x) \in F[x]\text{,}\) then \(p(x) \in E[x]\text{.}\)

6.2.18.

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 \((a_0, a_1, \ldots, a_{n - 1})\text{.}\)

7 Groups
7.6 Exercises

7.6.2.

Hint.
(a) \(X_{(1)} = \{1, 2, 3 \}\text{,}\) \(X_{(1 \, 2)} = \{3 \}\text{,}\) \(X_{(1 \, 3)} = \{ 2 \}\text{,}\) \(X_{(2 \, 3)} = \{1 \}\text{,}\) \(X_{(1 \, 2 \, 3)} = X_{(1 \, 3 \, 2)} = \emptyset\text{.}\) \(G_1 = \{ (1), (2 \, 3) \}\text{,}\) \(G_2 = \{(1), (1 \, 3) \}\text{,}\) \(G_3 = \{ (1), (1 \, 2)\}\text{.}\)

7.6.3.

Hint.
(a) \({\mathcal O}_1 = {\mathcal O}_2 = {\mathcal O}_3 = \{ 1, 2, 3\}\text{.}\)

8 Galois Theory
8.3 Exercises

8.3.1.

Hint.
(a) \({\mathbb Z}_2\text{;}\) (c) \({\mathbb Z}_2 \times {\mathbb Z}_2 \times {\mathbb Z}_2\text{.}\)

8.3.2.

Hint.
(a) Separable over \(\mathbb Q\) since \(x^3 + 2 x^2 - x - 2 = (x - 1)(x + 1)(x + 2)\text{;}\) (c) not separable over \(\mathbb Z_3\) since \(x^4 + x^2 + 1 = (x + 1)^2 (x + 2)^2 \text{.}\)

8.3.3.

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 \(\alpha \in \gf(729)\text{.}\)

8.3.7.

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 \(S_3\text{.}\)

8.3.20.

Hint.
  1. 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 \(\Phi_p( \omega^i)\text{.}\)
  2. 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 \(G({\mathbb Q}(\omega)/{\mathbb Q})\text{.}\)
  3. 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 \(G( {\mathbb Q}( \omega ) / {\mathbb Q})\text{.}\)

10 The Sylow Theorems
10.3 Exercises

10.3.6.

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 \(1 + 3 + 6 + 6 + 8 = 24\text{.}\)

10.3.8.

Hint.
If \(|G| = 18 = 2 \cdot 3^2\text{,}\) then the order of a Sylow \(2\)-subgroup is \(2\text{,}\) and the order of a Sylow \(3\)-subgroup is \(9\text{.}\)

10.3.9.

Hint.
The four Sylow \(3\)-subgroups of \(S_4\) are \(P_1 = \{ (1), (1 \, 2 \, 3), (1 \, 3 \, 2) \}\text{,}\) \(P_2 = \{ (1), (1 \, 2 \, 4), (1 \, 4 \, 2) \}\text{,}\) \(P_3 = \{ (1), (1 \, 3 \, 4), (1 \, 4 \, 3) \}\text{,}\) \(P_4 = \{ (1), (2 \, 3 \, 4), (2 \, 4 \, 3) \}\text{.}\)

10.3.12.

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.

10.3.15.

Hint.
Show that \(G\) has a normal Sylow \(p\)-subgroup of order \(p^2\) and a normal Sylow \(q\)-subgroup of order \(q^2\text{.}\)

10.3.24.

Hint.
If \(G\) is abelian, then \(G\) is cyclic, since \(|G| = 3 \cdot 5 \cdot 17\text{.}\)

10.3.30.

Hint.
Define a mapping between the right cosets of \(N(H)\) in \(G\) and the conjugates of \(H\) in \(G\) by \(N(H) g \mapsto g^{-1} H g\text{.}\) Prove that this map is a bijection.

10.3.33.

Hint.
Let \(a G', b G' \in G/G'\text{.}\) Then \((a G')( b G') = ab G' = ab(b^{-1}a^{-1}ba) G' = (abb^{-1}a^{-1})ba G' = ba G'\text{.}\)

10.3.36.

Hint.
Use the fact that \(x \in g C(a) g^{-1}\) if and only if \(g^{-1}x g \in C(a)\text{.}\)

11 Applications of Galois Theory
11.4 Exercises

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