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Consider the following algorithm, which models an experiment in which we toss a fair coin repeatedly until it comes up heads:\n\nrepeat\n\n\[ b\overset{\phi }{ \leftarrow }\{ 0,1\} \]\n\n\[ \text{until}b = 1 \]\n\nFor each positive integer \( n \), let \( {\beta }_{n} \) be the probability that the algorithm executes a... | It is not hard to see that \( {\beta }_{n} = {2}^{-n + 1} \), and since \( {\beta }_{n} \rightarrow 0 \) as \( n \rightarrow \infty \), the algorithm halts with probability 1 , even though the loop is not guaranteed to terminate after any particular, finite number of steps. | Yes |
Consider the following algorithm:\n\n\( i \leftarrow 0 \)\n\nrepeat\n\n\[ i \leftarrow i + 1 \]\n\n\[ \sigma \overset{\phi }{ \leftarrow }\{ 0,1{\} }^{\times i} \]\n\nuntil \( \sigma = {0}^{\times i} \)\n\nFor each positive integer \( n \), let \( {\beta }_{n} \) be the probability that the algorithm executes at least ... | It is not hard to see that\n\n\[ {\beta }_{n} = \mathop{\prod }\limits_{{i = 1}}^{{n - 1}}\left( {1 - {2}^{-i}}\right) \geq \mathop{\prod }\limits_{{i = 1}}^{{n - 1}}{e}^{-{2}^{-i + 1}} = {e}^{-\mathop{\sum }\limits_{{i = 0}}^{{n - 2}}{2}^{-i}} \geq {e}^{-2}, \]\n\nwhere we have made use of the estimate (iii) in §A1. T... | Yes |
Consider the following probabilistic algorithm that takes as input a positive integer \( m \) . It models an experiment in which we toss a fair coin repeatedly until it comes up heads \( m \) times. | Let \( L \) be the random variable that represents the number of loop iterations executed the algorithm on a fixed input \( m \) . We claim that \( \mathrm{E}\left\lbrack L\right\rbrack = {2m} \) . To see this, define random variables \( {L}_{1},\ldots ,{L}_{m} \), where \( {L}_{1} \) is the number of loop iterations n... | Yes |
Consider the following algorithm:\n\n\\( n \\leftarrow 0 \\)\n\nrepeat \\( n \\leftarrow n + 1, b\\overset{\\phi }{ \\leftarrow }\\{ 0,1\\} \\) until \\( b = 1 \\)\n\nrepeat \\( \\sigma \\overset{\\phi }{ \\leftarrow }\\{ 0,1{\\} }^{\\times n} \\) until \\( \\sigma = {0}^{\\times n} \\)\n\nThe expected running time is ... | To see this, define random variables \\( {L}_{1} \\) and \\( {L}_{2} \\), where \\( {L}_{1} \\) is the number of iterations of the first loop, and \\( {L}_{2} \\) is the number of iterations of the second. As in Example 9.7, the distribution of \\( {L}_{1} \\) is a geometric distribution with associated success probabi... | Yes |
Theorem 9.3. Under the assumptions above, (i) \( L \) has a geometric distribution with associated success probability \( \mathrm{P}\left\lbrack {\mathcal{H}}_{1}\right\rbrack \) , and in particular, \( \mathrm{E}\left\lbrack L\right\rbrack = 1/\mathrm{P}\left\lbrack {\mathcal{H}}_{1}\right\rbrack \) ; | Proof. (i) is clear. | No |
Suppose \( T \) is a finite set, and \( {T}^{\prime } \) is a non-empty, finite subset of \( T \) . Consider the following generalization of Algorithm RN:\n\nrepeat\n\n\[ y\overset{\phi }{ \leftarrow }T \]\n\nuntil \( y \in {T}^{\prime } \)\n\noutput \( y \)\n\nHere, we assume that we have an algorithm to generate a ra... | Since \( {Y}_{1} \) is uniformly distributed over \( T \), and \( {\mathcal{H}}_{1} \) is the event that \( {Y}_{1} \in {T}^{\prime } \), we have \( \mathrm{P}\left\lbrack {\mathcal{H}}_{1}\right\rbrack = \left| {T}^{\prime }\right| /\left| T\right| \) . It follows that \( \mathrm{E}\left\lbrack L\right\rbrack = \left|... | Yes |
Let us analyze the following algorithm:\n\nrepeat\n\n\[ y\overset{\phi }{ \leftarrow }\{ 1,2,3,4\} \]\n\n\[ z\overset{\phi }{ \leftarrow }\{ 1,\ldots, y\} \]\n\nuntil \( z = 1 \)\n\noutput \( y \)\n\nWith each loop iteration, the algorithm chooses \( y \) uniformly at random, and then decides to halt with probability \... | Thus, \( \mathrm{E}\left\lbrack L\right\rbrack = {48}/{25} \) . For the output distribution, for \( t = 1,\ldots ,4 \), we have\n\n\[ \mathsf{P}\left\lbrack {Y = t}\right\rbrack = \mathsf{P}\left\lbrack {{Y}_{1} = t \mid {\mathcal{H}}_{1}}\right\rbrack = \mathsf{P}\left\lbrack {\left( {{Y}_{1} = t}\right) \cap {\mathca... | Yes |
Theorem 10.1. If \( n \) is prime, then \( {L}_{n} = {\mathbb{Z}}_{n}^{ * } \) . If \( n \) is composite and \( {L}_{n} \varsubsetneq {\mathbb{Z}}_{n}^{ * } \), then \( \left| {L}_{n}\right| \leq \left( {n - 1}\right) /2 \) . | Proof. Note that \( {L}_{n} \) is the kernel of the \( \left( {n - 1}\right) \) -power map on \( {\mathbb{Z}}_{n}^{ * } \), and hence is a subgroup of \( {\mathbb{Z}}_{n}^{ * } \) . If \( n \) is prime, then we know that \( {\mathbb{Z}}_{n}^{ * } \) is a group of order \( n - 1 \) . Since the order of a group element d... | Yes |
Theorem 10.2. Every Carmichael number \( n \) is of the form \( n = {p}_{1}\cdots {p}_{r} \), where the \( {p}_{i} \) ’s are distinct primes, \( r \geq 3 \), and \( \left( {{p}_{i} - 1}\right) \mid \left( {n - 1}\right) \) for \( i = 1,\ldots, r \) . | Proof. Let \( n = {p}_{1}^{{e}_{1}}\cdots {p}_{r}^{{e}_{r}} \) be a Carmichael number. By the Chinese remainder theorem, we have an isomorphism of \( {\mathbb{Z}}_{n}^{ * } \) with the group\n\n\[ \n{\mathbb{Z}}_{{p}_{1}^{{e}_{1}}}^{ * } \times \cdots \times {\mathbb{Z}}_{{p}_{r}^{{e}_{r}}}^{ * }\n\]\n\nand we know tha... | Yes |
Theorem 10.3. If \( n \) is prime, then \( {L}_{n}^{\prime } = {\mathbb{Z}}_{n}^{ * } \) . If \( n \) is composite, then \( \left| {L}_{n}^{\prime }\right| \leq \left( {n - 1}\right) /4 \) . | Proof. Let \( n - 1 = t{2}^{h} \), where \( t \) is odd.\n\nCase 1: \( n \) is prime. Let \( \alpha \in {\mathbb{Z}}_{n}^{ * } \) . Since \( {\mathbb{Z}}_{n}^{ * } \) is a group of order \( n - 1 \), and the order of a group element divides the order of the group, we know that \( {\alpha }^{t{2}^{h}} = {\alpha }^{n - 1... | Yes |
Theorem 10.4. We have\n\n\\[\n\\gamma \\left( {m,1}\\right) \\leq \\exp \\left\\lbrack {-\\left( {1 + o\\left( 1\\right) }\\right) \\log \\left( m\\right) \\log \\left( {\\log \\left( {\\log \\left( m\\right) }\\right) }\\right) /\\log \\left( {\\log \\left( m\\right) }\\right) }\\right\\rbrack .\n\\]\n | Proof. Literature-see §10.5. | No |
Theorem 10.5. For all real \( x \geq 0 \) and \( y \geq 0 \), we have\n\n\[ \left| {R\left( {x, y}\right) - x\mathop{\prod }\limits_{{p \leq y}}\left( {1 - 1/p}\right) }\right| \leq {2}^{\pi \left( y\right) }.\] | Proof. To simplify the notation, we shall use the Möbius function \( \mu \) (see §2.9). Also, for a real number \( u \), let us write \( u = \lfloor u\rfloor + \{ u\} \), where \( 0 \leq \{ u\} < 1 \) . Let \( Q \) be the product of the primes up to the bound \( y \).\n\nNow, there are \( \lfloor x\rfloor \) positive i... | Yes |
Theorem 10.6. For all \( \ell \geq 2 \), we have\n\n\[ \n{\gamma }^{\prime }\left( {\ell ,1}\right) \leq {\ell }^{2}{4}^{2 - \sqrt{\ell }} \n\] | Proof. Literature-see §10.5. | No |
Theorem 12.1. Let \( p \) be an odd prime, and let \( a, b \in \mathbb{Z} \) . Then we have:\n\n(i) \( \left( {a \mid p}\right) \equiv {a}^{\left( {p - 1}\right) /2}\left( {\;\operatorname{mod}\;p}\right) \) ; in particular, \( \left( {-1 \mid p}\right) = {\left( -1\right) }^{\left( {p - 1}\right) /2} \) ; | Part (i) of the theorem is just a restatement of Euler's criterion (Theorem 2.21). As was observed in Theorem 2.31, this implies that -1 is a quadratic residue modulo \( p \) if and only if \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . Thus, the quadratic residuosity of -1 modulo \( p \) is determined by th... | Yes |
Let us characterize those primes \( p \) modulo which 5 is a quadratic residue. | Since \( 5 \equiv 1\left( {\;\operatorname{mod}\;4}\right) \), the law of quadratic reciprocity tells us that \( \left( {5 \mid p}\right) = \left( {p \mid 5}\right) \). Now, among the numbers \( \pm 1, \pm 2 \), the quadratic residues modulo 5 are \( \pm 1 \). It follows that 5 is a quadratic residue modulo \( p \) if ... | Yes |
Let us characterize those primes \( p \) modulo which 3 is a quadratic residue. | Since \( 3 ≢ 1\left( {\;\operatorname{mod}\;4}\right) \), we must be careful in our application of the law of quadratic reciprocity. First, suppose that \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . Then \( \left( {3 \mid p}\right) = \left( {p \mid 3}\right) \) , and so 3 is a quadratic residue modulo \( p ... | Yes |
Theorem 12.2 (Gauss' lemma). Let \( p \) be an odd prime and let \( a \) be an integer not divisible by \( p \) . Define \( {\alpha }_{j} \mathrel{\text{:=}} {ja}{\;\operatorname{mod}\;p} \) for \( j = 1,\ldots ,\left( {p - 1}\right) /2 \), and let \( n \) be the number of indices \( j \) for which \( {\alpha }_{j} > p... | Proof. Let \( {r}_{1},\ldots ,{r}_{n} \) denote the values \( {\alpha }_{j} \) that exceed \( p/2 \), and let \( {s}_{1},\ldots ,{s}_{k} \) denote the remaining values \( {\alpha }_{j} \) . The \( {r}_{i} \) ’s and \( {s}_{i} \) ’s are all distinct and non-zero. We have \( 0 < p - {r}_{i} < p/2 \) for \( i = 1,\ldots, ... | Yes |
Theorem 12.3. If \( p \) is an odd prime and \( \gcd \left( {a,{2p}}\right) = 1 \), then \( \left( {a \mid p}\right) = {\left( -1\right) }^{t} \) where \( t = \mathop{\sum }\limits_{{j = 1}}^{{\left( {p - 1}\right) /2}}\lfloor {ja}/p\rfloor \) . Also, \( \left( {2 \mid p}\right) = {\left( -1\right) }^{\left( {{p}^{2} -... | Proof. Let \( a \) be an integer not divisible by \( p \), but which may be even, and let us adopt the same notation as in the statement and proof of Theorem 12.2; in particular, \( {\alpha }_{1},\ldots ,{\alpha }_{\left( {p - 1}\right) /2},{r}_{1},\ldots ,{r}_{n} \), and \( {s}_{1},\ldots ,{s}_{k} \) are as defined th... | Yes |
Theorem 12.4. If \( p \) and \( q \) are distinct odd primes, then\n\n\[ \left( {p \mid q}\right) \left( {q \mid p}\right) = {\left( -1\right) }^{\frac{p - 1}{2}\frac{q - 1}{2}}. \] | Proof. Let \( S \) be the set of pairs of integers \( \left( {x, y}\right) \) with \( 1 \leq x \leq \left( {p - 1}\right) /2 \) and \( 1 \leq y \leq \left( {q - 1}\right) /2 \) . Note that \( S \) contains no pair \( \left( {x, y}\right) \) with \( {qx} = {py} \), so let us partition \( S \) into two subsets: \( {S}_{1... | Yes |
Theorem 12.5. Let \( m, n \) be odd, positive integers, and let \( a, b \in \mathbb{Z} \) . Then we have:\n\n(i) \( \left( {{ab} \mid n}\right) = \left( {a \mid n}\right) \left( {b \mid n}\right) \) ;\n\n(ii) \( \left( {a \mid {mn}}\right) = \left( {a \mid m}\right) \left( {a \mid n}\right) \) ;\n\n(iii) \( a \equiv b\... | Proof. Parts (i)-(iii) follow directly from the definition (exercise).\n\nFor parts (iv) and (vi), one can easily verify (exercise) that for all odd integers \( {n}_{1},\ldots ,{n}_{k} \)\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{k}\left( {{n}_{i} - 1}\right) /2 \equiv \left( {{n}_{1}\cdots {n}_{k} - 1}\right) /2\left( {\... | No |
Theorem 13.2. If \( M \) is a module over \( R \), then for all \( c \in R,\alpha \in M \), and \( k \in \mathbb{Z} \) , we have:\n\n(i) \( {0}_{R} \cdot \alpha = {0}_{M} \) ;\n\n(ii) \( c \cdot {0}_{M} = {0}_{M} \) ;\n\n(iii) \( \left( {-c}\right) \alpha = - \left( {c\alpha }\right) = c\left( {-\alpha }\right) \) ;\n\... | Proof. Exercise. | No |
The set \( {R}^{\times n} \), which consists of all of \( n \) -tuples of elements of \( R \) , forms an \( R \) -module, with addition and scalar multiplication defined componentwise: for \( \alpha = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \in {R}^{\times n},\beta = \left( {{b}_{1},\ldots ,{b}_{n}}\right) \in {R}^{\ti... | \[ \alpha + \beta \mathrel{\text{:=}} \left( {{a}_{1} + {b}_{1},\ldots ,{a}_{n} + {b}_{n}}\right) \text{and}{c\alpha } \mathrel{\text{:=}} \left( {c{a}_{1},\ldots, c{a}_{n}}\right) \text{.} \] | Yes |
Let \( G \) be any group, written additively, whose exponent divides \( n \) . Then we may define a scalar multiplication that maps \( {\left\lbrack k\right\rbrack }_{n} \in {\mathbb{Z}}_{n} \) and \( \alpha \in G \) to \( {k\alpha } \) . That this map is unambiguously defined follows from the fact that \( G \) has exp... | It is easy to check that this scalar multiplication map indeed makes \( G \) into a \( {\mathbb{Z}}_{n} \)-module. | No |
If \( I \) is an arbitrary set, and \( M \) is an \( R \) -module, then \( \operatorname{Map}\left( {I, M}\right) \) , which is the set of all functions \( f : I \rightarrow M \), may be naturally viewed as an \( R \) - module, with point-wise addition and scalar multiplication: for \( f, g \in \operatorname{Map}\left(... | \[ \left( {f + g}\right) \left( i\right) \mathrel{\text{:=}} f\left( i\right) + g\left( i\right) \text{ and }\left( {cf}\right) \left( i\right) \mathrel{\text{:=}} {cf}\left( i\right) \text{ for all }i \in I. \] | Yes |
Suppose \( N \) is a submodule of an \( R \) -module \( M \) . Then the natural map (see Example 6.36)\n\n\[ \rho : \;M \rightarrow M/N \]\n\n\[ \alpha \mapsto {\left\lbrack \alpha \right\rbrack }_{N} \] | is not just a group homomorphism, it is also easily seen to be an \( R \) -linear map. | No |
Generalizing the previous example, let \( M \) be an \( R \) -module, and let \( {\alpha }_{1},\ldots ,{\alpha }_{k} \) be elements of \( M \) . The map \[ \rho : \;{R}^{\times k} \rightarrow M \] \[ \left( {{c}_{1},\ldots ,{c}_{k}}\right) \mapsto {c}_{1}{\alpha }_{1} + \cdots + {c}_{k}{\alpha }_{k} \] | is easily seen to be an \( R \) -linear map whose image is the submodule \( R{\alpha }_{1} + \cdots + R{\alpha }_{k} \) (i.e., the submodule generated by \( {\alpha }_{1},\ldots ,{\alpha }_{k} \) ). | Yes |
Suppose that \( {M}_{1},\ldots ,{M}_{k} \) are submodules of an \( R \) -module \( M \) . Then the map\n\n\[ \rho : \;{M}_{1} \times \cdots \times {M}_{k} \rightarrow M \]\n\n\[ \left( {{\alpha }_{1},\ldots ,{\alpha }_{k}}\right) \mapsto {\alpha }_{1} + \cdots + {\alpha }_{k} \]\n\nis easily seen to be an \( R \) -line... | \[ ▱ \] | No |
Let \( E \) and \( {E}^{\prime } \) be extension rings of \( R \), which may be viewed as \( R \) -modules as in Example 13.5. Suppose that \( \rho : E \rightarrow {E}^{\prime } \) is a ring homomorphism whose restriction to \( R \) is the identity map (i.e., \( \rho \left( c\right) = c \) for all \( c \in R \) ). Then... | Indeed, for every \( c \in R \) and \( \alpha ,\beta \in E \), we have \( \rho \left( {\alpha + \beta }\right) = \rho \left( \alpha \right) + \rho \left( \beta \right) \) and \( \rho \left( {c\alpha }\right) = \rho \left( c\right) \rho \left( \alpha \right) = {c\rho }\left( \alpha \right) \). | Yes |
Let \( G \) and \( {G}^{\prime } \) be abelian groups. As we saw in Example 13.6, \( G \) and \( {G}^{\prime } \) may be viewed as \( \mathbb{Z} \) -modules. In addition, every group homomorphism \( \rho : G \rightarrow {G}^{\prime } \) is also a \( \mathbb{Z} \) -linear map. | Since an \( R \) -module homomorphism is also a group homomorphism on the underlying additive groups, all of the statements in Theorem 6.19 apply. In particular, an \( R \) -linear map is injective if and only if the kernel is trivial (i.e., contains only the zero element). However, in the case of \( R \) -module homom... | No |
Theorem 13.5. Let \( \rho : M \rightarrow {M}^{\prime } \) be an \( R \) -linear map. Then:\n\n(i) for every submodule \( N \) of \( M,\rho \left( N\right) \) is a submodule of \( {M}^{\prime } \) ; in particular (setting \( N \mathrel{\text{:=}} M \) ), \( \operatorname{Im}\rho \) is a submodule of \( {M}^{\prime } \)... | Proof. Exercise. | No |
Theorem 13.9 (First isomorphism theorem). Let \( \rho : M \rightarrow {M}^{\prime } \) be an \( R \) -linear map with kernel \( K \) and image \( {N}^{\prime } \) . Then we have an \( R \) -module isomorphism\n\n\[ M/K \cong {N}^{\prime }\text{.} \] | Specifically, the map\n\n\[ \bar{\rho } : \;M/K \rightarrow {M}^{\prime }\n\]\n\[ {\left\lbrack \alpha \right\rbrack }_{K} \mapsto \rho \left( \alpha \right) \]\n\nis an injective \( R \) -linear map whose image is \( {N}^{\prime } \) . | Yes |
Theorem 13.11 (Internal direct product). Let \( M \) be an \( R \) -module with submodules \( {N}_{1},{N}_{2} \), where \( {N}_{1} \cap {N}_{2} = \left\{ {0}_{M}\right\} \). Then we have an \( R \) -module isomorphism | \[ {N}_{1} \times {N}_{2} \cong {N}_{1} + {N}_{2} \] given by the map \[ \rho : \;{N}_{1} \times {N}_{2} \rightarrow {N}_{1} + {N}_{2} \] \[ \left( {{\alpha }_{1},{\alpha }_{2}}\right) \mapsto {\alpha }_{1} + {\alpha }_{2} \] | Yes |
Consider again the \( R \) -module \( R\\left\\lbrack X\\right\\rbrack /\\left( f\\right) \) discussed in Example 13.4, where \( f \\in R\\left\\lbrack X\\right\\rbrack \) is of degree \( \\ell \\geq 0 \) and \( \\operatorname{lc}\\left( f\\right) \\in {R}^{ * } \) . As an \( R \) -module, \( R\\left\\lbrack X\\right\\... | Indeed, based on the observations in Example 7.39, the map \( \\rho : R{\\left\\lbrack X\\right\\rbrack }_{ < \\ell } \\rightarrow R\\left\\lbrack X\\right\\rbrack /\\left( f\\right) \) that sends a polynomial \( g \\in R\\left\\lbrack X\\right\\rbrack \) of degree less than \( \\ell \) to \( {\\left\\lbrack g\\right\\... | Yes |
Consider the \( R \) -module \( {R}^{\times n} \) . Define \( {\alpha }_{1},\ldots ,{\alpha }_{n} \in {R}^{\times n} \) as follows:\n\n\[ \n{\alpha }_{1} \mathrel{\text{:=}} \left( {1,0,\ldots ,0}\right) ,{\alpha }_{2} \mathrel{\text{:=}} \left( {0,1,0,\ldots ,0}\right) ,\ldots ,{\alpha }_{n} \mathrel{\text{:=}} \left(... | Indeed, for all \( {c}_{1},\ldots ,{c}_{n} \in R \), we have\n\n\[ \n{c}_{1}{\alpha }_{1} + \cdots + {c}_{n}{\alpha }_{n} = \left( {{c}_{1},\ldots ,{c}_{n}}\right) , \n\]\n\nfrom which it is clear that \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) spans \( {R}^{\times n} \) and is linearly independent. The family ... | Yes |
Consider the \( \mathbb{Z} \) -module \( {\mathbb{Z}}^{\times 3} \) . In addition to the standard basis, which consists of the tuples\n\n\[ \left( {1,0,0}\right) ,\left( {0,1,0}\right) ,\left( {0,0,1}\right) ,\]\n\nthe tuples\n\n\[ {\alpha }_{1} \mathrel{\text{:=}} \left( {1,1,1}\right) ,{\alpha }_{2} \mathrel{\text{:=... | To see this, first observe that for all \( {c}_{1},{c}_{2},{c}_{3},{d}_{1},{d}_{2},{d}_{3} \in \mathbb{Z} \), we have\n\n\[ \left( {{d}_{1},{d}_{2},{d}_{3}}\right) = {c}_{1}{\alpha }_{1} + {c}_{2}{\alpha }_{2} + {c}_{3}{\alpha }_{3} \]\n\nif and only if\n\n\[ {d}_{1} = {c}_{1} + 2{c}_{3},{d}_{2} = {c}_{1} + {c}_{2}\tex... | Yes |
Theorem 13.14. If \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) is a basis for an \( R \) -module \( M \), then the map\n\n\[ \varepsilon : \;{R}^{\times n} \rightarrow M \]\n\n\[ \left( {{c}_{1},\ldots ,{c}_{n}}\right) \mapsto {c}_{1}{\alpha }_{1} + \cdots + {c}_{n}{\alpha }_{n} \]\n\nis an \( R \) -module isomor... | Proof. We already saw that \( \varepsilon \) is an \( R \) -linear map in Example 13.21. Since \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) is linearly independent, it follows that the kernel of \( \varepsilon \) is trivial, so that \( \varepsilon \) is injective. That \( \varepsilon \) is surjective follows imme... | Yes |
Theorem 13.16. Let \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) be a basis for an \( R \) -module \( M \), and let \( \rho : M \rightarrow {M}^{\prime } \) be an \( R \) -linear map. Then:\n\n(i) \( \rho \) is surjective if and only if \( {\left\{ \rho \left( {\alpha }_{i}\right) \right\} }_{i = 1}^{n} \) spans \... | Proof. By the previous theorem, we know that every element of \( M \) can be written uniquely as \( \mathop{\sum }\limits_{i}{c}_{i}{\alpha }_{i} \), where the \( {c}_{i} \) ’s are in \( R \) . Therefore, every element in \( \operatorname{Im}\rho \) can be expressed as \( \rho \left( {\mathop{\sum }\limits_{i}{c}_{i}{\... | Yes |
Theorem 13.17. Suppose that \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) is a linearly independent family of elements that spans a subspace \( W \varsubsetneq V \), and that \( {\alpha }_{n + 1} \in V \smallsetminus W \) . Then \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n + 1} \) is also linearly independent. | Proof. Suppose we have a linear relation\n\n\[ \n{0}_{V} = {c}_{1}{\alpha }_{1} + \cdots + {c}_{n}{\alpha }_{n} + {c}_{n + 1}{\alpha }_{n + 1}, \n\]\n\nwhere the \( {c}_{i} \) ’s are in \( F \) . We want to show that all the \( {c}_{i} \) ’s are zero. If \( {c}_{n + 1} \neq 0 \), then we have\n\n\[ \n{\alpha }_{n + 1} ... | Yes |
Theorem 13.18. Suppose \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) is a family of elements that spans \( V \) . Then for some subset \( J \subseteq \{ 1,\ldots, n\} \), the subfamily \( {\left\{ {\alpha }_{j}\right\} }_{j \in J} \) is a basis for \( V \) . | Proof. We prove this by induction on \( n \) . If \( n = 0 \), the theorem is clear, so assume \( n > 0 \) . Consider the subspace \( W \) of \( V \) spanned by \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n - 1} \) . By the induction hypothesis, for some \( K \subseteq \{ 1,\ldots, n - 1\} \), the subfamily \( {\left\... | Yes |
If \( V \) is spanned by some family of \( n \) elements of \( V \), then every family of \( n + 1 \) elements of \( V \) is linearly dependent. | We prove this by induction on \( n \) . If \( n = 0 \), the theorem is clear, so assume that \( n > 0 \) . Let \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) be a family that spans \( V \), and let \( {\left\{ {\beta }_{i}\right\} }_{i = 1}^{n + 1} \) be an arbitrary family of elements of \( V \) . We wish to show ... | Yes |
Theorem 13.20. If \( V \) is finitely generated, then any two bases for \( V \) have the same size. | Proof. If one basis had more elements than another, then Theorem 13.19 would imply that the first basis was linearly dependent, which contradicts the definition of a basis. | Yes |
Theorem 13.22. Suppose that \( {\dim }_{F}\left( V\right) = n \), and that \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) is a family of \( n \) elements of \( V \) . The following are equivalent:\n\n(i) \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) is linearly independent;\n\n(ii) \( {\left\{ {\alpha }_{i}\ri... | Proof. Let \( W \) be the subspace of \( V \) spanned by \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) .\n\nFirst, let us show that (i) implies (ii). Suppose \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) is linearly independent. Also, by way of contradiction, suppose that \( W \varsubsetneq V \), and choose \... | Yes |
Theorem 13.23. Suppose that \( V \) is finite dimensional and \( W \) is a subspace of \( V \) . Then \( W \) is also finite dimensional, with \( {\dim }_{F}\left( W\right) \leq {\dim }_{F}\left( V\right) \) . Moreover, \( {\dim }_{F}\left( W\right) = {\dim }_{F}\left( V\right) \) if and only if \( W = V \) . | Proof. Suppose \( {\dim }_{F}\left( V\right) = n \) . Consider the set \( \mathcal{S} \) of all linearly independent families of the form \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{m} \), where \( m \geq 0 \) and each \( {\alpha }_{i} \) is in \( W \) . The set \( \mathcal{S} \) is certainly non-empty, as it contains... | Yes |
Theorem 13.24. If \( V \) is finite dimensional, and \( W \) is a subspace of \( V \), then the quotient space \( V/W \) is also finite dimensional, and\n\n\[{\dim }_{F}\left( {V/W}\right) = {\dim }_{F}\left( V\right) - {\dim }_{F}\left( W\right)\] | Proof. Suppose that \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) spans \( V \) . Then it is clear that \( {\left\{ {\left\lbrack {\alpha }_{i}\right\rbrack }_{W}\right\} }_{i = 1}^{n} \) spans \( V/W \) . By Theorem 13.18, we know that \( V/W \) has a basis of the form \( {\left\{ {\left\lbrack {\alpha }_{i}\righ... | Yes |
Theorem 13.25. If \( V \) is finite dimensional, then every linearly independent family of elements of \( V \) can be extended to form a basis for \( V \) . | Proof. One can prove this by generalizing the proof of Theorem 13.18. Alternatively, we can adapt the proof of the previous theorem. Let \( {\left\{ {\beta }_{j}\right\} }_{j = 1}^{m} \) be a linearly independent family of elements that spans a subspace \( W \) of \( V \) . As in the proof of the previous theorem, if \... | No |
Theorem 13.26. If \( V \) is of finite dimension \( n \), and \( V \) is isomorphic to \( {V}^{\prime } \), then \( {V}^{\prime } \) is also of finite dimension \( n \) . | Proof. If \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) is a basis for \( V \), then by Theorem 13.16, \( {\left\{ \rho \left( {\alpha }_{i}\right) \right\} }_{i = 1}^{n} \) is a basis for \( {V}^{\prime } \) . \( ▱ \) | Yes |
Theorem 13.27. If \( \rho : V \rightarrow {V}^{\prime } \) is an \( F \) -linear map, and if \( V \) and \( {V}^{\prime } \) are finite dimensional with \( {\dim }_{F}\left( V\right) = {\dim }_{F}\left( {V}^{\prime }\right) \), then we have:\n\n\( \rho \) is injective if and only if \( \rho \) is surjective. | Proof. Let \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) be a basis for \( V \) . Then\n\n\( \rho \) is injective \( \Leftrightarrow {\left\{ \rho \left( {\alpha }_{i}\right) \right\} }_{i = 1}^{n} \) is linearly independent (by Theorem 13.16)\n\n\[ \Leftrightarrow {\left\{ \rho \left( {\alpha }_{i}\right) \right\... | Yes |
Theorem 13.28. If \( V \) is finite dimensional, and \( \rho : V \rightarrow {V}^{\prime } \) is an \( F \) -linear map, then \( \operatorname{Im}\rho \) is a finite dimensional vector space, and\n\n\[{\dim }_{F}\left( V\right) = {\dim }_{F}\left( {\operatorname{Im}\rho }\right) + {\dim }_{F}\left( {\operatorname{Ker}\... | Proof. As the reader may verify, this follows immediately from Theorem 13.24, together with Theorems 13.26 and 13.9. | No |
Theorem 14.1. With addition and scalar multiplication as defined above, \( {R}^{m \times n} \) is an \( R \) -module. The matrix \( {0}_{R}^{m \times n} \) is the additive identity, and the additive inverse of a matrix \( A \in {R}^{m \times n} \) is the \( m \times n \) matrix whose \( \left( {i, j}\right) \) entry is... | Proof. To prove this, one first verifies that matrix addition is associative and commutative, which follows from the associativity and commutativity of addition in \( R \) . The claims made about the additive identity and additive inverses are also easily verified. These observations establish that \( {R}^{m \times n} ... | No |
Matrix multiplication is associative; that is, \( A\left( {BC}\right) = \left( {AB}\right) C \) for all \( A \in {R}^{m \times n}, B \in {R}^{n \times p} \), and \( C \in {R}^{p \times q} \) . | All of these are trivial, except for (i), which requires just a bit of computation to show that the \( \left( {i,\ell }\right) \) entry of both \( A\left( {BC}\right) \) and \( \left( {AB}\right) C \) is equal to (as the reader may verify)\n\n\[ \mathop{\sum }\limits_{\substack{{1 \leq j \leq n} \\ {1 \leq k \leq p} }}... | No |
Theorem 14.3. If \( A, B \in {R}^{m \times n}, C \in {R}^{n \times p} \), and \( c \in R \), then:\n\n(i) \( {\left( A + B\right) }^{\top } = {A}^{\top } + {B}^{\top } \) ;\n\n(ii) \( {\left( cA\right) }^{\top } = c{A}^{\top } \) ;\n\n(iii) \( {\left( {A}^{\top }\right) }^{\top } = A \) ;\n\n(iv) \( {\left( AC\right) }... | Proof. Exercise. | No |
Consider the quotient ring \( E = R\left\lbrack X\right\rbrack /\left( f\right) \), where \( f \in R\left\lbrack X\right\rbrack \) with \( \deg \left( f\right) = \ell > 0 \) and \( \operatorname{lc}\left( f\right) \in {R}^{ * } \) . Let \( \xi \mathrel{\text{:=}} {\left\lbrack X\right\rbrack }_{f} \in E \) . As an \( R... | \[ A = \left( \begin{matrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ & & & \ddots & \\ 0 & 0 & 0 & \cdots & 1 \\ - {c}_{0}/{c}_{\ell } & - {c}_{1}/{c}_{\ell } & - {c}_{2}/{c}_{\ell } & \cdots & - {c}_{\ell - 1}/{c}_{\ell } \end{matrix}\right) ,\] | Yes |
Theorem 14.5. Let \( A \in {R}^{n \times n} \), and let \( {\lambda }_{A} : {R}^{1 \times n} \rightarrow {R}^{1 \times n} \) be the corresponding \( R \) -linear map. Then \( A \) is invertible if and only if \( {\lambda }_{A} \) is bijective, in which case \( {\lambda }_{{A}^{-1}} = {\lambda }_{A}^{-1} \) . | Proof. Suppose \( A \) is invertible, and that \( B \) is its inverse. We have \( {AB} = {BA} = I \) , and hence \( {\lambda }_{AB} = {\lambda }_{BA} = {\lambda }_{I} \), from which it follows (see (14.1)) that \( {\lambda }_{B} \circ {\lambda }_{A} = \) \( {\lambda }_{A} \circ {\lambda }_{B} = {\lambda }_{I} \) . Sinc... | Yes |
Theorem 14.6. Let \( A \in {R}^{n \times n} \) . The following are equivalent:\n\n(i) \( A \) is invertible;\n\n(ii) \( {\left\{ {\operatorname{Row}}_{i}\left( A\right) \right\} }_{i = 1}^{n} \) is a basis for \( {R}^{1 \times n} \) ;\n\n(iii) \( {\left\{ {\operatorname{Col}}_{j}\left( A\right) \right\} }_{j = 1}^{n} \... | Proof. We first prove the equivalence of (i) and (ii). By the previous theorem, \( A \) is invertible if and only if \( {\lambda }_{A} \) is bijective. Also, in the previous section, we observed that \( {\lambda }_{A} \) is surjective if and only if \( {\left\{ {\operatorname{Row}}_{i}\left( A\right) \right\} }_{i = 1}... | Yes |
The following \( 4 \times 6 \) matrix \( B \) over the rational numbers is in reduced row echelon form: | \[ B = \left( \begin{matrix} 0 & 1 & - 2 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 & 1 & - 4 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{matrix}\right) \] The pivot sequence of \( B \) is \( \left( {2,4,5}\right) \). Notice that the first three rows of \( B \) form a linearly independent family of vectors, that columns 2... | Yes |
Theorem 14.7. If \( B \) is a matrix in reduced row echelon form with pivot sequence \( \left( {{p}_{1},\ldots ,{p}_{r}}\right) \), then:\n\n(i) rows \( 1,2,\ldots, r \) of \( B \) form a linearly independent family of vectors;\n\n(ii) columns \( {p}_{1},\ldots ,{p}_{r} \) of \( B \) form a linearly independent family ... | ## Proof. Exercise-just look at the matrix! | No |
Consider the execution of the Gaussian elimination algorithm on input\n\n\[ A = \left( \begin{array}{lll} \left\lbrack 0\right\rbrack & \left\lbrack 1\right\rbrack & \left\lbrack 1\right\rbrack \\ \left\lbrack 2\right\rbrack & \left\lbrack 1\right\rbrack & \left\lbrack 2\right\rbrack \\ \left\lbrack 2\right\rbrack & \l... | After copying \( A \) into \( B \), the algorithm transforms \( B \) as follows:\n\n\[ \left( \begin{matrix} \left\lbrack 0\right\rbrack & \left\lbrack 1\right\rbrack & \left\lbrack 1\right\rbrack \\ \left\lbrack 2\right\rbrack & \left\lbrack 1\right\rbrack & \left\lbrack 2\right\rbrack \\ \left\lbrack 2\right\rbrack &... | Yes |
Continuing with Example 14.4, the execution of the extended Gaussian elimination algorithm initializes \( X \) to the identity matrix, and then transforms \( X \) as follows: | \[ \left( \begin{matrix} \left\lbrack 1\right\rbrack & \left\lbrack 0\right\rbrack & \left\lbrack 0\right\rbrack \\ \left\lbrack 0\right\rbrack & \left\lbrack 1\right\rbrack & \left\lbrack 0\right\rbrack \\ \left\lbrack 0\right\rbrack & \left\lbrack 0\right\rbrack & \left\lbrack 1\right\rbrack \end{matrix}\right) \xrig... | Yes |
Theorem 15.1. Let \( y \) be a function of \( x \) such that\n\n\[ \frac{y}{\log x} \rightarrow \infty \text{ and }u \mathrel{\text{:=}} \frac{\log x}{\log y} \rightarrow \infty \]\n\nas \( x \rightarrow \infty \) . Then\n\n\[ \Psi \left( {y, x}\right) \geq x \cdot \exp \left\lbrack {\left( {-1 + o\left( 1\right) }\rig... | Proof. Let us write \( u = \lfloor u\rfloor + \delta \), where \( 0 \leq \delta < 1 \) . Let us split the primes up to \( y \) into two sets: the set \( V \) of \ | No |
Lemma 15.4. For \( i = 1,\ldots, k + 2 \), we have \( \mathrm{E}\left\lbrack {L}_{i}\right\rbrack \leq {\sigma }^{-1} \) . | Proof. We first compute \( \mathrm{E}\left\lbrack {L}_{1}\right\rbrack \) . As \( \delta \) is chosen uniformly from \( {\mathbb{Z}}_{n}^{ * } \) and independent of \( {\alpha }_{1} \), at each attempt to find a relation, \( {\alpha }_{1}^{2}\delta \) is uniformly distributed over \( {\mathbb{Z}}_{n}^{ * } \), and henc... | Yes |
Theorem 15.7. Let \( y \) be a function of \( x \) such that for some \( \varepsilon > 0 \), we have\n\n\[ y = \Omega \left( {\left( \log x\right) }^{1 + \varepsilon }\right) \text{ and }u \mathrel{\text{:=}} \frac{\log x}{\log y} \rightarrow \infty \]\n\nas \( x \rightarrow \infty \) . Then\n\n\[ \Psi \left( {y, x}\ri... | Proof. See §15.5. | No |
Theorem 16.2. If \( E \) is an \( R \) -algebra, then the map\n\n\[ \n\tau : \;R \rightarrow E \]\n\n\[ \nc \mapsto c \cdot {1}_{E} \]\n\nis a ring homomorphism, and \( {c\alpha } = \tau \left( c\right) \alpha \) for all \( c \in R \) and \( \alpha \in E \) . | Proof. Exercise. | No |
Let \( E \) be an \( R \) -algebra and let \( I \) be an ideal of \( E \) . Then it is easily verified that \( I \) is also a submodule of \( E \) . This means that the quotient ring \( E/I \) may also be viewed as an \( R \) -module, and indeed, it is an \( R \) -algebra, called the quotient algebra (over \( R \) ) of... | For \( \alpha ,\beta \in E \) and \( c \in R \), addition, multiplication, and scalar multiplication in \( E \) are defined as follows:\n\n\[ \n{\left\lbrack \alpha \right\rbrack }_{I} + {\left\lbrack \beta \right\rbrack }_{I} \mathrel{\text{:=}} {\left\lbrack \alpha + \beta \right\rbrack }_{I},\;{\left\lbrack \alpha \... | Yes |
The ring of polynomials \( R\left\lbrack X\right\rbrack \) is an \( R \) -algebra via inclusion. Let \( f \in R\left\lbrack X\right\rbrack \) be a non-zero polynomial with \( \operatorname{lc}\left( f\right) \in {R}^{ * } \). We may form the quotient ring \( E \mathrel{\text{:=}} R\left\lbrack X\right\rbrack /\left( f\... | \[ \tau : \;R \rightarrow E \] \[ c \mapsto c \cdot {1}_{E} \] from Theorem 16.2. By definition, \( \tau \left( c\right) = {\left\lbrack c\right\rbrack }_{f} \). As discussed in Example 7.55, the map \( \tau \) is a natural embedding of rings, and so by identifying \( R \) with its image in \( E \) under \( \tau \), we... | Yes |
Theorem 16.4. If \( E \) is an \( R \) -algebra via inclusion, and \( S \) is a subring of \( E \), then \( S \) is a subalgebra if and only if \( S \) contains \( R \) . More generally, if \( E \) is an arbitrary \( R \) -algebra, and \( S \) is a subring of \( E \), then \( S \) is a subalgebra of \( E \) if and only... | Proof. Exercise. | No |
Theorem 16.5. If \( E \) and \( {E}^{\prime } \) are \( R \) -algebras via inclusion, and \( \rho : E \rightarrow {E}^{\prime } \) is a ring homomorphism, then \( \rho \) is an \( R \) -algebra homomorphism if and only if the restriction of \( \rho \) to \( R \) is the identity map. More generally, if \( E \) and \( {E... | Proof. Exercise. | No |
If \( E \) is an \( R \) -algebra and \( I \) is an ideal of \( E \), then as observed in Example 16.6, \( I \) is also a submodule of \( E \), and we may form the quotient algebra \( E/I \). | The natural map\n\n\[ \rho : \;E \rightarrow E/I \]\n\n\[ \alpha \mapsto {\left\lbrack \alpha \right\rbrack }_{I} \]\n\n is both a ring homomorphism and an \( R \) -linear map, and hence is an \( R \) -algebra homomorphism. | Yes |
Theorem 16.6. Let \( E \) be an \( R \) -algebra, and let \( \rho : E \rightarrow E \) be an \( R \) -algebra homomorphism. Then the set \( S \mathrel{\text{:=}} \{ \alpha \in E : \rho \left( \alpha \right) = \alpha \} \) is a subalgebra of \( E \) , called the subalgebra of \( E \) fixed by \( \rho \) . Moreover, if \... | Proof. Let us verify that \( S \) is closed under addition. If \( \alpha ,\beta \in S \), then we have\n\n\[ \rho \left( {\alpha + \beta }\right) = \rho \left( \alpha \right) + \rho \left( \beta \right) \text{ (since }\rho \text{ is a group homomorphism) } \]\n\n\[ = \alpha + \beta \text{ (since }\alpha ,\beta \in S\te... | Yes |
Theorem 16.7. Let \( \rho : E \rightarrow {E}^{\prime } \) be an \( R \) -algebra homomorphism. Then for all \( g \in R\left\lbrack X\right\rbrack \) and \( \alpha \in E \), we have\n\n\[ \rho \left( {g\left( \alpha \right) }\right) = g\left( {\rho \left( \alpha \right) }\right) . \] | Proof. Let \( g = \mathop{\sum }\limits_{i}{a}_{i}{X}^{i} \in R\left\lbrack X\right\rbrack \) . Then we have\n\n\[ \rho \left( {g\left( \alpha \right) }\right) = \rho \left( {\mathop{\sum }\limits_{i}{a}_{i}{\alpha }^{i}}\right) = \mathop{\sum }\limits_{i}\rho \left( {{a}_{i}{\alpha }^{i}}\right) = \mathop{\sum }\limit... | Yes |
Let \( f \in R\left\lbrack X\right\rbrack \) be a non-zero polynomial with \( \operatorname{lc}\left( f\right) \in {R}^{ * } \). As in Example 16.7, we may form the quotient algebra \( E \mathrel{\text{:=}} R\left\lbrack X\right\rbrack /\left( f\right) \). Let \( \xi \mathrel{\text{:=}} {\left\lbrack X\right\rbrack }_{... | Now let \( {E}^{\prime } \) be any \( R \) -algebra, and suppose that \( \rho : E \rightarrow {E}^{\prime } \) is an \( R \) -algebra homomorphism, and let \( {\xi }^{\prime } \mathrel{\text{:=}} \rho \left( \xi \right) \). By the previous theorem, \( \rho \) sends \( g\left( \xi \right) \) to \( g\left( {\xi }^{\prime... | Yes |
Lemma 16.8. For all \( \left( {{a}_{1},{b}_{1}}\right) ,\left( {{a}_{2},{b}_{2}}\right) ,\left( {{a}_{3},{b}_{3}}\right) \in S \), we have\n\n(i) \( \left( {{a}_{1},{b}_{1}}\right) \sim \left( {{a}_{1},{b}_{1}}\right) \) ;\n\n(ii) \( \left( {{a}_{1},{b}_{1}}\right) \sim \left( {{a}_{2},{b}_{2}}\right) \) implies \( \le... | Proof. (i) and (ii) are rather trivial, and we do not comment on these any further. As for (iii), assume that \( {a}_{1}{b}_{2} = {a}_{2}{b}_{1} \) and \( {a}_{2}{b}_{3} = {a}_{3}{b}_{2} \) . Multiplying the first equation by \( {b}_{3} \), we obtain \( {a}_{1}{b}_{2}{b}_{3} = {a}_{2}{b}_{1}{b}_{3} \) and substituting ... | Yes |
Lemma 16.9. Let \( \left( {{a}_{1},{b}_{1}}\right) ,\left( {{a}_{1}^{\prime },{b}_{1}^{\prime }}\right) ,\left( {{a}_{2},{b}_{2}}\right) ,\left( {{a}_{2}^{\prime },{b}_{2}^{\prime }}\right) \in S \), where \( \left( {{a}_{1},{b}_{1}}\right) \sim \left( {{a}_{1}^{\prime },{b}_{1}^{\prime }}\right) \) and \( \left( {{a}_... | Proof. This is a straightforward calculation. Since \( {a}_{1}{b}_{1}^{\prime } = {a}_{1}^{\prime }{b}_{1} \) and \( {a}_{2}{b}_{2}^{\prime } = {a}_{2}^{\prime }{b}_{2} \) , we have\n\n\[ \left( {{a}_{1}{b}_{2} + {a}_{2}{b}_{1}}\right) {b}_{1}^{\prime }{b}_{2}^{\prime } = {a}_{1}{b}_{2}{b}_{1}^{\prime }{b}_{2}^{\prime ... | Yes |
Lemma 16.10. With addition and multiplication as defined above, \( K \) is a ring, with additive identity \( \left\lbrack {{0}_{D},{1}_{D}}\right\rbrack \) and multiplicative identity \( \left\lbrack {{1}_{D},{1}_{D}}\right\rbrack \) . | Proof. Exercise. | No |
Every non-zero polynomial \( f \in F\left\lbrack X\right\rbrack \) can be expressed as\n\n\[ f = c \cdot {p}_{1}^{{e}_{1}}\cdots {p}_{r}^{{e}_{r}} \]\n\nwhere \( c \in {F}^{ * },{p}_{1},\ldots ,{p}_{r} \) are distinct monic irreducible polynomials, and \( {e}_{1},\ldots ,{e}_{r} \) are positive integers. Moreover, this... | To prove this theorem, we may assume that \( f \) is monic, since the non-monic case trivially reduces to the monic case.\n\nThe proof of the existence part of Theorem 16.11 is just as for Theorem 1.3. If \( f \) is 1 or a monic irreducible, we are done. Otherwise, there exist \( g, h \in F\left\lbrack X\right\rbrack \... | Yes |
Theorem 16.12. Let \( I \) be an ideal of \( F\left\lbrack X\right\rbrack \) . Then there exists a unique polynomial \( d \in F\left\lbrack X\right\rbrack \) such that \( I = {dF}\left\lbrack X\right\rbrack \) and \( d \) is either zero or monic. | Proof. We first prove the existence part of the theorem. If \( I = \{ 0\} \), then \( d = 0 \) does the job, so let us assume that \( I \neq \{ 0\} \) . Since \( I \) contains non-zero polynomials, it must contain monic polynomials, since if \( g \) is a non-zero polynomial in \( I \), then its monic associate \( \oper... | Yes |
Theorem 16.13. For all \( g, h \in F\left\lbrack X\right\rbrack \), there exists a unique greatest common divisor \( d \) of \( g \) and \( h \), and moreover, \( {gF}\left\lbrack X\right\rbrack + {hF}\left\lbrack X\right\rbrack = {dF}\left\lbrack X\right\rbrack \) . | Proof. We apply the previous theorem to the ideal \( I \mathrel{\text{:=}} {gF}\left\lbrack X\right\rbrack + {hF}\left\lbrack X\right\rbrack \) . Let \( d \in F\left\lbrack X\right\rbrack \) with \( I = {dF}\left\lbrack X\right\rbrack \), as in that theorem. Note that \( g, h, d \in I \) and \( d \) is monic or zero.\n... | Yes |
Theorem 16.14. For \( f, g, h \in F\left\lbrack X\right\rbrack \) such that \( f \mid {gh} \) and \( \gcd \left( {f, g}\right) = 1 \), we have \( f \mid h \) . | Proof. Suppose that \( f \mid {gh} \) and \( \gcd \left( {f, g}\right) = 1 \) . Then since \( \gcd \left( {f, g}\right) = 1 \), by Theorem 16.13 we have \( {fs} + {gt} = 1 \) for some \( s, t \in F\left\lbrack X\right\rbrack \) . Multiplying this equation by \( h \), we obtain \( {fhs} + {ght} = h \) . Since \( f \mid ... | Yes |
Theorem 16.15. Let \( p \in F\left\lbrack X\right\rbrack \) be irreducible, and let \( g, h \in F\left\lbrack X\right\rbrack \) . Then \( p \mid {gh} \) implies that \( p \mid g \) or \( p \mid h \) . | Proof. Assume that \( p \mid {gh} \) . The only divisors of \( p \) are associate to 1 or \( p \) . Thus, \( \gcd \left( {p, g}\right) \) is either 1 or the monic associate of \( p \) . If \( p \mid g \), we are done; otherwise, if \( p \nmid g \), we must have \( \gcd \left( {p, g}\right) = 1 \), and by the previous t... | Yes |
Theorem 16.16. There are infinitely many monic irreducible polynomials in \( F\left\lbrack X\right\rbrack \) . | If \( F \) is infinite, then this theorem is true simply because there are infinitely many monic, linear polynomials; in any case, one can easily prove this theorem by mimicking the proof of Theorem 1.11 (as the reader may verify). | No |
Let \( f \in F\left\lbrack X\right\rbrack \) be a polynomial of degree 2 or 3 . Then it is easy to see that \( f \) is irreducible if and only if \( f \) has no roots in \( F \) . | Indeed, if \( f \) is reducible, then it must have a factor of degree 1 , which we can assume is monic; thus, we can write \( f = \left( {X - x}\right) g \), where \( x \in F \) and \( g \in F\left\lbrack X\right\rbrack \), and so \( f\left( x\right) = \left( {x - x}\right) g\left( x\right) = 0 \) . Conversely, if \( x... | Yes |
As a special case of the previous example, consider the polynomials \( f \mathrel{\text{:=}} {X}^{2} - 2 \in \mathbb{Q}\left\lbrack X\right\rbrack \) and \( g \mathrel{\text{:=}} {X}^{3} - 2 \in \mathbb{Q}\left\lbrack X\right\rbrack \) . We claim that as polynomials over \( \mathbb{Q}, f \) and \( g \) are irreducible. | Indeed, neither of them have integer roots, and so neither of them have rational roots (see Exercise 1.26); therefore, they are irreducible. | No |
Theorem 16.19 (Chinese remainder theorem). Let \( {\left\{ {f}_{i}\right\} }_{i = 1}^{k} \) be a pairwise relatively prime family of non-zero polynomials in \( F\left\lbrack X\right\rbrack \), and let \( {g}_{1},\ldots ,{g}_{k} \) be arbitrary polynomials in \( F\left\lbrack X\right\rbrack \) . Then there exists a solu... | Let us recall the formula for the solution \( g \) (see proof of Theorem 2.6). We have\n\n\[ g \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 1}}^{k}{g}_{i}{e}_{i} \]\n\nwhere\n\n\[ {e}_{i} \mathrel{\text{:=}} {f}_{i}^{ * }{t}_{i},\;{f}_{i}^{ * } \mathrel{\text{:=}} f/{f}_{i},\;{t}_{i} \mathrel{\text{:=}} {\left( {f}_... | Yes |
The polynomial \( {X}^{2} + 1 \) is irreducible over \( \mathbb{R} \) | since if it were not, it would have a root in \( \mathbb{R} \) (see Example 16.12), which is clearly impossible, since -1 is not the square of any real number. It follows immediately that \( \mathbb{C} = \mathbb{R}\left\lbrack X\right\rbrack /\left( {{X}^{2} + 1}\right) \) is a field, without having to explicitly calcu... | Yes |
Consider the polynomial \( f \mathrel{\text{:=}} {X}^{4} + {X}^{3} + 1 \) over \( {\mathbb{Z}}_{2} \). We claim that \( f \) is irreducible. It suffices to show that \( f \) has no irreducible factors of degree 1 or 2. | If \( f \) had a factor of degree 1, then it would have a root; however, \( f\left( 0\right) = 0 + 0 + 1 = 1 \) and \( f\left( 1\right) = 1 + 1 + 1 = 1 \). So \( f \) has no factors of degree 1.\n\nDoes \( f \) have a factor of degree 2 ? The polynomials of degree 2 are \( {X}^{2},{X}^{2} + X \), \( {X}^{2} + 1 \), and... | Yes |
Consider the real numbers \( \sqrt{2} \) and \( \sqrt[3]{2} \). We claim that \( {X}^{2} - 2 \) is the minimal polynomial of \( \sqrt{2} \) over \( \mathbb{Q} \). | To see this, first observe that \( \sqrt{2} \) is a root of \( {X}^{2} - 2 \). Thus, the minimal polynomial of \( \sqrt{2} \) divides \( {X}^{2} - 2 \). However, as we saw in Example 16.13, the polynomial \( {X}^{2} - 2 \) is irreducible over \( \mathbb{Q} \), and hence must be equal to the minimal polynomial of \( \sq... | Yes |
Theorem 16.20. Suppose \( E \) is an \( F \) -algebra, and that as an \( F \) -vector space, \( E \) has finite dimension \( n \) . Then every \( \alpha \in E \) has a non-zero minimal polynomial of degree at most \( n \) . | Proof. Indeed, the family of elements\n\n\[ \n{1}_{E},\alpha ,\ldots ,{\alpha }^{n} \n\]\n\nmust be linearly dependent (as must any family of \( n + 1 \) elements of a vector space\n\nof dimension \( n \) ), and hence there exist \( {c}_{0},\ldots ,{c}_{n} \in F \), not all zero, such that\n\n\[ \n{c}_{0}{1}_{E} + {c}_... | Yes |
Theorem 16.22. Suppose \( E \) is a finite extension of a field \( K \), with a basis \( {\left\{ {\beta }_{j}\right\} }_{j = 1}^{m} \) over \( K \), and \( K \) is a finite extension of \( F \), with a basis \( {\left\{ {\alpha }_{i}\right\} }_{i = 1}^{n} \) over \( F \) . Then the elements\n\n\[ \n{\alpha }_{i}{\beta... | Now suppose that \( E \) is a finite extension of a field \( F \) . Let \( K \) be an intermediate field, that is, a subfield of \( E \) containing \( F \) . Then evidently, \( E \) is a finite extension of \( K \) (since any basis for \( E \) over \( F \) also spans \( E \) over \( K \) ), and \( K \) is a finite exte... | Yes |
Theorem 16.25. Let \( F \) be a field, and \( f \in F\left\lbrack X\right\rbrack \) a non-zero polynomial of degree \( n \) . Then there exists a finite extension \( E \) of \( F \) over which \( f \) factors as\n\n\[ f = c\left( {X - {\alpha }_{1}}\right) \left( {X - {\alpha }_{2}}\right) \cdots \left( {X - {\alpha }_... | Proof. We may assume that \( f \) is monic. We prove the existence of \( E \) by induction on the degree \( n \) of \( f \) . If \( n = 0 \), then the theorem is trivially true. Otherwise, let \( h \) be an irreducible factor of \( f \), and set \( K \mathrel{\text{:=}} F\left\lbrack X\right\rbrack /\left( h\right) \),... | Yes |
Theorem 16.26. We have:\n\n(i) \( \\mathbf{D}\\left( c\\right) = 0 \) for all \( c \\in R \) ;\n\n(ii) \( \\mathbf{D}\\left( X\\right) = 1 \) ;\n\n(iii) \( \\mathbf{D}\\left( {g + h}\\right) = \\mathbf{D}\\left( g\\right) + \\mathbf{D}\\left( h\\right) \) for all \( g, h \\in R\\left\\lbrack X\\right\\rbrack \) ;\n\n(i... | Proof. Parts (i) and (ii) are immediate from the definition. Parts (iii) and (iv) follow from the definition by a simple calculation. Suppose\n\n\[ g\\left( {X + Y}\\right) \\equiv g + {g}_{1}Y\\left( {\\;\\operatorname{mod}\\{Y}^{2}}\\right) \\text{ and }h\\left( {X + Y}\\right) \\equiv h + {h}_{1}Y\\left( {\\;\\opera... | Yes |
Theorem 16.27. Let \( g = \mathop{\sum }\limits_{{i = 0}}^{\infty }{a}_{i}{X}^{i} \in R\llbracket X\rrbracket \) . Then \( g \in {\left( R\llbracket X\rrbracket \right) }^{ * } \) if and only if \( {a}_{0} \in {R}^{ * } \) . | Proof. If \( {a}_{0} \) is not a unit, then it is clear that \( g \) is not a unit, since the constant term of a product of formal power series is equal to the product of the constant terms.\n\nConversely, if \( {a}_{0} \) is a unit, we show how to define the coefficients of the inverse \( h = \mathop{\sum }\limits_{{i... | Yes |
Theorem 16.28. If \( D \) is an integral domain, then \( D\left( \left( X\right) \right) \) is an integral domain. | Proof. Let \( g = \mathop{\sum }\limits_{{i = m}}^{\infty }{a}_{i}{X}^{i} \) and \( h = \mathop{\sum }\limits_{{i = n}}^{\infty }{b}_{i}{X}^{i} \), where \( {a}_{m} \neq 0 \) and \( {b}_{n} \neq 0 \) . Then \( {gh} = \mathop{\sum }\limits_{{i = m + n}}^{\infty }{c}_{i}{X}^{i} \), where \( {c}_{m + n} = {a}_{m}{b}_{n} \... | Yes |
Theorem 16.29. Let \( g \in R\left( \left( X\right) \right) \), and suppose that \( g \neq 0 \) and \( g = \mathop{\sum }\limits_{{i = m}}^{\infty }{a}_{i}{X}^{i} \) with \( {a}_{m} \in {R}^{ * } \) . Then \( g \) has a multiplicative inverse in \( R\left( \left( X\right) \right) \). | Proof. We can write \( g = {X}^{m}{g}^{\prime } \), where \( {g}^{\prime } \) is a formal power series whose constant term is a unit, and hence there is a formal power series \( h \) such that \( {g}^{\prime }h = 1 \) . Thus, \( {X}^{-m}h \) is the multiplicative inverse of \( g \) in \( R\left( \left( X\right) \right)... | Yes |
Theorem 16.31. For \( g, h \in R\left( \left( {X}^{-1}\right) \right) \), we have \( \deg \left( {gh}\right) \leq \deg \left( g\right) + \deg \left( h\right) \), where equality holds unless both \( \operatorname{lc}\left( g\right) \) and \( \operatorname{lc}\left( h\right) \) are zero divisors. Furthermore, if \( h \ne... | Proof. Exercise. | No |
Theorem 16.32. Let \( g, h \in R\left\lbrack X\right\rbrack \) with \( h \neq 0 \) and \( \operatorname{lc}\left( h\right) \in {R}^{ * } \), and using the usual division with remainder property for polynomials, write \( g = {hq} + r \), where \( q, r \in R\left\lbrack X\right\rbrack \) with \( \deg \left( r\right) < \d... | Proof. Multiplying the equation \( g = {hq} + r \) by \( {h}^{-1} \), we obtain \( g{h}^{-1} = q + r{h}^{-1} \) , and \( \deg \left( {r{h}^{-1}}\right) < 0 \), from which it follows that \( \left\lfloor {g{h}^{-1}}\right\rfloor = q \) . | Yes |
Consider the subring \( \mathbb{Z}\left\lbrack \sqrt{-3}\right\rbrack \) of the complex numbers, which consists of all complex numbers of the form \( a + b\sqrt{-3} \), where \( a, b \in \mathbb{Z} \). As this is a subring of the field \( \mathbb{C} \), it is an integral domain (one may also view \( \mathbb{Z}\left\lbr... | Let us first determine the units in \( \mathbb{Z}\left\lbrack \sqrt{-3}\right\rbrack \) . For \( a, b \in \mathbb{Z} \), we have \( N\left( {a + b\sqrt{-3}}\right) = \) \( {a}^{2} + 3{b}^{2} \), where \( N \) is the usual norm map on \( \mathbb{C} \) (see Example 7.5). If \( \alpha \in \mathbb{Z}\left\lbrack \sqrt{-3}\... | Yes |
Theorem 16.34. Suppose \( D \) satisfies part (i) of Definition 16.33, and that \( D/{pD} \) is an integral domain for every irreducible \( p \in D \) . Then \( D \) is a UFD. | Proof. Exercise. | No |
Both \( \mathbb{Z} \) and \( F\left\lbrack X\right\rbrack \) are Euclidean domains. | In \( \mathbb{Z} \), we can take the ordinary absolute value function \( \left| \cdot \right| \) as a size function, and for \( F\left\lbrack X\right\rbrack \), the function \( \deg \left( \cdot \right) \) will do. | Yes |
Let us show that this is a Euclidean domain, using the usual norm map \( N \) on complex numbers (see Example 7.5) for the size function. Let \( \alpha ,\beta \in \mathbb{Z}\left\lbrack i\right\rbrack \), with \( \beta \neq 0 \) . We want to show the existence of \( \kappa ,\rho \in \mathbb{Z}\left\lbrack i\right\rbrac... | Suppose that in the field \( \mathbb{C} \), we compute \( \alpha {\beta }^{-1} = r + {si} \), where \( r, s \in \mathbb{Q} \) . Let \( m, n \) be integers such that \( \left| {m - r}\right| \leq 1/2 \) and \( \left| {n - s}\right| \leq 1/2 \) -such integers \( m \) and \( n \) always exist, but may not be uniquely dete... | Yes |
Theorem 16.36. If \( D \) is a Euclidean domain and \( I \) is an ideal of \( D \), then there exists \( d \in D \) such that \( I = {dD} \) . | Proof. If \( I = \{ 0\} \), then \( d = 0 \) does the job, so let us assume that \( I \neq \{ 0\} \) . Let \( d \) be any non-zero element of \( I \) such that \( S\left( d\right) \) is minimal, where \( S \) is a size function that makes \( D \) into a Euclidean domain. We claim that \( I = {dD} \) .\n\nIt will suffic... | Yes |
Theorem 16.38. If \( D \) is a PID, and \( {I}_{1} \subseteq {I}_{2} \subseteq \cdots \) are ideals of \( D \), then there exists an integer \( k \) such that \( {I}_{k} = {I}_{k + 1} = \cdots \) . | Proof. Let \( I \mathrel{\text{:=}} \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{I}_{i} \), which is an ideal of \( D \) (see Exercise 7.37). Thus, \( I = {dD} \) for some \( d \in D \) . But \( d \in \mathop{\bigcup }\limits_{{i = 1}}^{\infty }{I}_{i} \) implies that \( d \in {I}_{k} \) for some \( k \), which shows t... | No |
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