Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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Theorem 5.13 (Mertens' theorem). We have\n\n\[ \mathop{\prod }\limits_{{p \leq x}}\left( {1 - 1/p}\right) = \Theta \left( {1/\log x}\right) \] | Proof. Using parts (i) and (iii) of §A1, for any fixed prime \( p \), we have\n\n\[ - \frac{1}{{p}^{2}} \leq \frac{1}{p} + \log \left( {1 - 1/p}\right) \leq 0. \]\n\n(5.9)\n\nMoreover, since\n\n\[ \mathop{\sum }\limits_{{p \leq x}}\frac{1}{{p}^{2}} \leq \mathop{\sum }\limits_{{i \geq 2}}\frac{1}{{i}^{2}} < \infty \]\n\... | No |
Theorem 5.14 (Prime number theorem). We have\n\n\[ \pi \left( x\right) \sim x/\log x. \] | Proof. Literature-see §5.6. | No |
Theorem 5.15. Let \( \kappa \left( x\right) \mathrel{\text{:=}} {\left( \log x\right) }^{3/5}{\left( \log \log x\right) }^{-1/5} \) . Then for some \( c > 0 \), we have\n\n\[ \pi \left( x\right) = \operatorname{li}\left( x\right) + O\left( {x{e}^{-{c\kappa }\left( x\right) }}\right) . \] | Proof. Literature-see §5.6. | No |
Theorem 5.17 (Euler’s identity). For every real number \( s > 1 \), we have\n\n\[ \zeta \left( s\right) = \mathop{\prod }\limits_{p}{\left( 1 - {p}^{-s}\right) }^{-1}, \]\n\nwhere the product is over all primes \( p \) . | Proof. The rigorous interpretation of the infinite product on the right-hand side of (5.13) is as a limit of finite products. Thus, if \( {p}_{i} \) denotes the \( i \) th prime, for \( i = 1,2,\ldots \), then we are really proving that\n\n\[ \zeta \left( s\right) = \mathop{\lim }\limits_{{r \rightarrow \infty }}\matho... | Yes |
Theorem 5.19. We have:\n\n(i) \( \frac{x}{\log x}\left( {1 + \frac{1}{2\log x}}\right) < \pi \left( x\right) < \frac{x}{\log x}\left( {1 + \frac{3}{2\log x}}\right) \), for \( x \geq {59} \) ;\n\n(ii) \( n\left( {\log n + \log \log n - 3/2}\right) < {p}_{n} < n\left( {\log n + \log \log n - 1/2}\right) \), for \( n \ge... | Proof. Literature-see §5.6. | No |
Theorem 5.20 (Dirichlet’s theorem). Let \( a, d \in \mathbb{Z} \) with \( d > 0 \) and \( \gcd \left( {a, d}\right) = 1 \). Then there are infinitely many primes \( p \equiv a\left( {\;\operatorname{mod}\;d}\right) \). | Proof. Literature-see §5.6. | No |
Theorem 5.21. Let \( a, d \in \mathbb{Z} \) with \( d > 0 \) and \( \gcd \left( {a, d}\right) = 1 \) . Then\n\n\[ \pi \left( {x;d, a}\right) \sim \frac{x}{\varphi \left( d\right) \log x}. \] | Proof. Literature-see §5.6. | No |
Theorem 5.23. There exists a constant \( c \) such that for all \( a, d \in \mathbb{Z} \) with \( d \geq 2 \) and \( \gcd \left( {a, d}\right) = 1 \), the least prime \( p \equiv a\left( {\;\operatorname{mod}\;d}\right) \) is at most \( c{d}^{{11}/2} \) . | Proof. Literature-see §5.6. | No |
Theorem 6.2. Let \( G \) be an abelian group with binary operation \( \star \) . Then we have:\n\n(i) \( G \) contains only one identity element;\n\n(ii) every element of \( G \) has only one inverse. | Proof. Suppose \( e,{e}^{\prime } \) are both identities. Then we have\n\n\[ e = e \star {e}^{\prime } = {e}^{\prime } \]\n\nwhere we have used part (ii) of Definition 6.1, once with \( {e}^{\prime } \) as the identity, and once with \( e \) as the identity. That proves part (i) of the theorem.\n\nTo prove part (ii) of... | Yes |
The set \( {\mathcal{F}}^{ * } \) of all arithmetic functions \( f \), such that \( f\left( 1\right) \neq 0 \), and with the Dirichlet product as the binary operation (see §2.9) forms an abelian group. | The special function \( I \) is the identity, and inverses are guaranteed by Exercise 2.54. | No |
The set of all finite bit strings under concatenation does not form an abelian group. | Although concatenation is associative and the empty string acts as an identity element, inverses do not exist (except for the empty string), nor is concatenation commutative. | Yes |
Theorem 6.3. Let \( G \) be an abelian group. Then for all \( a, b, c \in G \), we have:\n\n(i) if \( a + b = a + c \), then \( b = c \) ; | Proof. These statements all follow easily from Definition 6.1 and Theorem 6.2. For (i), just add \( - a \) to both sides of the equation \( a + b = a + c \) . | Yes |
Theorem 6.4. Let \( G \) be an abelian group. Then for all \( a, b \in G \) and \( k,\ell \in \mathbb{Z} \), we have:\n\n(i) \( k\left( {\ell a}\right) = \left( {k\ell }\right) a = \ell \left( {ka}\right) \) ;\n\n(ii) \( \left( {k + \ell }\right) a = {ka} + \ell a \) ;\n\n(iii) \( k\left( {a + b}\right) = {ka} + {kb} \... | Proof. The proof of this is easy, but tedious. We leave the details as an exercise to the reader. | No |
If \( {G}_{1},\ldots ,{G}_{k} \) are abelian groups, we can form the direct product \( H \mathrel{\text{:=}} {G}_{1} \times \cdots \times {G}_{k} \), which consists of all \( k \) -tuples \( \left( {{a}_{1},\ldots ,{a}_{k}}\right) \) with \( {a}_{1} \in {G}_{1} \) , \( \ldots ,{a}_{k} \in {G}_{k} \). We can view \( H \... | \[ \left( {{a}_{1},\ldots ,{a}_{k}}\right) + \left( {{b}_{1},\ldots ,{b}_{k}}\right) \mathrel{\text{:=}} \left( {{a}_{1} + {b}_{1},\ldots ,{a}_{k} + {b}_{k}}\right) . \] Of course, the groups \( {G}_{1},\ldots ,{G}_{k} \) may be different, and the group operation applied in the \( i \) th component corresponds to the g... | No |
Let \( G \) be an abelian group. An element \( \left( {{a}_{1},\ldots ,{a}_{k}}\right) \) of \( {G}^{\times k} \) may be identified with the function \( f : \{ 1,\ldots, k\} \rightarrow G \) given by \( f\left( i\right) = {a}_{i} \) for \( i = 1,\ldots, k \) . We can generalize this, replacing \( \{ 1,\ldots, k\} \) by... | \[ \left( {f + g}\right) \left( i\right) \mathrel{\text{:=}} f\left( i\right) + g\left( i\right) \text{ for all }i \in I. \] Again, we leave it to the reader to verify that \( \operatorname{Map}\left( {I, G}\right) \) is an abelian group, where the identity element is the function that maps each \( i \in I \) to \( {0}... | No |
Theorem 6.6. If \( G \) is an abelian group, and \( H \) is a subgroup of \( G \), then \( H \) contains \( {0}_{G} \) ; moreover, the binary operation of \( G \), when restricted to \( H \), yields a binary operation that makes \( H \) into an abelian group whose identity is \( {0}_{G} \). | Proof. First, to see that \( {0}_{G} \in H \), just pick any \( a \in H \), and using both properties of the definition of a subgroup, we see that \( {0}_{G} = a + \left( {-a}\right) \in H \). Next, note that by property (i) of Definition 6.5, \( H \) is closed under addition, which means that the restriction of the bi... | No |
Theorem 6.7. Let \( G \) be an abelian group, and let \( m \) be an integer. Then\n\n\[ \n{mG} \mathrel{\text{:=}} \{ {ma} : a \in G\} \n\]\n\nis a subgroup of \( G \) . | Proof. The set \( {mG} \) is non-empty, since \( {0}_{G} = m{0}_{G} \in {mG} \) . For \( {ma},{mb} \in {mG} \), we have \( {ma} + {mb} = m\left( {a + b}\right) \in {mG} \), and \( - \left( {ma}\right) = m\left( {-a}\right) \in {mG} \) . | Yes |
Theorem 6.8. Let \( G \) be an abelian group, and let \( m \) be an integer. Then\n\n\[ G\{ m\} \mathrel{\text{:=}} \left\{ {a \in G : {ma} = {0}_{G}}\right\} \]\n\nis a subgroup of \( G \) . | Proof. The set \( G\{ m\} \) is non-empty, since \( m{0}_{G} = {0}_{G} \), and so \( G\{ m\} \) contains \( {0}_{G} \) . If \( {ma} = {0}_{G} \) and \( {mb} = {0}_{G} \), then \( m\left( {a + b}\right) = {ma} + {mb} = {0}_{G} + {0}_{G} = {0}_{G} \) and \( m\left( {-a}\right) = - \left( {ma}\right) = - {0}_{G} = {0}_{G}... | Yes |
Let \( n \) be a positive integer, let \( m \in \mathbb{Z} \), and consider the subgroup \( m{\mathbb{Z}}_{n} \) of the additive group \( {\mathbb{Z}}_{n} \). Now, for every residue class \( \left\lbrack z\right\rbrack \in {\mathbb{Z}}_{n} \), we have \( m\left\lbrack z\right\rbrack = \left\lbrack {mz}\right\rbrack \).... | By part (i) of Theorem 2.5, such a \( z \) exists if and only if \( d \mid b \), where \( d \mathrel{\text{:=}} \gcd \left( {m, n}\right) \). Thus, \( m{\mathbb{Z}}_{n} \) consists precisely of the \( n/d \) distinct residue classes \[ \left\lbrack {i \cdot d}\right\rbrack \left( {i = 0,\ldots, n/d - 1}\right) ,\] and ... | Yes |
Theorem 6.9. If \( G \) is a subgroup of \( \mathbb{Z} \), then there exists a unique non-negative integer \( m \) such that \( G = m\mathbb{Z} \) . Moreover, for two non-negative integers \( {m}_{1} \) and \( {m}_{2} \) , we have \( {m}_{1}\mathbb{Z} \subseteq {m}_{2}\mathbb{Z} \) if and only if \( {m}_{2} \mid {m}_{1... | Proof. Actually, we have already proven this. One only needs to observe that a subset \( G \) of \( \mathbb{Z} \) is a subgroup if and only if it is an ideal of \( \mathbb{Z} \), as defined in \( §{1.2} \) (see Exercise 1.8). The first statement of the theorem then follows from Theorem 1.6. The second statement follows... | No |
Theorem 6.10. If \( G \) is a subgroup of \( {\mathbb{Z}}_{n} \), then there exists a unique positive integer \( d \) dividing \( n \) such that \( G = d{\mathbb{Z}}_{n} \) . Also, for all positive divisors \( {d}_{1},{d}_{2} \) of \( n \), we have \( {d}_{1}{\mathbb{Z}}_{n} \subseteq {d}_{2}{\mathbb{Z}}_{n} \) if and ... | Proof. Note that the second statement implies the uniqueness part of the first statement, so it suffices to prove just the existence part of the first statement and the second statement.\n\nLet \( G \) be an arbitrary subgroup of \( {\mathbb{Z}}_{n} \), and let \( H \mathrel{\text{:=}} \{ z \in \mathbb{Z} : \left\lbrac... | Yes |
Consider the group \( {\mathbb{Z}}_{15}^{ * } \). We can enumerate its elements as \[ \left\lbrack {\pm 1}\right\rbrack ,\left\lbrack {\pm 2}\right\rbrack ,\left\lbrack {\pm 4}\right\rbrack ,\left\lbrack {\pm 7}\right\rbrack \text{.} \] Therefore, the elements of \( {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{2} \) are \... | Going further, one sees that \( {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{4} = \{ \left\lbrack 1\right\rbrack \} \). Thus, \( {\alpha }^{4} = \left\lbrack 1\right\rbrack \) for all \( \alpha \in {\mathbb{Z}}_{15}^{ * } \). By direct calculation, one can determine that \( {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{3} = {... | Yes |
Consider the group \( {\mathbb{Z}}_{5}^{ * } = \{ \left\lbrack {\pm 1}\right\rbrack ,\left\lbrack {\pm 2}\right\rbrack \} \) . The elements of \( {\left( {\mathbb{Z}}_{5}^{ * }\right) }^{2} \) are | \[ {\left\lbrack 1\right\rbrack }^{2} = \left\lbrack 1\right\rbrack ,{\left\lbrack 2\right\rbrack }^{2} = \left\lbrack 4\right\rbrack = \left\lbrack {-1}\right\rbrack \] thus, \( {\left( {\mathbb{Z}}_{5}^{ * }\right) }^{2} = \{ \left\lbrack {\pm 1}\right\rbrack \} \) and has order 2 . | Yes |
Theorem 6.11. If \( {H}_{1} \) and \( {H}_{2} \) are subgroups of an abelian group \( G \), then so is\n\n\[ \n{H}_{1} + {H}_{2} \mathrel{\text{:=}} \left\{ {{a}_{1} + {a}_{2} : {a}_{1} \in {H}_{1},{a}_{2} \in {H}_{2}}\right\} .\n\] | Proof. It is evident that \( {H}_{1} + {H}_{2} \) is non-empty, as it contains \( {0}_{G} + {0}_{G} = {0}_{G} \) . Consider two elements in \( {H}_{1} + {H}_{2} \), which we can write as \( {a}_{1} + {a}_{2} \) and \( {b}_{1} + {b}_{2} \) , where \( {a}_{1},{b}_{1} \in {H}_{1} \) and \( {a}_{2},{b}_{2} \in {H}_{2} \) .... | Yes |
Theorem 6.12. If \( {H}_{1} \) and \( {H}_{2} \) are subgroups of an abelian group \( G \), then so is \( {H}_{1} \cap {H}_{2} \) | Proof. It is evident that \( {H}_{1} \cap {H}_{2} \) is non-empty, as both \( {H}_{1} \) and \( {H}_{2} \) contain \( {0}_{G} \) , and hence so does their intersection. If \( a \in {H}_{1} \cap {H}_{2} \) and \( b \in {H}_{1} \cap {H}_{2} \), then since \( a, b \in {H}_{1} \), we have \( a + b \in {H}_{1} \), and since... | Yes |
Theorem 6.13. Let \( G \) be an abelian group and \( H \) a subgroup of \( G \) . For all \( a, b, c \in G \), we have:\n\n(i) \( a \equiv a\left( {\;\operatorname{mod}\;H}\right) \) ;\n\n(ii) \( a \equiv b\left( {\;\operatorname{mod}\;H}\right) \) implies \( b \equiv a\left( {\;\operatorname{mod}\;H}\right) \) ;\n\n(i... | Proof. For (i), observe that \( H \) contains \( {0}_{G} = a - a \) . For (ii), observe that if \( H \) contains \( a - b \), then it also contains \( - \left( {a - b}\right) = b - a \) . For (iii), observe that if \( H \) contains \( a - b \) and \( b - c \), then it also contains \( \left( {a - b}\right) + \left( {b ... | Yes |
Theorem 6.14. Let \( G \) be an abelian group and \( H \) a subgroup of \( G \). For all \( a, b \in G \), the function\n\n\[ f : G \rightarrow G \]\n\n\[ x \mapsto b - a + x \]\n\n is a bijection, which, when restricted to the coset \( {\left\lbrack a\right\rbrack }_{H} \), yields a bijection from \( {\left\lbrack a\r... | Proof. First, we claim that \( f \) is a bijection. Indeed, if \( f\left( x\right) = f\left( {x}^{\prime }\right) \), then \( b - a + x = b - a + {x}^{\prime } \), and subtracting \( b \) and adding \( a \) to both sides of this equation yields \( x = {x}^{\prime } \). That proves that \( f \) is injective. To prove th... | Yes |
Theorem 6.15 (Lagrange’s theorem). If \( G \) is a finite abelian group, and \( H \) is a subgroup of \( G \), then the order of \( H \) divides the order of \( G \) . | Proof. This is an immediate consequence of the previous theorem, and the fact that the cosets of \( H \) in \( G \) partition \( G \) . | No |
Theorem 6.16. Suppose \( G \) is an abelian group and \( H \) is a subgroup of \( G \) . For all \( a,{a}^{\prime }, b,{b}^{\prime } \in G \), if \( a \equiv {a}^{\prime }\left( {\;\operatorname{mod}\;H}\right) \) and \( b \equiv {b}^{\prime }\left( {\;\operatorname{mod}\;H}\right) \), then we have \( a + b \equiv {a}^... | Proof. Now, \( a \equiv {a}^{\prime }\left( {\;\operatorname{mod}\;H}\right) \) and \( b \equiv {b}^{\prime }\left( {\;\operatorname{mod}\;H}\right) \) means that \( a = {a}^{\prime } + x \) and \( b = {b}^{\prime } + y \) for some \( x, y \in H \) . Therefore, \( a + b = \left( {{a}^{\prime } + x}\right) + \left( {{b}... | Yes |
Theorem 6.17. Suppose \( G \) is a finite abelian group and \( H \) is a subgroup of \( G \) . Then \( \left\lbrack {G : H}\right\rbrack = \left| G\right| /\left| H\right| \) . Moreover, if \( K \) is a subgroup of \( H \), then\n\n\[ \left\lbrack {G : K}\right\rbrack = \left\lbrack {G : H}\right\rbrack \left\lbrack {H... | Proof. The fact that \( \left\lbrack {G : H}\right\rbrack = \left| G\right| /\left| H\right| \) follows directly from Theorem 6.14. The fact that \( \left\lbrack {G : K}\right\rbrack = \left\lbrack {G : H}\right\rbrack \left\lbrack {H : K}\right\rbrack \) follows from a simple calculation:\n\n\[ \left\lbrack {G : H}\ri... | Yes |
Example 6.32. Continuing with Example 6.30, let \( G \mathrel{\text{:=}} {\mathbb{Z}}_{6} \) and \( H \mathrel{\text{:=}} {3G} = \) \( \{ \left\lbrack 0\right\rbrack ,\left\lbrack 3\right\rbrack \} \) . The quotient group \( G/H \) has order 3, and consists of the cosets\n\n\[ \alpha \mathrel{\text{:=}} \{ \left\lbrack... | If we write out an addition table for \( G \), grouping together elements in cosets of \( H \) in \( G \), then we also get an addition table for the quotient group \( G/H \) :\n\n<table><thead><tr><th>\( + \)</th><th>[0]</th><th>[3]</th><th>[1]</th><th>[4]</th><th>[2]</th><th>[5]</th></tr></thead><tr><td>[0]</td><td>[... | Yes |
Let us return to Example 6.26. The multiplicative group \( {\mathbb{Z}}_{15}^{ * } \), as we saw, is of order 8 . The subgroup \( {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{2} \) of \( {\mathbb{Z}}_{15}^{ * } \) has order 2 . Therefore, the quotient group \( {\mathbb{Z}}_{15}^{ * }/{\left( {\mathbb{Z}}_{15}^{ * }\right)... | Indeed, the cosets are\n\n\[ \n{\alpha }_{00} \mathrel{\text{:=}} {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{2} = \{ \left\lbrack 1\right\rbrack ,\left\lbrack 4\right\rbrack \} ,\;{\alpha }_{01} \mathrel{\text{:=}} \left\lbrack {-1}\right\rbrack {\left( {\mathbb{Z}}_{15}^{ * }\right) }^{2} = \{ \left\lbrack {-1}\right\r... | Yes |
Example 6.34. As we saw in Example 6.27, \( {\left( {\mathbb{Z}}_{5}^{ * }\right) }^{2} = \{ \left\lbrack {\pm 1}\right\rbrack \} \) . Therefore, the quotient group \( {\mathbb{Z}}_{5}^{ * }/{\left( {\mathbb{Z}}_{5}^{ * }\right) }^{2} \) has order 2 . The cosets of \( {\left( {\mathbb{Z}}_{5}^{ * }\right) }^{2} \) in \... | \[ \begin{matrix} \cdot & {\alpha }_{0} & {\alpha }_{1} \\ {\alpha }_{0} & {\alpha }_{0} & {\alpha }_{1} \\ {\alpha }_{1} & {\alpha }_{1} & {\alpha }_{0} \end{matrix} \] We see that the quotient group is essentially just a \ | Yes |
Suppose \( H \) is a subgroup of an abelian group \( G \) . We define the map\n\n\[ \rho : G \rightarrow G/H \]\n\n\[ a \mapsto {\left\lbrack a\right\rbrack }_{H} \] | It is not hard to see that this is a group homomorphism. Indeed, this follows almost immediately from the way we defined addition in the quotient group \( G/H \) :\n\n\[ \rho \left( {a + b}\right) = {\left\lbrack a + b\right\rbrack }_{H} = {\left\lbrack a\right\rbrack }_{H} + {\left\lbrack b\right\rbrack }_{H} = \rho \... | Yes |
Suppose \( G \) is an abelian group and \( m \) is an integer. The map \[ \rho : \;G \rightarrow G \] \[ a \mapsto {ma} \] is a group homomorphism. | \[ \rho \left( {a + b}\right) = m\left( {a + b}\right) = {ma} + {mb} = \rho \left( a\right) + \rho \left( b\right) .\] | Yes |
Suppose \( G \) is an abelian group and \( a \) is an element of \( G \) . It is easy to see that the map\n\n\[ \rho : \;\mathbb{Z} \rightarrow G \]\n\n\[ z \mapsto {za} \]\nis a group homomorphism, since | \[\rho \left( {z + {z}^{\prime }}\right) = \left( {z + {z}^{\prime }}\right) a = {za} + {z}^{\prime }a = \rho \left( z\right) + \rho \left( {z}^{\prime }\right) .\] | Yes |
Theorem 6.19. Let \( \rho \) be a group homomorphism from \( G \) to \( {G}^{\prime } \). Then:\n\n(i) \( \rho \left( {0}_{G}\right) = {0}_{{G}^{\prime }} \);\n\n(ii) \( \rho \left( {-a}\right) = - \rho \left( a\right) \) for all \( a \in G \);\n\n(iii) \( \rho \left( {na}\right) = {n\rho }\left( a\right) \) for all \(... | Proof. These are all straightforward calculations.\n\n(i) We have\n\n\[ \n{0}_{{G}^{\prime }} + \rho \left( {0}_{G}\right) = \rho \left( {0}_{G}\right) = \rho \left( {{0}_{G} + {0}_{G}}\right) = \rho \left( {0}_{G}\right) + \rho \left( {0}_{G}\right) .\n\] \n\nNow cancel \( \rho \left( {0}_{G}\right) \) from both sides... | Yes |
Theorem 6.20. If \( \rho : G \rightarrow {G}^{\prime } \) and \( {\rho }^{\prime } : {G}^{\prime } \rightarrow {G}^{\prime \prime } \) are group homomorphisms, then so is their composition \( {\rho }^{\prime } \circ \rho : G \rightarrow {G}^{\prime \prime } \) . | Proof. For all \( a, b \in G \), we have\n\n\[ \n{\rho }^{\prime }\left( {\rho \left( {a + b}\right) }\right) = {\rho }^{\prime }\left( {\rho \left( a\right) + \rho \left( b\right) }\right) = {\rho }^{\prime }\left( {\rho \left( a\right) }\right) + {\rho }^{\prime }\left( {\rho \left( b\right) }\right) .\n\] | Yes |
Theorem 6.21. Let \( {\rho }_{i} : G \rightarrow {G}_{i}^{\prime } \), for \( i = 1,\ldots, k \), be group homomorphisms. Then the map\n\n\[ \rho : \;G \rightarrow {G}_{1}^{\prime } \times \cdots \times {G}_{k}^{\prime } \]\n\n\[ a \mapsto \left( {{\rho }_{1}\left( a\right) ,\ldots ,{\rho }_{k}\left( a\right) }\right) ... | Proof. For all \( a, b \in G \), we have\n\n\[ \rho \left( {a + b}\right) = \left( {{\rho }_{1}\left( {a + b}\right) ,\ldots ,{\rho }_{k}\left( {a + b}\right) }\right) = \left( {{\rho }_{1}\left( a\right) + {\rho }_{1}\left( b\right) ,\ldots ,{\rho }_{k}\left( a\right) + {\rho }_{k}\left( b\right) }\right) \]\n\n\[ = \... | Yes |
Theorem 6.22. If \( \rho \) is a group isomorphism of \( G \) with \( {G}^{\prime } \), then the inverse function \( {\rho }^{-1} \) is a group isomorphism of \( {G}^{\prime } \) with \( G \) . | Proof. For all \( {a}^{\prime },{b}^{\prime } \in {G}^{\prime } \), we have\n\n\[ \rho \left( {{\rho }^{-1}\left( {a}^{\prime }\right) + {\rho }^{-1}\left( {b}^{\prime }\right) }\right) = \rho \left( {{\rho }^{-1}\left( {a}^{\prime }\right) }\right) + \rho \left( {{\rho }^{-1}\left( {b}^{\prime }\right) }\right) = {a}^... | Yes |
Theorem 6.23 (First isomorphism theorem). Let \( \rho : G \rightarrow {G}^{\prime } \) be a group homomorphism with kernel \( K \) and image \( {H}^{\prime } \). Then we have a group isomorphism\n\n\[ G/K \cong {H}^{\prime } \]\n\nSpecifically, the map\n\n\[ \bar{\rho } : \;G/K \rightarrow {G}^{\prime } \]\n\n\[ {\left... | Proof. Using part (vi) of Theorem 6.19, we see that for all \( a, b \in G \), we have\n\n\[ {\left\lbrack a\right\rbrack }_{K} = {\left\lbrack b\right\rbrack }_{K} \Leftrightarrow a \equiv b\left( {\;\operatorname{mod}\;K}\right) \Leftrightarrow \rho \left( a\right) = \rho \left( b\right) .\n\nThis immediately implies ... | Yes |
Theorem 6.24. Let \( \rho : G \rightarrow {G}^{\prime } \) be a group homomorphism. Then for every subgroup \( H \) of \( G \) with \( H \subseteq \operatorname{Ker}\rho \), we may define a group homomorphism\n\n\[ \bar{\rho } : \;G/H \rightarrow {G}^{\prime } \]\n\n\[ {\left\lbrack a\right\rbrack }_{H} \mapsto \rho \l... | Proof. Using the assumption that \( H \subseteq \operatorname{Ker}\rho \), we see that \( \bar{\rho } \) is unambiguously defined, since for all \( a, b \in G \), we have\n\n\[ {\left\lbrack a\right\rbrack }_{H} = {\left\lbrack b\right\rbrack }_{H} \Rightarrow a \equiv b\left( {\;\operatorname{mod}\;H}\right) \Rightarr... | Yes |
Theorem 6.25 (Internal direct product). Let \( G \) be an abelian group with subgroups \( {H}_{1},{H}_{2} \), where \( {H}_{1} \cap {H}_{2} = \left\{ {0}_{G}\right\} \) . Then we have a group isomorphism\n\n\[ \n{H}_{1} \times {H}_{2} \cong {H}_{1} + {H}_{2} \n\]\n\ngiven by the map\n\n\[ \n\rho : \;{H}_{1} \times {H}_... | Proof. We already saw that \( \rho \) is a surjective group homomorphism in Example 6.45. To see that \( \rho \) is injective, it suffices to show that \( \operatorname{Ker}\rho \) is trivial; that is, it suffices to show that for all \( {a}_{1} \in {H}_{1} \) and \( {a}_{2} \in {H}_{2} \), if \( {a}_{1} + {a}_{2} = {0... | Yes |
We can use the general theory developed so far to get a quick-and-dirty proof of the Chinese remainder theorem (Theorem 2.6). Let \( {\left\{ {n}_{i}\right\} }_{i = 1}^{k} \) be a pairwise relatively prime family of positive integers, and let \( n \mathrel{\text{:=}} \mathop{\prod }\limits_{{i = 1}}^{k}{n}_{i} \) . Con... | It is easy to see that this map is a group homomorphism; indeed, it is the map constructed in Theorem 6.21 applied with the natural maps \( {\rho }_{i} : \mathbb{Z} \rightarrow {\mathbb{Z}}_{{n}_{i}} \), for \( i = 1,\ldots, k \) . Evidently, \( a \in \operatorname{Ker}\rho \) if and only if \( {n}_{i} \mid a \) for \(... | Yes |
Let \( {n}_{1},{n}_{2} \) be positive integers with \( {n}_{1} \mid {n}_{2} \) . Consider the natural map \( \rho : \mathbb{Z} \rightarrow {\mathbb{Z}}_{{n}_{1}} \) . This is a surjective group homomorphism with \( \operatorname{Ker}\rho = {n}_{1}\mathbb{Z} \) . Since \( H \mathrel{\text{:=}} {n}_{2}\mathbb{Z} \subsete... | \[ \bar{\rho } : \;{\mathbb{Z}}_{{n}_{2}} \rightarrow {\mathbb{Z}}_{{n}_{1}} \] \[ {\left\lbrack a\right\rbrack }_{{n}_{2}} \mapsto {\left\lbrack a\right\rbrack }_{{n}_{1}}. \] | Yes |
Let \( n \) be a positive integer, and let \( m \) be any integer. Let \( {\rho }_{1} : \mathbb{Z} \rightarrow {\mathbb{Z}}_{n} \) be the natural map, and let \( {\rho }_{2} : {\mathbb{Z}}_{n} \rightarrow {\mathbb{Z}}_{n} \) be the \( m \) -multiplication map. The composed map \( \rho \mathrel{\text{:=}} {\rho }_{2} \c... | Theorem 6.23 therefore implies that the map\n\n\[ \bar{\rho } : \;{\mathbb{Z}}_{n/d} \rightarrow m{\mathbb{Z}}_{n} \]\n\n\[ {\left\lbrack z\right\rbrack }_{n/d} \mapsto m{\left\lbrack z\right\rbrack }_{n} \] \n\nis a group isomorphism. | Yes |
Consider the group \( {\mathbb{Z}}_{p}^{ * } \) where \( p \) is an odd prime, and let \( \rho : {\mathbb{Z}}_{p}^{ * } \rightarrow {\mathbb{Z}}_{p}^{ * } \) be the squaring map. By definition, \( \operatorname{Im}\rho = {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} \), and we proved in Theorem 2.18 that \( \operatorname... | Let \( H \mathrel{\text{:=}} {\left( {\mathbb{Z}}_{p}^{ * }\right) }^{2} \), and consider the quotient group \( {\mathbb{Z}}_{p}^{ * }/H \) . Since \( \left| H\right| = \left( {p - 1}\right) /2 \) , we know that \( \left| {{\mathbb{Z}}_{p}^{ * }/H}\right| = \left| {\mathbb{Z}}_{p}^{ * }\right| /\left| H\right| = 2 \), ... | Yes |
Theorem 6.26. Let \( G \) and \( {G}^{\prime } \) be abelian groups, and consider the group of functions \( \operatorname{Map}\left( {G,{G}^{\prime }}\right) \) . Then\n\n\[ \operatorname{Hom}\left( {G,{G}^{\prime }}\right) \mathrel{\text{:=}} \left\{ {\sigma \in \operatorname{Map}\left( {G,{G}^{\prime }}\right) : \sig... | Proof. First, observe that \( \operatorname{Hom}\left( {G,{G}^{\prime }}\right) \) is non-empty, as it contains the map that sends everything in \( G \) to \( {0}_{{G}^{\prime }} \) (this is the identity element of \( \operatorname{Map}\left( {G,{G}^{\prime }}\right) \) ).\n\nNext, we have to show that if \( \sigma \) ... | Yes |
Consider the additive group \( \mathbb{Z} \). This is a cyclic group, with 1 being a generator: | \[ \langle 1\rangle = \{ z \cdot 1 : z \in \mathbb{Z}\} = \{ z : z \in \mathbb{Z}\} = \mathbb{Z}. \] For every \( m \in \mathbb{Z} \), we have \[ \langle m\rangle = \{ {zm} : z \in \mathbb{Z}\} = \{ {mz} : z \in \mathbb{Z}\} = m\mathbb{Z}. \] It follows that the only elements of \( \mathbb{Z} \) that generate \( \mathb... | Yes |
For \( n > 0 \), consider the additive group \( {\mathbb{Z}}_{n} \). This is a cyclic group, with [1] being a generator: | \[ \langle \left\lbrack 1\right\rbrack \rangle = \{ z\left\lbrack 1\right\rbrack : z \in \mathbb{Z}\} = \{ \left\lbrack z\right\rbrack : z \in \mathbb{Z}\} = {\mathbb{Z}}_{n}. \] For every \( m \in \mathbb{Z} \), we have \[ \langle \left\lbrack m\right\rbrack \rangle = \{ z\left\lbrack m\right\rbrack : z \in \mathbb{Z}... | Yes |
Theorem 6.27. Let \( G \) be a cyclic group generated by \( a \) . Then for every \( m \in \mathbb{Z} \) , we have\n\n\[ \langle {ma}\rangle = {mG}\text{.} \] | Proof. We have\n\n\[ \langle {ma}\rangle = \{ z\left( {ma}\right) : z \in \mathbb{Z}\} = \{ m\left( {za}\right) : z \in \mathbb{Z}\} = m\langle a\rangle = {mG}. \] | Yes |
Consider the additive group \( G \mathrel{\text{:=}} \mathbb{Z} \times \mathbb{Z} \). Set\n\n\[ {\alpha }_{1} \mathrel{\text{:=}} \left( {1,0}\right) \in G\text{ and }{\alpha }_{2} \mathrel{\text{:=}} \left( {0,1}\right) \in G. \]\n\nIt is not hard to see that \( G = \left\langle {{\alpha }_{1},{\alpha }_{2}}\right\ran... | To see this, let \( \beta = \left( {{b}_{1},{b}_{2}}\right) \) be an arbitrary element of \( G \). We claim that one of \( {\alpha }_{1} \) or \( {\alpha }_{2} \) does not belong to \( \langle \beta \rangle \). Suppose to the contrary that both \( {\alpha }_{1} \) and \( {\alpha }_{2} \) belong to \( \langle \beta \ran... | Yes |
Consider the additive group \( G \mathrel{\text{:=}} {\mathbb{Z}}_{{n}_{1}} \times {\mathbb{Z}}_{{n}_{2}} \). Set \[ {\alpha }_{1} \mathrel{\text{:=}} \left( {{\left\lbrack 1\right\rbrack }_{{n}_{1}},{\left\lbrack 0\right\rbrack }_{{n}_{2}}}\right) \in G\text{ and }{\alpha }_{2} \mathrel{\text{:=}} \left( {{\left\lbrac... | If \( d = 1 \), then \( G \) is cyclic, with \( \alpha \mathrel{\text{:=}} \left( {{\left\lbrack 1\right\rbrack }_{{n}_{1}},{\left\lbrack 1\right\rbrack }_{{n}_{2}}}\right) \) being a generator. One can see this easily using the Chinese remainder theorem: for all \( {z}_{1},{z}_{2} \in \mathbb{Z} \), there exists \( z ... | Yes |
Theorem 6.28. Let \( \rho : G \rightarrow {G}^{\prime } \) be a group isomorphism.\n\n(i) For all \( a \in G \), we have \( \rho \left( {\langle a\rangle }\right) = \langle \rho \left( a\right) \rangle \) . | Proof. For all \( a \in G \), we have\n\n\[ \rho \left( {\langle a\rangle }\right) = \{ \rho \left( {za}\right) : z \in \mathbb{Z}\} = \{ {z\rho }\left( a\right) : z \in \mathbb{Z}\} = \langle \rho \left( a\right) \rangle . \]\n\nThat proves (i). | Yes |
Consider again the additive group \( G \mathrel{\text{:=}} {\mathbb{Z}}_{{n}_{1}} \times {\mathbb{Z}}_{{n}_{2}} \), discussed in Example 6.56. If \( \gcd \left( {{n}_{1},{n}_{2}}\right) = 1 \), then one can also see that \( G \) is cyclic as follows: | by the discussion in Example 6.48, we know that \( G \) is isomorphic to \( {\mathbb{Z}}_{{n}_{1}{n}_{2}} \), and since \( {\mathbb{Z}}_{{n}_{1}{n}_{2}} \) is cyclic, so is \( \mathbf{G} \). | Yes |
Consider again the subgroup \( m{\mathbb{Z}}_{n} \) of \( {\mathbb{Z}}_{n} \), discussed in Example 6.54. One can also see that this is cyclic of order \( n/d \), where \( d \mathrel{\text{:=}} \gcd \left( {m, n}\right) \). | in Example 6.50, we constructed an isomorphism between \( {\mathbb{Z}}_{n/d} \) and \( m{\mathbb{Z}}_{n} \) , and this implies \( m{\mathbb{Z}}_{n} \) is cyclic of order \( n/d \) . | Yes |
Theorem 6.30. Let \( G \) be a finite abelian group and let \( a \in G \) . Then \( \left| G\right| a = {0}_{G} \) and the order of a divides \( \left| G\right| \) . | Proof. Since \( \langle a\rangle \) is a subgroup of \( G \), by Lagrange’s theorem (Theorem 6.15), the order of \( a \) divides \( \left| G\right| \) . It then follows by Theorem 6.29 that \( \left| G\right| a = {0}_{G} \) . | Yes |
The group \( {\mathbb{Z}}_{5}^{ * } \) is cyclic, with [2] being a generator: | \[ {\left\lbrack 2\right\rbrack }^{2} = \left\lbrack 4\right\rbrack = \left\lbrack {-1}\right\rbrack ,\;{\left\lbrack 2\right\rbrack }^{3} = \left\lbrack {-2}\right\rbrack ,\;{\left\lbrack 2\right\rbrack }^{4} = \left\lbrack 1\right\rbrack . \] | Yes |
Consider again the additive group \( G \mathrel{\text{:=}} {\mathbb{Z}}_{{n}_{1}} \times {\mathbb{Z}}_{{n}_{2}} \), discussed in Example 6.56. If \( d \mathrel{\text{:=}} \gcd \left( {{n}_{1},{n}_{2}}\right) > 1 \), then one can also see that \( G \) is not cyclic as follows: | for every \( \beta \in G \), we have \( \left( {{n}_{1}{n}_{2}/d}\right) \beta = {0}_{G} \), and hence by Theorem 6.29, the order of \( \beta \) divides \( {n}_{1}{n}_{2}/d \) . | Yes |
Theorem 6.31. Let \( G \) be a cyclic group of infinite order.\n\n(i) \( G \) is isomorphic to \( \mathbb{Z} \) .\n\n(ii) There is a one-to-one correspondence between the non-negative integers and the subgroups of \( G \), where each such integer \( m \) corresponds to the cyclic group \( {mG} \) .\n\n(iii) For every t... | Proof. That \( G \cong \mathbb{Z} \) was established in our classification of cyclic groups, and so it suffices to prove the other statements of the theorem for \( G = \mathbb{Z} \) . As we saw in Example 6.53, for every integer \( m \), the subgroup \( m\mathbb{Z} \) is cyclic, as it is generated by \( m \) . This fac... | No |
Theorem 6.32. Let \( G \) be a cyclic group of finite order \( n \) . (i) \( G \) is isomorphic to \( {\mathbb{Z}}_{n} \) . | Proof. That \( G \cong {\mathbb{Z}}_{n} \) was established in our classification of cyclic groups, and so it suffices to prove the other statements of the theorem for \( G = {\mathbb{Z}}_{n} \) . | No |
Theorem 6.33. If \( G \) is an abelian group of prime order, then \( G \) is cyclic. | Proof. Let \( \left| G\right| = p \), which, by hypothesis, is prime. Let \( a \in G \) with \( a \neq {0}_{G} \), and let \( k \) be the order of \( a \) . As the order of an element divides the order of the group, we have \( k \mid p \), and so \( k = 1 \) or \( k = p \) . Since \( a \neq {0}_{G} \), we must have \( ... | Yes |
Theorem 6.34. If \( {G}_{1} \) and \( {G}_{2} \) are finite cyclic groups of relatively prime order, then \( {G}_{1} \times {G}_{2} \) is also cyclic. In particular, if \( {G}_{1} \) is generated by \( {a}_{1} \) and \( {G}_{2} \) is generated by \( {a}_{2} \), then \( {G}_{1} \times {G}_{2} \) is generated by \( \left... | Proof. We give a direct proof, based on Theorem 6.29. Let \( {n}_{1} \mathrel{\text{:=}} \left| {G}_{1}\right| \) and \( {n}_{2} \mathrel{\text{:=}} \left| {G}_{2}\right| \), where \( \gcd \left( {{n}_{1},{n}_{2}}\right) = 1 \) . Also, let \( {a}_{1} \in {G}_{1} \) have order \( {n}_{1} \) and \( {a}_{2} \in {G}_{2} \)... | Yes |
Theorem 6.35. Let \( G \) be a cyclic group. Then for every subgroup \( H \) of \( G \), both \( H \) and \( G/H \) are cyclic. | Proof. The fact that \( H \) is cyclic follows from part (ii) of Theorem 6.31 in the case where \( G \) is infinite, and part (ii) of Theorem 6.32 in the case where \( G \) is finite. If \( G \) is generated by \( a \), then it is easy to see that \( G/H \) is generated by \( {\left\lbrack a\right\rbrack }_{H} \) . | No |
Theorem 6.36. Let \( G \) be an abelian group, let \( a \in G \) be of finite order \( n \), and let \( m \) be an arbitrary integer. Then the order of \( {ma} \) is \( n/\gcd \left( {m, n}\right) \) . | Proof. Let \( H \mathrel{\text{:=}} \langle a\rangle \), and \( d \mathrel{\text{:=}} \gcd \left( {m, n}\right) \) . By Theorem 6.27, we have \( \langle {ma}\rangle = {mH} \) , and by Theorem 6.32, we have \( {mH} = {dH} \), which has order \( n/d \) .\n\nThat proves the theorem. Alternatively, we can give a direct pro... | Yes |
Theorem 6.37. Suppose that \( a \) is an element of an abelian group, and for some prime \( p \) and integer \( e \geq 1 \), we have \( {p}^{e}a = {0}_{G} \) and \( {p}^{e - 1}a \neq {0}_{G} \) . Then \( a \) has order \( {p}^{e} \) . | Proof. If \( m \) is the order of \( a \), then since \( {p}^{e}a = {0}_{G} \), we have \( m \mid {p}^{e} \) . So \( m = {p}^{f} \) for some \( f = 0,\ldots, e \) . If \( f < e \), then \( {p}^{e - 1}a = {0}_{G} \), contradicting the assumption that \( {p}^{e - 1}a \neq {0}_{G} \). | Yes |
Theorem 6.38. Suppose \( G \) is an abelian group with \( {a}_{1},{a}_{2} \in G \) such that \( {a}_{1} \) is of finite order \( {n}_{1},{a}_{2} \) is of finite order \( {n}_{2} \), and \( \gcd \left( {{n}_{1},{n}_{2}}\right) = 1 \) . Then the order of \( {a}_{1} + {a}_{2} \) is \( {n}_{1}{n}_{2} \) . | Proof. Let \( {H}_{1} \mathrel{\text{:=}} \left\langle {a}_{1}\right\rangle \) and \( {H}_{2} \mathrel{\text{:=}} \left\langle {a}_{2}\right\rangle \) so that \( \left| {H}_{1}\right| = {n}_{1} \) and \( \left| {H}_{2}\right| = {n}_{2} \). First, we claim that \( {H}_{1} \cap {H}_{2} = \left\{ {0}_{G}\right\} \) . To s... | Yes |
Theorem 6.39. Let \( G \) be an abelian group of exponent \( m \). (i) For every integer \( k, k \) kills \( G \) if and only if \( m \mid k \). | Proof. Exercise. | No |
Theorem 6.40. If \( {G}_{1} \) and \( {G}_{2} \) are abelian groups of exponents \( {m}_{1} \) and \( {m}_{2} \), then the exponent of \( {G}_{1} \times {G}_{2} \) is \( \operatorname{lcm}\left( {{m}_{1},{m}_{2}}\right) \) . | Proof. Exercise. | No |
If an abelian group \( G \) has non-zero exponent \( m \), then \( G \) contains an element of order \( m \). | The second statement follows immediately from the first. For the first statement, let \( m = \mathop{\prod }\limits_{{i = 1}}^{r}{p}_{i}^{{e}_{i}} \) be the prime factorization of \( m \) .\n\nFirst, we claim that for each \( i = 1,\ldots, r \), there exists \( {a}_{i} \in G \) such that \( \left( {m/{p}_{i}}\right) {a... | Yes |
Theorem 6.42. Let \( G \) be a finite abelian group of order \( n \) . If \( p \) is a prime dividing \( n \), then \( G \) contains an element of order \( p \) . | Proof. We can prove this by induction on \( n \) .\n\nIf \( n = 1 \), then the theorem is vacuously true.\n\nNow assume \( n > 1 \) and that the theorem holds for all groups of order strictly less than \( n \) . Let \( a \) be any non-zero element of \( G \), and let \( m \) be the order of \( a \) . Since \( a \) is n... | Yes |
Theorem 6.43. Let \( G \) be a finite abelian group. Then the primes dividing the exponent of \( G \) are the same as the primes dividing its order. | Proof. Since the exponent divides the order, every prime dividing the exponent must divide the order. Conversely, if a prime \( p \) divides the order, then since there is an element of order \( p \) in the group, the exponent must be divisible by \( p \) . | Yes |
Theorem 6.44 (Fundamental theorem of finite abelian groups). A finite abelian group (with more than one element) is isomorphic to a direct product of cyclic groups | \[ {\mathbb{Z}}_{{p}_{1}^{{e}_{1}}} \times \cdots \times {\mathbb{Z}}_{{p}_{r}^{{e}_{r}}} \] where the \( {p}_{i} \) ’s are primes (not necessarily distinct) and the \( {e}_{i} \) ’s are positive integers. This direct product of cyclic groups is unique up to the order of the factors. | Yes |
Lemma 6.47. Suppose that \( G \mathrel{\text{:=}} {\mathbb{Z}}_{{m}_{1}} \times \cdots \times {\mathbb{Z}}_{{m}_{t}} \) and \( H \mathrel{\text{:=}} {\mathbb{Z}}_{{n}_{1}} \times \cdots \times {\mathbb{Z}}_{{n}_{t}} \) are isomorphic, where the \( {m}_{i} \) ’s and \( {n}_{i} \) ’s are positive integers (possibly 1) su... | Proof. Clearly, \( \mathop{\prod }\limits_{i}{m}_{i} = \left| G\right| = \left| H\right| = \mathop{\prod }\limits_{i}{n}_{i} \) . We prove the lemma by induction on the order of the group. If the group order is 1, then clearly all the \( {m}_{i} \) ’s and \( {n}_{i} \) ’s must be 1, and we are done. Otherwise, let \( p... | Yes |
Theorem 7.2. Let \( R \) be a ring. Then:\n\n(i) the multiplicative identity \( {1}_{R} \) is unique; | Proof. Part (i) may be proved using the same argument as was used to prove part (i) of Theorem 6.2. | No |
The set \( \mathbb{C} \) of complex numbers under the usual rules of multiplication and addition forms a ring. | Every \( \alpha \in \mathbb{C} \) can be written (uniquely) as \( \alpha = a + {bi} \) , where \( a, b \in \mathbb{R} \) and \( i = \sqrt{-1} \) . If \( {\alpha }^{\prime } = {a}^{\prime } + {b}^{\prime }i \) is another complex number, with \( {a}^{\prime },{b}^{\prime } \in \mathbb{R} \), then\n\n\[ \alpha + {\alpha }... | No |
Consider the set \( \mathcal{F} \) of all arithmetic functions, that is, functions mapping positive integers to reals. Let us define addition of arithmetic functions point-wise (i.e., \( \left( {f + g}\right) \left( n\right) = f\left( n\right) + g\left( n\right) \) for all positive integers \( n \) ) and multiplication... | The reader should verify that with addition and multiplication so defined, \( \mathcal{F} \) forms a ring, where the all-zero function is the additive identity, and the special function \( I \) defined in \( §{2.9} \) is the multiplicative identity. | No |
Generalizing Example 6.19, if \( I \) is an arbitrary set and \( R \) is a ring, then \( \operatorname{Map}\left( {I, R}\right) \), which is the set of all functions \( f : I \rightarrow R \), may be naturally viewed as a ring, with addition and multiplication defined point-wise: for \( f, g \in \operatorname{Map}\left... | We leave it to the reader to verify that \( \operatorname{Map}\left( {I, R}\right) \) is indeed a ring, where the additive identity is the all-zero function, and the multiplicative identity is the all-one function. | No |
For non-zero \( \alpha = a + {bi} \in \mathbb{C} \), with \( a, b \in \mathbb{R} \), we have \( c \mathrel{\text{:=}} N\left( \alpha \right) = \) \( {a}^{2} + {b}^{2} > 0 \). It follows that the complex number \( \bar{\alpha }{c}^{-1} = \left( {a{c}^{-1}}\right) + \left( {-b{c}^{-1}}\right) i \) is the multiplicative i... | since \( \alpha \cdot \bar{\alpha }{c}^{-1} = \left( {\alpha \bar{\alpha }}\right) {c}^{-1} = 1 \) . Hence, every non-zero element of \( \mathbb{C} \) is a unit, and so \( \mathbb{C} \) is a field. | Yes |
For rings \( {R}_{1},\ldots ,{R}_{k} \), it is easy to see that the multiplicative group of units of the direct product \( {R}_{1} \times \cdots \times {R}_{k} \) is equal to \( {R}_{1}^{ * } \times \cdots \times {R}_{k}^{ * } \). | Indeed, by definition, \( \left( {{a}_{1},\ldots ,{a}_{k}}\right) \) has a multiplicative inverse if and only if each individual \( {a}_{i} \) does. | Yes |
Consider the ring \( \mathcal{F} \) of arithmetic functions defined in Example 7.6. | By the result of Exercise 2.54, \( {\mathcal{F}}^{ * } = \{ f \in \mathcal{F} : f\left( 1\right) \neq 0\} \) | No |
For \( n > 1,{\mathbb{Z}}_{n} \) is an integral domain if and only if \( n \) is prime. | In particular, if \( n \) is composite, so \( n = {ab} \) with \( 1 < a < n \) and \( 1 < b < n \), then \( {\left\lbrack a\right\rbrack }_{n} \) and \( {\left\lbrack b\right\rbrack }_{n} \) are zero divisors: \( {\left\lbrack a\right\rbrack }_{n}{\left\lbrack b\right\rbrack }_{n} = {\left\lbrack 0\right\rbrack }_{n} \... | No |
Theorem 7.3. If \( R \) is a ring, and \( a, b, c \in R \) such that \( a \neq 0 \) and \( a \) is not a zero divisor, then \( {ab} = {ac} \) implies \( b = c \) . | Proof. \( {ab} = {bc} \) implies \( a\left( {b - c}\right) = 0 \) . The fact that \( a \neq 0 \) and \( a \) is not a zero divisor implies that we must have \( b - c = 0 \), and so \( b = c \) . | Yes |
Theorem 7.4. Let \( R \) be a ring.\n\n(i) Suppose \( a, b \in R \), and that either \( a \) or \( b \) is not a zero divisor. Then \( a \mid b \) and \( b \mid a \) if and only if \( {ar} = b \) for some \( r \in {R}^{ * } \). | Proof. For the first statement, if \( {ar} = b \) for some \( r \in {R}^{ * } \), then we also have \( b{r}^{-1} = a \) ; thus, \( a\left| {b\text{and}b}\right| a \) . For the converse, suppose that \( a\left| {b\text{and}b}\right| a \) . We may assume that \( b \) is not a zero divisor (otherwise, exchange the roles o... | Yes |
Theorem 7.5. The characteristic of an integral domain is either zero or a prime. | Proof. By way of contradiction, suppose that \( D \) is an integral domain with characteristic \( m \) that is neither zero nor prime. Since, by definition, \( D \) is not a trivial ring, we cannot have \( m = 1 \), and so \( m \) must be composite. Say \( m = {st} \), where \( 1 < s < m \) and \( 1 < t < m \) . Since ... | Yes |
Theorem 7.6. Every finite integral domain is a field. | Proof. Let \( D \) be a finite integral domain, and let \( a \) be any non-zero element of \( D \) . Consider the \( a \) -multiplication map that sends \( b \in D \) to \( {ab} \), which is a group homomorphism on the additive group of \( D \) . Since \( a \) is not a zero-divisor, it follows that the kernel of the \(... | Yes |
Theorem 7.7. Every finite field \( F \) must be of cardinality \( {p}^{w} \), where \( p \) is prime, \( w \) is a positive integer, and \( p \) is the characteristic of \( F \) . | Proof. By Theorem 7.5, the characteristic of \( F \) is either zero or a prime, and since \( F \) is finite, it must be prime. Let \( p \) denote the characteristic. By definition, \( p \) is the exponent of the additive group of \( F \), and by Theorem 6.43, the primes dividing the exponent are the same as the primes ... | Yes |
Example 7.24. \( \mathbb{R} \) is a subring of \( \mathbb{C} \) . | Note that for all \( \alpha \mathrel{\text{:=}} a + {bi} \in \mathbb{C} \), with \( a, b \in \mathbb{R} \) , we have \( \bar{\alpha } = \alpha \Leftrightarrow a + {bi} = a - {bi} \Leftrightarrow b = 0 \) . That is, \( \bar{\alpha } = \alpha \Leftrightarrow \alpha \in \mathbb{R} \) . | No |
Let us determine the units of \( \mathbb{Z}\left\lbrack i\right\rbrack \) . Suppose \( \alpha \in \mathbb{Z}\left\lbrack i\right\rbrack \) is a unit, so that there exists \( {\alpha }^{\prime } \in \mathbb{Z}\left\lbrack i\right\rbrack \) such that \( \alpha {\alpha }^{\prime } = 1 \) . | Taking norms, we obtain\n\n\[ 1 = N\left( 1\right) = N\left( {\alpha {\alpha }^{\prime }}\right) = N\left( \alpha \right) N\left( {\alpha }^{\prime }\right) . \]\n\nSince the norm of any Gaussian integer is itself a non-negative integer, and since \( N\left( \alpha \right) N\left( {\alpha }^{\prime }\right) = 1 \), we ... | Yes |
Let \( m \) be a positive integer, and let \( {\mathbb{Q}}^{\left( m\right) } \) be the set of rational numbers which can be written as \( a/b \), where \( a \) and \( b \) are integers, and \( b \) is relatively prime to \( m \). Then \( {\mathbb{Q}}^{\left( m\right) } \) is a subring of \( \mathbb{Q} \). | since for all \( a, b, c, d \in \mathbb{Z} \) with \( \gcd \left( {b, m}\right) = 1 \) and \( \gcd \left( {d, m}\right) = 1 \), we have\n\n\[ \frac{a}{b} + \frac{c}{d} = \frac{{ad} + {bc}}{bd}\text{ and }\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd} \]\n\nand since \( \gcd \left( {{bd}, m}\right) = 1 \), it follows tha... | Yes |
Let us define a few polynomials over the ring \( \mathbb{Z} \) :\n\n\[ a \mathrel{\text{:=}} 3 + {X}^{2}, b \mathrel{\text{:=}} 1 + {2X} - {X}^{3}, c \mathrel{\text{:=}} 5, d \mathrel{\text{:=}} 1 + X, e \mathrel{\text{:=}} X, f \mathrel{\text{:=}} 4{X}^{3}. \] | We have\n\n\[ a + b = 4 + {2X} + {X}^{2} - {X}^{3}, a \cdot b = 3 + {6X} + {X}^{2} - {X}^{3} - {X}^{5},{cd} + {ef} = 5 + {5X} + 4{X}^{4}. \] | Yes |
Theorem 7.9. Let \( D \) be an integral domain. Then:\n\n(i) for all \( g, h \in D\left\lbrack X\right\rbrack \), we have \( \deg \left( {gh}\right) = \deg \left( g\right) + \deg \left( h\right) \) ;\n\n(ii) \( D\left\lbrack X\right\rbrack \) is an integral domain;\n\n(iii) \( {\left( D\left\lbrack X\right\rbrack \righ... | Proof. Exercise. | No |
Theorem 7.10 (Division with remainder property). Let \( R \) be a ring. For all \( g, h \in R\left\lbrack X\right\rbrack \) with \( h \neq 0 \) and \( \operatorname{lc}\left( h\right) \in {R}^{ * } \), there exist unique \( q, r \in R\left\lbrack X\right\rbrack \) such that \( g = {hq} + r \) and \( \deg \left( r\right... | Proof. Consider the set \( S \mathrel{\text{:=}} \{ g - {ht} : t \in R\left\lbrack X\right\rbrack \} \) . Let \( r = g - {hq} \) be an element of \( S \) of minimum degree. We must have \( \deg \left( r\right) < \deg \left( h\right) \), since otherwise, we could subtract an appropriate multiple of \( h \) from \( r \) ... | Yes |
Let \( R \) be a subring of a ring \( E \) . For all \( g, h \in R\left\lbrack X\right\rbrack \) and \( \alpha \in E \), if \( s \mathrel{\text{:=}} g + h \in R\left\lbrack X\right\rbrack \) and \( p \mathrel{\text{:=}} {gh} \in R\left\lbrack X\right\rbrack \), then we have\n\n\[ s\left( \alpha \right) = g\left( \alpha... | The statement about evaluating a constant polynomial is clear from the definitions. The proof of the statements about evaluating the sum or product of polynomials is really just symbol pushing. Indeed, suppose \( g = \mathop{\sum }\limits_{i}{a}_{i}{X}^{i} \) and \( h = \mathop{\sum }\limits_{i}{b}_{i}{X}^{i} \) . Then... | Yes |
If \( E = R\left\lbrack X\right\rbrack \), then evaluating a polynomial \( g \in R\left\lbrack X\right\rbrack \) at a point \( \alpha \in E \) amounts to polynomial composition. For example, if \( g \mathrel{\text{:=}} {X}^{2} + X \) and \( \alpha \mathrel{\text{:=}} X + 1 \), then | \[ g\left( \alpha \right) = g\left( {X + 1}\right) = {\left( X + 1\right) }^{2} + \left( {X + 1}\right) = {X}^{2} + {3X} + 2. \] | Yes |
Theorem 7.12. Let \( R \) be a ring, \( g \in R\left\lbrack X\right\rbrack \), and \( x \in R \) . Then there exists a unique polynomial \( q \in R\left\lbrack X\right\rbrack \) such that \( g = \left( {X - x}\right) q + g\left( x\right) \) . In particular, \( x \) is a root of \( g \) if and only if \( \left( {X - x}\... | Proof. If \( R \) is the trivial ring, there is nothing to prove, so assume that \( R \) is nontrivial. Using the division with remainder property for polynomials, there exist unique \( q, r \in R\left\lbrack X\right\rbrack \) such that \( g = \left( {X - x}\right) q + r \), with \( q, r \in R\left\lbrack X\right\rbrac... | Yes |
Theorem 7.13. Let \( D \) be an integral domain, and let \( {x}_{1},\ldots ,{x}_{k} \) be distinct elements of \( D \) . Then for every polynomial \( g \in D\left\lbrack X\right\rbrack \), the elements \( {x}_{1},\ldots ,{x}_{k} \) are roots of \( g \) if and only if the polynomial \( \mathop{\prod }\limits_{{i = 1}}^{... | Proof. One direction is trivial: if \( \mathop{\prod }\limits_{{i = 1}}^{k}\left( {X - {x}_{i}}\right) \) divides \( g \), then it is clear that each \( {x}_{i} \) is a root of \( g \) . We prove the converse by induction on \( k \) . The base case \( k = 1 \) is just Theorem 7.12. So assume \( k > 1 \), and that the s... | Yes |
Theorem 7.14. Let \( D \) be an integral domain, and suppose that \( g \in D\left\lbrack X\right\rbrack \), with \( \deg \left( g\right) = k \geq 0 \) . Then \( g \) has at most \( k \) distinct roots. | Proof. If \( g \) had \( k + 1 \) distinct roots \( {x}_{1},\ldots ,{x}_{k + 1} \), then by the previous theorem, the polynomial \( \mathop{\prod }\limits_{{i = 1}}^{{k + 1}}\left( {X - {x}_{i}}\right) \), which has degree \( k + 1 \), would divide \( g \), which has degree \( k \) -an impossibility. | Yes |
Theorem 7.15 (Lagrange interpolation). Let \( F \) be a field, let \( {x}_{1},\ldots ,{x}_{k} \) be distinct elements of \( F \), and let \( {y}_{1},\ldots ,{y}_{k} \) be arbitrary elements of \( F \) . Then there exists a unique polynomial \( g \in F\left\lbrack X\right\rbrack \) with \( \deg \left( g\right) < k \) su... | Proof. For the existence part of the theorem, one just has to verify that \( g\left( {x}_{i}\right) = {y}_{i} \) for the given \( g \), which clearly has degree less than \( k \) . This is easy to see: for \( i = 1,\ldots, k \), evaluating the \( i \) th term in the sum defining \( g \) at \( {x}_{i} \) yields \( {y}_{... | Yes |
Let \( R \) be a ring and \( x \in R \) . Consider the set\n\n\[ I \mathrel{\text{:=}} \{ g \in R\left\lbrack X\right\rbrack : g\left( x\right) = 0\} . \] | It is not hard to see that \( I \) is an ideal of \( R\left\lbrack X\right\rbrack \) . Indeed, for all \( g, h \in I \) and \( q \in R\left\lbrack X\right\rbrack \) , we have\n\n\[ \left( {g + h}\right) \left( x\right) = g\left( x\right) + h\left( x\right) = 0 + 0 = 0\text{ and }\left( {gq}\right) \left( x\right) = g\l... | Yes |
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