Homework Archive¶

Question

(i) Write out an addition table for \(\mathbf{Z}_2 \oplus \mathbf{Z}_2\). \(\mathbf{Z}_2 \oplus \mathbf{Z}_2\) is called the Klein four group.

\(\star\)(ii) Proof the following conditions on a group \(G\) are equivalent:
(a) \(G\) is Abelian;
(b) \((ab)^2=a^2b^2, \forall a, b \in G\);
(c) \((ab)^{-1}=a^{-1}b^{-1}, \forall a, b \in G\);
(d) \((ab)^n=a^nb^n, \forall n \in \mathbf{Z}, a, b \in G\);
(e) \((ab)^n=a^nb^n\) for three consectuive integers \(n\) and all \(a, b \in G\).
And show that \((e) \Rightarrow (a)\) is false if "three" is replaced by "two".

(iii) Let \(Q_8\) be the group (under ordinary matrix multiplication) generated by the complex matrices \(A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\) and \(B = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}\), where \(i^2=-1\). Show that \(Q_8\) is a nonabelian group of order 8. \(Q_8\) is called the quaternion group.[Hint: Observe that \(BA = A^3B\), whence every element of \(Q_8\) is of the form \(A^iB^j\). Note also that \(A^4=B^4=I\), where \(I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) is the identity element of \(Q_8\).]

\(\star\)(iv) List all subgroups of \(\mathbf{Z}_2 \oplus \mathbf{Z}_2\). Is \(\mathbf{Z}_2 \oplus \mathbf{Z}_2\) isomorphic to \(\mathbf{Z}_4\)?

\(\star\)(v) Let \(G\) be an abelian group containing elements \(a\) and \(b\) of orders \(m\) and \(n\) respectively. Show that \(G\) contains an element whose order is the least common multiple of \(m\) and \(n\). [Hint: First try the case when \((m,n) = 1\).]

(vi) Let \(G\) be the multiplicative group of all nonsingular \(2 \times 2\) matrices with rational entries. Show that \(a = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) has order 4 and \(b = \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}\) has order 3, but \(ab\) has infinite order. Conversely, show that the additive group \(\mathbf{Z}_2 \oplus \mathbf{Z}\) contains nonzero elements \(a, b\) of infinite order such that \(a+b\) has finite order.

Question

(i) (Euler-Fermat) Let \(a\) be a integer and \(p\) a prime such that \(p \nmid a\). Then \(a^{p-1} \equiv 1 \pmod p\).[Hint: Consider \(\overline{a} \in \mathbf{Z}_p\) and the multiplicative group of nonzero elements of \(\mathbf{Z}_p\).] It follows that \(a^p \equiv a \pmod p\) for any integer \(a\).

(ii) Let \(G\) be a group of order \(2n\); then \(G\) contains an element of order 2. If \(n\) is odd and \(G\) abelian, there is only one element of order 2.

(iii) (a) If \(G\) is a group, then the center of \(G\) is a normal subgroup of \(G\).(Center: \(C=\{a \in G \mid ax=xa, \forall x \in G\}\).)
(b) The center of \(S_n\) is the identity subgroup for all \(n>2\).

(iv) If \(f: G \rightarrow H\) is a homomorphism with kernel \(N\) and \(K<G\), then prove that \(f^{-1}(f(K))=KN\). Hence \(f^{-1}(f(K))=K\) iff \(N<K\).

(v) Show that \(N=\{(1), (12)(34), (13)(24), (14)(23)\}\) is a normal subgroup of \(S_4\) contained in \(A_4\) such that \(S_4/N \cong S_3\) and \(A_4/N \cong Z_3\).

(vi) Find all normal subgroups of \(D_n\).

Question

(i) A subset \(X\) of an abelian group \(F\) is said to be linearly independent if \(n_1x_1 + \cdots + n_kx_k = 0\) always implies \(n_i = 0\) for all \(i\) (where \(n_i \in \mathbf{Z}\) and \(x_1, \ldots ,x_k\) are distinct elements of \(X\)).
(a) \(X\) is linearly independent if and only if every nonzero element of the subgroup \(\langle X \rangle\) may be written uniquely in the form \(n_1x_1 + \cdots + n_kx_k\) (\(n_i \in \mathbf{Z}, n_i \neq 0, x_1, \ldots ,x_k\) distinct elements of \(X\)).
(b) If \(F\) is free abelian of finite rank \(n\), it is not true that every linearly independent subset of \(n\) elements is a basis [Hint: consider \(F = \mathbf{Z}\)].
(c) If \(F\) is free abelian, it is not true that every linearly independent subset of \(F\) may be extended to a basis of \(F\).
(d) If \(F\) is free abelian, it is not true that every generating set of \(F\) contains a basis of \(F\). However, if \(F\) is also finitely generated by \(n\) elements, \(F\) has rank \(m \leqslant n\).

\(\star\)(ii) (a) Show that the additive group of rationals \(\mathbf{Q}\) is not finitely generated.
(b) Show that \(\mathbf{Q}\) is not free.
(c) Conclude that Exercise 9 is false if the hypothesis "finitely generated" is omitted.
Exercise 9: Let \(G\) be a finitely generated abelian group in which no element (except 0) has finite order. Then \(G\) is a free abelian group.

(iii) Let \(k, m \in N^*\). If \((k,m) = 1\), then \(k\mathbf{Z}_m = \mathbf{Z}_m\) and \(\mathbf{Z}_m[k] = 0\). If \(k \mid m\), say \(m = kd\), then \(k\mathbf{Z}_m \cong \mathbf{Z}_d\) and \(\mathbf{Z}_m[k] \cong \mathbf{Z}_k\)

\(\star\)(iv) (a) What are the elementary divisors of the group \(\mathbf{Z}_2 \oplus \mathbf{Z}_9 \oplus \mathbf{Z}_{35}\); what are its invariant factors? Do the same for \(\mathbf{Z}_{26} \oplus \mathbf{Z}_{42} \oplus \mathbf{Z}_{49} \oplus \mathbf{Z}_{200} \oplus \mathbf{Z}_{1000}\).
(b) Determine up to isomorphism all abelian groups of order 64; do the same for order 96.
(c) Determine all abelian groups of order \(n\) for \(n \leqslant 20\).

Question

\(\star\) (i) If \(\lvert G \rvert = p^nq\), with \(p > q\) primes, then \(G\) contains a unique normal subgroup of index \(q\).

(ii) Every group \(G\) of order \(p^2\) (\(p\) prime) is abelian [Hint: Exercise 4.9 and Corollary 5.4].
Exercise 4.9: If \(G/C(G)\) is cyclic, then \(G\) is abelian.
Corollary 5.4: The center \(C(G)\) of a nontrivial finite \(p\)-group \(G\) contains more than one element.

(iii) Consider the set \(G = \{\pm 1, \pm i, \pm j, \pm k\}\) with multiplication given by \(i^2 = j^2 = k^2 = -1\); \(ij = k; jk = i, ki = j\); \(ji = -k, kj = -i, ik = -j\), and the usual rules for multiplying by \(\pm 1\). Show that \(G\) is a group isomorphic to the quaternion group \(Q_8\).

(iv) (a) Show that there is a nonabelian subgroup \(T\) of \(S_3 \times \mathbf{Z}_4\) of order 12 generated by elements \(a, b\) such that \(\lvert a \rvert = 6\), \(a^3 = b^2\), \(ba = a^{-1}b\).
(b) Any group of order 12 with generators \(a, b\) such that \(\lvert a \rvert = 6\), \(a^3 = b^2\), \(ba = a^{-1}b\) is isomorphic to \(T\).

Question

(i) (a) \(A_4\) is not the direct product of its Sylow subgroups, but \(A_4\) does have the property: \(mn = 12\) and \((m, n) = 1\) imply there is a subgroup of order \(m\).
(b) \(S_3\) has subgroups of orders 1, 2, 3, and 6 but is not the direct product of its Sylow subgroups.

(ii) If \(D_n\) is the dihedral group with generators \(a\) of order \(n\) and \(b\) of order \(2\), then
(a) \(a^2 \in D_n'\).
(b) If \(n\) is odd, \(D_n' \cong Z_n\).
(c) If \(n\) is even, \(Dn' \cong Z_m\), where \(2m = n\).
(d) \(D_n\) is nilpotent if and only if \(n\) is a power of 2.

(iii) (a) Find a normal series of \(D_4\) consisting of 4 subgroups.
(b) Find all composition series of the group \(D_4\).
(c) Do part (b) for the group \(A_4\).
(d) Do part (b) for the group \(S_3 \times \mathbf{Z}_2\).
(e) Find all composition factors of \(S_4\) and \(D_6\).

(iv) (a) Let \(G\) be an (additive) abelian group. Define an operation of multiplication in \(G\) by \(ab = 0\) (for all \(a, b \in G\)). Then \(G\) is a ring.
(b) Let \(S\) be the set of all subsets of some fixed set \(U\). For \(A, B \in S\), define \(A + B = (A - B) \cup (B - A)\) and \(AB = A \cap B\). Then \(S\) is a ring. Is \(S\) commutative? Does it have an identity?

(v) Let \(R\) be the set of all \(2 \times 2\) matrices over the complex field \(\mathbf{C}\) of the form

where \(\overline{z}, \overline{w}\) are the complex conjugates of \(z\) and \(w\) respectively (that is, \(c = a+b\sqrt{-1} \Leftrightarrow \overline{c} = a-b\sqrt{-1}\)). Then \(R\) is a division ring that is isomorphic to the division ring \(K\) of real quaternions. Hint: Define an isomorphism $\varphi K \rightarrow R $ by letting the images of \(1, i, j, k \in K\) be respectively the matrices

(vi) (a) The subset \(G = \{1, -1, i, -i, j, -j, k, -k\}\) of the division ring \(K\) of real quaternions forms a group under multiplication.
(b) \(G\) is isomorphic to the quaternion group (Exercises I.4.14 and I.2.3).
(c) What is the difference between the ring \(K\) and the group ring \(R(G)\) (\(R\) the field of real numbers)?

Question

\(\star\)(i) (a) The center of the ring \(S\) of all \(2 \times 2\) matrices over a field \(F\) consists of all matrices of the form \(\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}\).
(b) The center of \(S\) is not an ideal in \(S\).
(c) What is the center of the ring of all \(n \times n\) matrices over a division ring?

(ii) (a) Show that \(\mathbf{Z}\) is a principal ideal ring [see Theorem I.3.1].
(b) Every homomorphic image of a principal ideal ring is also a principal ideal ring.
(c) \(Z_m\) is a principal ideal ring for every \(m > 0\).

(iii) Determine all prime and maximal ideals in the ring \(\mathbf{Z}_m\).

(iv) An element \(e\) in a ring \(R\) is said to be idempotent if \(e^2 = e\). An element of the center of the ring \(R\) is said to be central. If \(e\) is a central idempotent in a ring \(R\) with identity, then
(a) \(1_R-e\) is a central idempotent;
(b) \(eR\) and \((1_R-e)R\) are ideals in \(R\) such that \(R = eR \times (1_R-e)R\).

\(\star\)(v) Let \(R\) be the subring \(\{a+b\sqrt{10} \mid a, b \in \mathbf{Z}\}\) of the field of real numbers.
(a) The map \(N: R \rightarrow \mathbf{Z}\) given by \(a+b\sqrt{10} \mapsto (a+b\sqrt{10})(a-b\sqrt{10}) = a^2-10b^2\) is such that \(N(uv) = N(u)N(v)\) for all \(u,v \in R\) and \(N(u) = 0\) if and only if \(u = 0\).
(b) \(u\) is a unit in \(R\) if and only if \(N(u) = \pm 1\).
(c) \(2, 3, 4+\sqrt{10}\) and \(4-\sqrt{10}\) are irreducible elements of \(R\).
(d) \(2, 3, 4+\sqrt{10}\) and \(4-\sqrt{10}\) are not prime elements of \(R\). [Hint: \(3 \cdot 2 = 6 = (4+\sqrt{10})(4-\sqrt{10})\)].

(vi) Show that in the integral domain of Exercise 3 every element can be factored into a product of irreducibles, but this factorization need not be unique (in the sense of Definition 3.5 (ii)).

Question

(i) (a) If \(a\) and \(n\) are integers, \(n>0\), then there exist integers \(q\) and \(r\) such that \(a = qn+r\), where \(\lvert r \rvert < n/2\).
(b) The Gaussian integers \(\mathbf{Z}[\sqrt{-1}]\) form a Euclidean domain with \(\varphi(a+b\sqrt{-1}) = a^2 + b^2\) [Hint: To show that Definition 3.8(ii) holds, first let \(y = a+b\sqrt{-1}\) and assume \(x\) is a positive integer. By part (a) there are integers such that \(a = q_1x+r_1\) and \(b = q_2x+r_2\), with \(\lvert r_1 \rvert < x/2\), \(\lvert r_2 \rvert < x/2\). Let \(q = q_1 + q_2\sqrt{-1}\) and \(r = r_1+r_2\sqrt{-1}\); then \(y = qx+r\), with \(r = 0\) or \(\varphi(r) < \varphi(x)\). In the general case, observe that for \(x = c+d\sqrt{-1} \neq 0\) and \(\overline{x} = c-d\sqrt{-1}\), \(x\overline{x} > 0\). There are \(q, r_0 \in \mathbf{Z}[\sqrt{-1}]\) such that \(y\overline{x} = q(x\overline{x})+r_0\), with \(r_0 = 0\) or \(\varphi(r_0) < \varphi(x\overline{x})\). Let \(r = y-qx\); then \(y = qx+r\) and \(r = 0\) or \(\varphi(r) < \varphi(x)\).]

(ii) (a) The set \(E\) of positive even integers is a multiplicative subset of \(\mathbf{Z}\) such that \(E^{-1}\mathbf{Z}\) is the field of rational numbers.
(b) State and prove condition(s) on a multiplicative subset \(S\) of \(\mathbf{Z}\) which insure that \(S^{-1}\mathbf{Z}\) is the field of rationals.

(iii) If \(S = \{2, 4\}\) and \(R = \mathbf{Z}_6\), then \(S^{-1}R\) is isomorphic to the field \(\mathbf{Z}_3\). Consequently, the converse of Theorem 4.3(ii) is false.

(iv) (a) If \(R\) is the ring of all \(2 \times 2\) matrices over \(\mathrm{Z}\), then for any \(A in R\),

(b) There exist \(C, A \in R\) such that \((C + A)(C - A) \neq C^2 - A^2\). Therefore, Corollary 5.6 is false if the rings involved are not commutative.

\(\star\)(v) (a) If \(F\) is a field then every nonzero element of \(F[[x]]\) is of the form \(x^ku\) with \(u \in F[[x]]\) a unit.
(b) \(F[[x]]\) is a principal ideal domain whose only ideals are \(0, F[[x]] = (1_F) = (x^0)\) and \((x^k)\) for each \(k > 1\).

Question

(i) (a) If \(D\) is an integral domain and \(c\) is an irreducible element in \(D\), then \(D[x]\) is not a principal ideal domain. [Hint: consider the ideal \((x,c)\) generated by $x4 and \(c\).]
(b) \(\mathrm{Z}[x]\) is not a principal ideal domain.
(c) If \(F\) is a field and \(n > 2\), then \(F[x_1, \ldots , x_n]\) is not a principal ideal domain. [Hint: show that \(x_1\) is irreducible in \(F[x_1, \ldots , x_{n-1}]\).]

(ii) Let \(f\) be a polynomial of positive degree over an integral domain \(D\).
(a) If \(\mathrm{char} D = 0\), then \(f' \neq 0\).
(b) If \(\mathrm{char} D = p \neq 0\), then \(f' = 0\) if and only if \(f\) is a polynomial in \(x^p\) (that is, \(f = a_o + a_px^p + a_{2p}x^{2p} + \cdots + a_{ip}x^{ip}\)).

(iii) (a) Let \(D\) be an integral domain and \(c \in D\). Let \(f(x) = \sum_{i=0}^n a_ix^i \in D[x]\) and \(f(x - c) = \sum_{i=0}^n a_i(x-c)^i \in D[x]\). Then \(f(x)\) is irreducible in \(D[x]\) iff \(f(x - c)\) is irreducible.
(b) For each prime \(p\), the cyclotomic polynomial（分圆多项式） \(f = x^{p-1} + x^{p-2} + \cdots + x + 1\) is irreducible in \(\mathrm{Z}[x]\). [Hint: observe that \(f = \dfrac{x^p-1}{x-1}\), whence \(f(x + 1) = \dfrac{(x + 1)^p - 1}{x}\). Use the Binomial Theorem 1 .6 and Eisenstein's Criterion to show that \(f(x + 1)\) is irreducible in \(\mathrm{Z}[x]\).]

(iv) Let \(I\) be a left ideal of a ring \(R\) and \(A\) an \(R\)-module.
(a) If \(S\) is a nonempty subset of \(A\), then \(IS = \{\sum_{i=1}^nr_ia_i \mid n \in N^*, r_i \in I, a_i \in S\}\) is a submodule of \(A\). Note that if \(S = \{a\}\) , then \(IS = Ia = \{ra \mid r \in I\}\).
(b) If \(I\) is a two-sided ideal, then \(A/IA\) is an \(R/I\)-module with the action of \(R/I\) given by \((r + I)(a + IA) = ra + IA\).

\(\star\)(v) (a) If \(A\) and \(B\) are \(R\)-modules, then the set \(\mathrm{Hom}_R(A,B)\) of all \(R\)-module homomorphisms \(A \rightarrow B\) is an abelian group with \(f + g\) given on \(a \in A\) by \((f + g)(a) = f(a) + g(a) \in B\). The identity element is the zero map.
(b) \(\mathrm{Hom}_R(A,A)\) is a ring with identity, where multiplication is composition of functions. \(\mathrm{Hom}_R(A,A)\) is called the endomorphism ring（自同态环） of \(A\).
(c) \(A\) is a left \(\mathrm{Hom}_R(A,A)\)-module with \(fa\) defined to be \(f(a)\) (\(a \in A, f \in \mathrm{Hom}_R(A,A)\)).

Question

(Consider all the rings as communative rings with identity, and all division rings as fields.)

(i) (The five lemma) Let

(ii) (a) If \(0 \rightarrow A \rightarrow B \xrightarrow{f} C \rightarrow 0\) and \(0 \rightarrow C \xrightarrow{g} D \rightarrow E \rightarrow 0\) are exact sequences of \(R\)-modules, then the sequence \(0 \rightarrow A \rightarrow B \xrightarrow{gf} D \rightarrow E \rightarrow 0\) is exact.
(b) Show that every exact sequence may be obtained by splicing together suitable short exact sequences as in (a).

\(\star\)(iii) Let \(R\) be a principal ideal domain, \(A\) a unitary left R-module, and \(p \in R\) a prime (= irreducible). Let \(pA = \{pa \mid a \in A\}\) and \(A[p] = \{a \in A \mid pa = 0\}\).
(a) \(R/(p)\) is a field(Theorems III.2.20 and III.3.4);
(b) \(pA\) and \(A[p]\) are submodules of \(A\);
(c) \(A/pA\) is a vector space over \(R/(p)\), with scalar multiplication given by \((r + (p))(a + pA) = ra + pA\);
(d) \(A[p]\) is a vector space over \(R/(p)\), with scalar multiplication given by \((r + (p))(a) = ra\).

(iv) The following conditions on a ring \(R\) [with identity] are equivalent:
(a) Every [unitary] R-module is projective.
(b) Every short exact sequence of [unitary] R-modules is split exact.
(c) Every [unitary] R-module is injective.

Question

(i) Without using Lemma 3.9 prove that:
(a) Every homomorphic image of a divisible abelian group is divisible.
(b) Every direct summand (Exercise I.8.12) of a divisible abelian group is divisible.
(c) A direct sum of divisible abelian groups is divisible.

(Lemma 3.9: An abelian group \(D\) is divisible if and only if \(D\) is an injective (unitary) \(\mathbf{Z}\)-module.
Exercise I.8.12: A normal subgroup \(H\) of a group \(G\) is said to be a direct factor (direct summand if \(G\) is additive abelian) if there exists a (normal) subgroup \(K\) of \(G\) such that \(G = H \times K\).)

(ii)
(a) For any abelian group \(A\) and positive integer \(m\), \(\operatorname{Hom}(\mathbf{Z}_m, A) \cong A[m] = \{a \in A \mid ma = 0\}\).
(b) \(\operatorname{Hom}(\mathbf{Z}_m, \mathbf{Z}_n) \cong \mathbf{Z}_{(m, n)}\).
(c) The \(\mathbf{Z}\)-module \(\mathbf{Z}_m\) has \(\mathbf{Z}_m^* = 0\).
(d) For each \(k > 1\), \(\mathbf{Z}_m\) is a \(\mathbf{Z}_{mk}\)-module (Exercise 1.1); as a \(\mathbf{Z}_{mk}\)-module, \(\mathbf{Z}_m^* \cong \mathbf{Z}_m\).

(iii) Let \(A\) and \(B\) be abelian groups.
(a) For each \(m > 0\), \(A \otimes \mathbf{Z}_m \cong A/mA\).
(b) \(\mathbf{Z}_m \otimes \mathbf{Z}_n \cong \mathbf{Z}_{(m, n)}\).
(c) Describe \(A \otimes B\), when \(A\) and \(B\) are finitely generated.

(iv) If \(A\) is a torsion abelian group and \(\mathbf{Q}\) the (additive) group of rationals, then
(a) \(A \otimes \mathbf{Q} = 0\).
(b) \(\mathbf{Q} \otimes \mathbf{Q} \cong \mathbf{Q}\).

Question

(i) If \(R\) is a nonzero commutative ring with identity and every submodule of every free \(R\)-module is free, then \(R\) is a principal ideal domain. [Hint: Every ideal \(I\) of \(R\) is a free \(R\)-module. If \(u, v \in I\) (\(u \neq 0, v \neq 0\)), then \(uv + (-v)u = 0\), which implies that \(I\) has a basis of one element; that is, \(I\) is principal.]

(ii) Every free module over an arbitrary integral domain with identity is torsion-free. The converse is false (Exercise II.1.10).

\(\star\)(iii) Let \(A\) be a cyclic \(R\)-module of order \(r \in R\).
(a) If \(s \in R\) is relatively prime to \(r\), then \(sA = A\) and \(A[s] = 0\).
(b) If \(s\) divides \(r\), say \(sk = r\), then \(sA \cong R/(k)\) and \(A[s] \cong R/(s)\).

(iv) If \(A\) is a finitely generated torsion module, then \(\{r \in R \mid rA = 0\}\) is a nonzero ideal in \(R\), say \((r_1)\). \(r_1\) is called the minimal annihilator of \(A\). Let \(A\) be a finite abelian group with minimal annihilator \(m \in \mathbf{Z}\). Show that a cyclic subgroup of \(A\) of order properly dividing \(m\) need not be a direct summand of \(A\).

Question

\(\star\)(i) (a) \([F : K] = 1\) if and only if \(F = K\).
(b) If \([F : K]\) is prime, then there are no intermediate fields between \(F\) and \(K\).
(c) lf \(u \in F\) has degree \(n\) over \(K\), then \(n\) divides \([F : K]\).

\(\star\)(ii) (a) Consider the extension \(\mathbf{Q}(u) of \mathbf{Q}\) generated by a real root \(u\) of \(x^3 - 6x^2+ 9x + 3\). (Why is this irreducible?) Express each of the following elements in terms of the basis \(\{1, u, u^2\}\): \(u^4\); \(u^5\); \(3u^5 - u^4 + 2\); \((u + 1)^{-1}\); \((u^2 - 6u + 8)^{-1}\).
(b) Do the same with respect to the basis \(\{1, u, u^2, u^3, u^4\}\) of \(\mathbf{Q}(u)\) where \(u\) is a real root of \(x^5 + 2x + 2\) and the elements in question are: \((u^2 + 2)(u^3 + 3u)\); \(u^{-1}\); \(u^4(u^4 + 3u^2 + 7u + 5)\); \((u + 2)(u^2 + 3)^{-1}\).

(iii) A complex number is said to be an algebraic number if it is algebraic over \(\mathbf{Q}\) and an algebraic integer if it is the root of a monic polynomial in \(\mathbf{Z}[x]\).
(a) If \(u\) is an algebraic number, there exists an integer \(n\) such that \(nu\) is an algebraic integer.
(b) If \(r \in \mathbf{Q}\) is an algebraic integer, then \(r \in \mathbf{Z}\).
(c) If \(u\) is an algebraic integer and \(n \in \mathbf{Z}\), then \(u + n\) and \(nu\) are algebraic integers.
(d) The sum and product of two algebraic integers are algebraic integers.

(iv) Let \(c,d\) be constructible real numbers.
(a) \(c + d\) and \(c - d\) are constructible.
(b) If \(d \neq 0\), then \(c/d\) is constructible. [Hint: If \((x, 0)\) is the intersection of the \(x\) axis and the straight line through \((0, 1)\) that is parallel the line through \((0, d)\) and \((c, 0)\), then \(x = c/d\).]
(c) \(cd\) is constructible [Hint: use (b)].
(d) The constructible real numbers form a subfield containing \(\mathbf{Q}\).
(e) If \(c > 0\), then \(\sqrt{c}\) is constructible. [Hint: If \(y\) is the length of the straight line segment perpendicular to the \(x\) axis that joins \((1,0)\) with the (upper half of the) circle with center \((\frac{c + 1}{2}, 0)\) and radius \(\frac{c + 1}{2}\) then \(y = \sqrt{c}\).]