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The graph of a linear function \( f\left( {x, y}\right) = {ax} + {by} \) is the plane \( z = {ax} + {by} \) . The gradient is the vector \( \nabla f = \left\lbrack \begin{array}{l} a \\ b \end{array}\right\rbrack \) of partial derivatives. The length of that vector is \( \parallel \nabla \mathbf{f}\parallel = \sqrt{{a}... | That steepest direction is perpendicular to the level direction. The level direction \( z = \) constant has \( {ax} + {by} = \) constant. It is the safe direction to walk, perpendicular to \( \nabla f \) . The component of \( \nabla f \) in that flat direction is zero. Figure VI. 7 shows the two perpendicular direction... | Yes |
Example 1.1 (Recording Maps by Words) Given two finite sets \( X = \{ 1,2,\ldots, n\} \) , \( Y = \{ 1,2,\ldots, m\} \), every map \( f : X \rightarrow Y \) can be represented by a sequence of its values \( w\left( f\right) \overset{\text{ def }}{ = }\left( {f\left( 1\right), f\left( 2\right) ,\ldots, f\left( n\right) ... | \[ w : \operatorname{Hom}\left( {X, Y}\right) \simeq \{ \left| X\right| - \text{ letter words in the alphabet }\mathrm{Y}\} ,\;f \mapsto w\left( f\right) . \] This map takes monomorphisms to words without duplicate letters. Epimorphisms go to words containing the whole alphabet. Isomorphisms go to words in which every ... | Yes |
Proposition 1.1 \( \\left| {\\operatorname{Hom}\\left( {X, Y}\\right) }\\right| = {\\left| Y\\right| }^{\\left| X\\right| } \) for all finite sets \( X, Y \) . | Proof Fix an arbitrary point \( x \\in X \) and consider the evaluation map\n\n\[ \n{\\operatorname{ev}}_{x} : \\operatorname{Hom}\\left( {X, Y}\\right) \\rightarrow Y,\\;f \\mapsto f\\left( x\\right) ,\n\]\n\nwhich takes the map \( f : X \\rightarrow Y \) to its value at \( x \) . The maps \( X \\rightarrow Y \) with ... | Yes |
Proposition 1.2 Let \( \left| X\right| = \left| Y\right| = n \) . We write \( \operatorname{Isom}\left( {X, Y}\right) \subset \operatorname{Hom}\left( {X, Y}\right) \) for the set of all bijections \( X \simeq Y \) . Then \( \left| {\operatorname{Isom}\left( {X, Y}\right) }\right| = n! \), where \( n!\overset{\text{ de... | Proof For every \( x \in X \), the restriction of the evaluation map (1.3) to the subset of bijections assigns the surjective map \( {\mathrm{{ev}}}_{x} : \operatorname{Isom}\left( {X, Y}\right) \rightarrow Y, f \mapsto f\left( x\right) \) . The bijections \( f : X \rightarrow Y \) with a prescribed value \( y = f\left... | Yes |
Example 1.4 (Residue Classes) Fix a nonzero \( n \in \mathbb{Z} \) and write \( \mathbb{Z}/\left( n\right) \) for the quotient of \( \mathbb{Z} \) modulo the congruence relation (1.15). The elements of \( \mathbb{Z}/\left( n\right) \) are called residue classes modulo \( n \) . The class of a number \( z \in \mathbb{Z}... | The factorization map\n\n\[ \mathbb{Z} \rightarrow \mathbb{Z}/\left( n\right) ,\;z \mapsto {\left\lbrack z\right\rbrack }_{n} \]\n\nis called reduction modulo \( n \) . The set \( \mathbb{Z}/\left( n\right) \) consists of the \( n \) elements \( {\left\lbrack 0\right\rbrack }_{n},{\left\lbrack 1\right\rbrack }_{n},\ldo... | Yes |
Example 1.5 (Fractions) The set of rational numbers \( \mathbb{Q} \) is usually introduced as the set of fractions \( a/b \), where \( a, b \in \mathbb{Z}, b \neq 0 \) . By definition, such a fraction is an equivalence class of the pair \( \left( {a, b}\right) \in \mathbb{Z} \times \left( {\mathbb{Z} \smallsetminus 0}\... | Hence, relations (1.19) give a complete explicit description for the equivalence generated by relations (1.18). | Yes |
Example 1.6 (Endomorphisms of a Two-Element Set) The set \( X = \{ 1,2\} \) has four endomorphisms. Let us record maps \( f : X \rightarrow X \) by two-letter words \( \left( {f\left( 1\right), f\left( 2\right) }\right) \) as in Example 1.1 on p. 3. Then the four endomorphisms \( X \) are \( \left( {1,1}\right) ,\left(... | \[ \begin{array}{l} f \smallsetminus g \mid \left( {1,1}\right) \left( {1,2}\right) \left( {2,1}\right) \left( {2,2}\right) \\ \left( {1,1}\right) \left( {1,1}\right) \left( {1,1}\right) \left( {1,1}\right) \left( {1,1}\right) \end{array} \] \[ \left( {1,2}\right) \mid \left( {1,1}\right) \left( {1,2}\right) \left( {2,... | Yes |
Lemma 1.1 (Left Inverse Map) The following conditions on a map \( f : X \rightarrow Y \) are equivalent:\n\n1. \( f \) is injective;\n\n2. there exists a map \( g : Y \rightarrow X \) such that \( {gf} = {\operatorname{Id}}_{X} \) (any such \( g \) is called a left inverse to \( f \) );\n\n3. for any two maps \( {g}_{1... | Proof We verify the implications \( \left( 1\right) \Rightarrow \left( 2\right) \Rightarrow \left( 3\right) \Rightarrow \left( 1\right) \) . Let \( f \) be injective. For \( y = f\left( x\right) \), put \( g\left( y\right) = x \) . For \( y \notin \operatorname{im}f \), define \( g\left( y\right) \) arbitrarily. Then \... | Yes |
Proposition 1.3 The following properties of a map \( f : X \rightarrow Y \) are equivalent:\n\n(1) \( f \) is bijective;\n\n(2) there exists a map \( g : Y \rightarrow X \) such that \( f \circ g = {\operatorname{Id}}_{Y} \) and \( g \circ f = {\operatorname{Id}}_{X} \) ;\n\n(3) there exist maps \( {g}^{\prime },{g}^{\... | Proof If (1) holds, then \( g = {f}^{-1} \) satisfies (2). Implication (2) \( \Rightarrow \) (3) is obvious. Conversely, if (3) holds, then \( {g}^{\prime } = {g}^{\prime } \circ {\operatorname{Id}}_{Y} = {g}^{\prime } \circ \left( {f \circ {g}^{\prime \prime }}\right) = \left( {{g}^{\prime } \circ f}\right) \circ {g}^... | Yes |
For \( X = \{ 1,2,\ldots, n\} \), the group \( \operatorname{Aut}\left( X\right) \) is denoted by \( {S}_{n} \) and called the \( n \) th symmetric group or the permutation group of \( n \) elements. By Proposition 1.2, \( \left| {S}_{n}\right| = n \) !. | We will indicate a permutation \( \sigma \in {S}_{n} \) by the row \( \left( {{\sigma }_{1},{\sigma }_{2},\ldots ,{\sigma }_{n}}\right) \) of its values \( {\sigma }_{i} = \sigma \left( i\right) \), as in Example 1.1. For example,\n\n\[ \sigma = \left( {3,4,2,1}\right) \text{ and }\tau = \left( {2,3,4,1}\right) \]\n\ne... | No |
Lemma 1.2 For every poset \( P \), there is no map \( \beta : \mathcal{W}\left( P\right) \rightarrow P \) sending each \( W \in \mathcal{W}\left( P\right) \) to some exterior upper bound of \( W \) . | Proof Let such a map \( \beta \) exist. We will say that \( W \in \mathcal{W}\left( P\right) \) is \( \beta \) -stable if \( \beta \left( {\lbrack y}\right) ) = y \) for all \( y \in W \) . For example, the set \( \{ \beta \left( \varnothing \right) ,\beta \left( {\{ \beta \left( \varnothing \right) \} }\right) ,\beta ... | No |
Corollary 1.1 (Zorn's Lemma I) Suppose that every well-ordered subset in a poset \( P \) has an upper bound, not necessarily exterior. Then there exists a maximal element in \( P \) . | Proof Assume the contrary. Then for all \( x \in P \) there exists \( y > x \) . Hence, the axiom of choice allows us to choose some exterior upper bound \( {}^{12}b\left( W\right) \) for every \( W \in \mathcal{W}\left( P\right) \) . The resulting map \( W \mapsto b\left( W\right) \) contradicts Lemma 1.2. | No |
Lemma 1.3 (Zorn's Lemma II) Every complete poset \( P \) has a maximal element. | Proof Every complete poset surely satisfies the assumption of Corollary 1.1. | No |
Problem 1.18 Show that Zorn's lemma, Lemma 1.3, is equivalent to the axiom of choice. More precisely, assume that Lemma 1.3 holds for every poset \( P \) and prove that every surjective map \( f : X \rightarrow Y \) admits a section. | Hint: consider the set of maps \( {g}_{U} : U \rightarrow X \) such that \( U \subset Y \) and \( f{g}_{U} = {\operatorname{Id}}_{U} \) ; equip it with a partial order, where \( {g}_{U} \leq {g}_{W} \) means that \( U \subset W \) and \( {\left. {g}_{W}\right| }_{U} = {g}_{U} \) ; verify that Lemma 1.3 can be applied; ... | No |
Example 2.2 (Rational Numbers) The field of rational numbers \( \mathbb{Q} \) is the main motivating example for Definition 2.1. As a set, \( \mathbb{Q} \) consists of fractions \( a/b \), which are equivalence classes \( {}^{3} \) of pairs \( \left( {a, b}\right) \), where \( a, b \in \mathbb{Z}, b \neq 0 \), modulo t... | This equivalence is exhausted by the relations\n\n\[ \left( {{a}_{1},{b}_{1}}\right) \sim \left( {{a}_{2},{b}_{2}}\right) \;\text{ for all }\;{a}_{1}{b}_{2} = {a}_{2}{b}_{1}, \]\n\nand each relation (2.12) can be achieved by at most a two-step chain of relations (2.11). Addition and multiplication of fractions are defi... | Yes |
Example 2.4 (Geometric Vectors) In the framework of Euclidean geometry as studied in high school, let us declare two directed segments to be equivalent if they are parallel displacements of each other. The equivalence classes of directed segments are called geometric vectors. The zero vector, i.e., the class of the emp... | Commutativity and associativity of addition are established by means of the parallelogram and quadrangle diagrams shown in Figs. 2.1 and 2.2:\n\n\n\nFig. 2.1 The parallelogram rule\n\n![459d24fc-b77c-4c8d-9d8f-b9b75b55... | Yes |
Lemma 2.1 In every abelian group \( A \), the neutral element is unique, and for each \( a \in A \), its opposite \( - a \) is uniquely determined by \( a \) . In particular, \( - \left( {-a}\right) = a \) . | Proof Let us write + for the operation in \( A \) . If there are two identity elements \( {0}_{1} \) and \( {0}_{2} \), then \( {0}_{1} = {0}_{1} + {0}_{2} = {0}_{2} \), where the first equality holds because \( {0}_{2} \) is an identity element, and the second holds because \( {0}_{1} \) is an identity element. If the... | Yes |
In every commutative ring \( K \), the equality \( 0 \cdot a = 0 \) holds for all \( a \in K \) . If \( K \) has a unit, then for every \( a \in A \), the product \( \left( {-1}\right) \cdot a \) equals the opposite element of \( a \) . | Let \( a \cdot 0 = b \) . Then \( b + a = a \cdot 0 + a = a \cdot 0 + a \cdot 1 = a\left( {0 + 1}\right) = a \cdot 1 = a \) . Adding \( \left( {-a}\right) \) to both sides, we get \( b = 0 \) . The second statement follows from the computation \( \left( {-1}\right) \cdot a + a = \left( {-1}\right) \cdot a + 1 \cdot a =... | Yes |
Lemma 2.3 Let \( K \) be an arbitrary commutative ring with unit and let \( a, b \in K \) be coprime. Then for every \( c \in K \) , \[ b \mid {ac}\; \Rightarrow \;b \mid c, \] (2.19) \[ a\left| {c\& b}\right| c \Rightarrow {ab} \mid c. \] (2.20) Furthermore, if \( a \in K \) is coprime to each of \( {b}_{1},{b}_{2},\l... | Proof Multiplying both sides of \( {ax} + {by} = 1 \) by \( c \), we get the equality \( c = {acx} + {bcy} \) , which gives both implications (2.19),(2.20). If for each \( i = 1,2,\ldots, n \), there exist \( {x}_{i},{y}_{i} \in K \) such that \( a{x}_{i} + {b}_{i}{y}_{i} = 1 \), then by multiplying all these equalitie... | Yes |
For a prime \( p \in \mathbb{N} \), there is a remarkable identity in the residue class field \( {\mathbb{F}}_{p} = \mathbb{Z}/\left( p\right) \), namely\n\n\[\n\underset{p\text{ times }}{\underbrace{1 + 1 + \cdots + 1}} = 0.\n\]\n\n(2.22)\n\nIt forces the sum of \( m \) ones to vanish as soon \( m \) is a multiple of ... | Indeed, by Lemma 2.3 on p. 26, for such \( k \), the number \( p \) is coprime to the product in the denominator. Then by the same lemma, the denominator divides the product \( \left( {p - 1}\right) \cdots \left( {p - k + 1}\right) \) . Therefore, the entire quotient is divisible by \( p \) .\n\nWe conclude that in \( ... | Yes |
Theorem 2.1 (Fermat’s Little Theorem) \( {a}^{p} \equiv a\left( {\;\operatorname{mod}\;p}\right) \) for every \( a \in \mathbb{Z} \) and prime \( p \in \mathbb{N} \) . | Proof We have to show that \( \left\lbrack {a}^{p}\right\rbrack = \left\lbrack a\right\rbrack \) in \( {\mathbb{F}}_{p} \) . This follows immediately from (2.23):\n\n\[ \n{\left\lbrack a\right\rbrack }^{p} = {\left( \underset{a\text{ times }}{\underbrace{\left\lbrack 1\right\rbrack + \left\lbrack 1\right\rbrack + \cdot... | Yes |
For every homomorphism of abelian groups \( \varphi : A \rightarrow B \), the equalities\n\n\[ \varphi \left( 0\right) = 0\;\text{ and }\;\varphi \left( {-a}\right) = - \varphi \left( a\right) \;\left( {\text{ for all }a \in A}\right) \]\n\nhold. In particular, \( \operatorname{im}\left( A\right) = \varphi \left( A\rig... | Proof Since \( \varphi \left( 0\right) = \varphi \left( {0 + 0}\right) = \varphi \left( 0\right) + \varphi \left( 0\right) \), subtraction of \( \varphi \left( 0\right) \) from both sides forces \( 0 = \varphi \left( 0\right) \) . The second equality is verified by the computation \( \varphi \left( a\right) + \varphi \... | Yes |
For every homomorphism of abelian groups \( \varphi : A \rightarrow B \) and every element \( b = f\left( a\right) \in \operatorname{im}A \) , \[ {f}^{-1}\left( b\right) = a + \ker \varphi = \left\{ {a + {a}^{\prime } \mid {a}^{\prime } \in \ker \varphi }\right\} ,\] i.e., the fiber of \( \varphi \) over \( b \) is a s... | The conditions \( \varphi \left( {a}_{1}\right) = \varphi \left( {a}_{2}\right) \) and \( \varphi \left( {{a}_{1} - {a}_{2}}\right) = 0 \) are equivalent. | No |
Lemma 2.5 Let \( \varphi : A \rightarrow B \) be a nonzero homomorphism of commutative rings with unit. If \( B \) has no zero divisors, then \( \varphi \left( 1\right) = 1 \) . | Proof Since \( \varphi \left( 1\right) = \varphi \left( {1 \cdot 1}\right) = \varphi \left( 1\right) \cdot \varphi \left( 1\right) \), the equality \( \varphi \left( 1\right) \left( {1 - \varphi \left( 1\right) }\right) = 0 \) holds in the integral domain \( B \) . Hence, either \( \varphi \left( 1\right) = 1 \) as req... | Yes |
Proposition 2.2 Every nonzero homomorphism of a field to a ring is injective. | Proof If \( \varphi \left( a\right) = 0 \) for some \( a \neq 0 \), then \( \varphi \left( b\right) = \varphi \left( {b{a}^{-1}a}\right) = \varphi \left( {b{a}^{-1}}\right) \varphi \left( a\right) = 0 \) for all \( b \) . Thus, a nonzero \( \varphi \) has zero kernel. | Yes |
To demonstrate the effectiveness of the above procedure, let us find the smallest positive integer with remainders \( {r}_{1} = 2,{r}_{2} = 7,{r}_{3} = {43} \) on division by \( {n}_{1} = {57},{n}_{2} = {91},{n}_{3} = {179} \) respectively. | We first invert \( {91} \cdot {179} \) modulo 57 . Since \( {91} \cdot {179} \equiv {34} \cdot 8 \equiv - {13}\left( {\;\operatorname{mod}\;{57}}\right) \), we can apply the Euclidean algorithm to \( {E}_{0} = {57},{E}_{1} = {13} \) . The output \( {22} \cdot {13} - 5 \cdot {57} = 1 \) (check!) means that \( - {22} \cd... | Yes |
Problem 2.18 (Euler’s Formula for \( \varphi \) ) A function \( f : \mathbb{N} \rightarrow \mathbb{C} \) is called a multiplicative character if \( f\left( {mn}\right) = f\left( m\right) f\left( n\right) \) for every coprime \( m, n \in \mathbb{Z} \) . Show that Euler’s function \( {}^{17}\varphi \) is a multiplicative... | \[ \varphi \left( n\right) = n \cdot \left( {1 - {p}_{1}^{-1}}\right) \cdots \left( {1 - {p}_{n}^{-1}}\right) . \] | Yes |
Proposition 3.1 A power series \( f\left( x\right) = {a}_{0} + {a}_{1}x + {a}_{2}{x}^{2} + \cdots \in K\llbracket x\rrbracket \) is invertible in \( K\llbracket x\rrbracket \) if and only if its constant term \( {a}_{0} \in K \) is invertible in \( K \) . The inversion map \( f \mapsto {f}^{-1} \) is a unary algebraic ... | Proof If there exists \( {f}^{-1}\left( x\right) = {b}_{0} + {b}_{1}x + {b}_{2}{x}^{2} + \cdots \) such that \( f\left( x\right) \cdot {f}^{-1}\left( x\right) = 1 \), then \( {a}_{0}{b}_{0} = 1 \), i.e., \( {a}_{0} \) is invertible. Conversely, let \( {a}_{0} \in K \) be invertible. A comparison of coefficients at the ... | Yes |
Example 3.1 (Series with Zero Derivative) Now assume that \( K \) is an integral domain. If the characteristic \( {}^{2} \) char \( K \) is equal to 0, then formula (3.7) implies that \( {f}^{\prime } = 0 \) if and only if \( f = \) const. However, if \( \operatorname{char}K = p > 0 \), the derivation kills exactly all... | Moreover, the same is true in the subring of polynomials \( K\left\lbrack x\right\rbrack \subset K\llbracket x\rrbracket \) . | No |
Lemma 3.1 Over any prime \( p \in \mathbb{N} \), the polynomials with zero derivative in \( {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) are exhausted by the pth powers \( {g}^{p}, g \in {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) . | Proof Since the Frobenius endomorphism \( {}^{3}{F}_{p} : {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \rightarrow {\mathbb{F}}_{p}\left\lbrack x\right\rbrack, h \mapsto {h}^{p} \), acts identically on the coefficients, for every \( g\left( x\right) = {b}_{0}{x}^{m} + {b}_{1}{x}^{m - 1} + \cdots + {b}_{m - 1}x + {b}_{m}... | Yes |
Proposition 3.2 (Differentiation Rules) For a commutative ring \( K \) with unit, the following equalities hold:\n\n\[{\left( \alpha f\right) }^{\prime } = \alpha \cdot {f}^{\prime }\;\text{ for every }\alpha \in K\text{ and }f, g \in K\llbracket x\rrbracket ,\]\n\n(3.8)\n\n\[{\left( f + g\right) }^{\prime } = {f}^{\pr... | Proof The first two equalities follow directly from (3.7). To verify the Leibniz rule (3.10), write\n\n\[f\left( {x + t}\right) = f\left( x\right) + t \cdot {f}^{\prime }\left( x\right) + \left( \right. \text{terms divisible by}{t}^{2}\text{),}\]\n\n\[g\left( {x + t}\right) = g\left( x\right) + t \cdot {g}^{\prime }\le... | Yes |
Proposition 3.3 (Division with Remainder) Let \( K \) be a commutative ring with unit and \( u \in K\left\lbrack x\right\rbrack \) a polynomial with invertible leading coefficient. Then for a given polynomial \( f \in K\left\lbrack x\right\rbrack \), there exist polynomials \( q, r \in K\left\lbrack x\right\rbrack \) s... | Proof Let \( u = {b}_{0}{x}^{k} + {b}_{1}{x}^{k - 1} + \cdots + {b}_{k - 1}x + {b}_{k} \) . If \( \deg f < k \), we can take \( q = 0 \) , \( r = f \) . For \( f = {a}_{0}{x}^{n} + {a}_{1}{x}^{n - 1} + \cdots + {a}_{n - 1}x + {a}_{n} \), where \( n \geq k \), assume inductively that \( q \) and \( r \) exist for all po... | Yes |
For a polynomial \( f\left( x\right) = {a}_{n}{x}^{n} + {a}_{n - 1}{x}^{n - 1} + \cdots + {a}_{1}x + {a}_{0} \in K\left\lbrack x\right\rbrack \), the remainder on division of \( f \) by a linear binomial \( x - \alpha \) has degree at most zero, i.e., is a constant. | Substitution of \( x = \alpha \) in the equality \( f\left( x\right) = \left( {x - \alpha }\right) \cdot q\left( x\right) + r \) leads to \( r = f\left( \alpha \right) \) . Thus, the value of \( f \) at \( \alpha \in K \) is equal to the remainder on division of \( f \) by \( x - \alpha \) . | Yes |
For a field \( \mathbb{k} \) and collection of polynomials \( {f}_{1},{f}_{2},\ldots ,{f}_{n} \in \mathbb{k}\left\lbrack x\right\rbrack \) , there exists a unique monic polynomial \( d \in \mathbb{k}\left\lbrack x\right\rbrack \) dividing all \( {f}_{i} \) and divisible by every common divisor of all the \( {f}_{i} \) ... | Proof Existence is established by the same arguments as in Sect. 2.4.3 on p.28. Write\n\n\[ \n\left( {{f}_{1},{f}_{2},\ldots ,{f}_{n}}\right) \overset{\text{ def }}{ = }\left\{ {{f}_{1}{h}_{1} + {f}_{2}{h}_{2} + \cdots + {f}_{n}{h}_{n} \mid {h}_{i} \in \mathbb{k}\left\lbrack x\right\rbrack }\right\} \n\]\n\nfor the set... | Yes |
Example 3.3 Let us carry out the Euclidean algorithm for \( {f}_{1}\left( x\right) = {x}^{7} + 3{x}^{6} + 4{x}^{5} + {x}^{4} + 3{x}^{3} + 5{x}^{2} + {3x} + 4 \) and \( {f}_{2}\left( x\right) = {x}^{5} + 5{x}^{4} + {11}{x}^{3} + {12}{x}^{2} + {7x} + 4 \) : | \[ {E}_{0} = {x}^{7} + 3{x}^{6} + 4{x}^{5} + {x}^{4} + 5{x}^{2} + 3{x}^{3} + {3x} + 4, \] \[ {E}_{1} = {x}^{5} + 5{x}^{4} + {11}{x}^{3} + {12}{x}^{2} + {7x} + 4, \] \[ {E}_{2} = - 4{x}^{4} - {13}{x}^{3} - {21}{x}^{2} - {10x} - 8 = {E}_{0} - \left( {{x}^{2} - {2x} + 3}\right) {E}_{1}; \] it is more convenient first to d... | Yes |
Proposition 3.5 Let \( K \) be an integral domain. If a polynomial \( f \in K\left\lbrack x\right\rbrack \) has \( s \) distinct roots \( {\alpha }_{1},{\alpha }_{2},\ldots ,{\alpha }_{s} \in K \), then \( f \) is divisible by \( \mathop{\prod }\limits_{i}\left( {x - {\alpha }_{i}}\right) \) in \( K\left\lbrack x\right... | Proof Write \( f \) as \( f\left( x\right) = \left( {x - {\alpha }_{1}}\right) \cdot q\left( x\right) \) and substitute \( x = {\alpha }_{2},{\alpha }_{3},\ldots ,{\alpha }_{s} \) in this equality. Since \( \left( {{\alpha }_{i} - {\alpha }_{1}}\right) \neq 0 \) for all \( i \neq 1 \) and \( K \) has no zero divisors, ... | Yes |
Corollary 3.3 Let \( K \) be an integral domain and suppose \( f, g \in K\left\lbrack x\right\rbrack \) are each of degree at most \( n \) . If \( f\left( {\alpha }_{i}\right) = g\left( {\alpha }_{i}\right) \) for more than \( n \) distinct \( {\alpha }_{i} \in K \), then \( f = g \) in \( K\left\lbrack x\right\rbrack ... | Proof Since \( f - g \) has more than \( n \) roots but \( \deg \left( {f - g}\right) \leq n \), it must be the zero polynomial. | Yes |
Proposition 3.6 Let \( \mathbb{k} \) be a field, \( f \in \mathbb{k}\left\lbrack x\right\rbrack \), and \( \alpha \in \mathbb{k} \) a root of \( f \) . Then \( \alpha \) is a multiple root if and only if \( {f}^{\prime }\left( \alpha \right) = 0 \) . | Proof If \( \alpha \) is multiple root, then \( f\left( x\right) = {\left( x - \alpha \right) }^{2}g\left( x\right) \) . Differentiation of both sides leads to \( {f}^{\prime }\left( x\right) = \left( {x - \alpha }\right) \left( {{2g}\left( x\right) + \left( {x - \alpha }\right) {g}^{\prime }\left( x\right) }\right) \)... | Yes |
Proposition 3.7 Let \( \operatorname{char}\mathbb{k} = 0 \) . A root \( \alpha \in \mathbb{k} \) of a polynomial \( f \in \mathbb{k}\left\lbrack x\right\rbrack \) has multiplicity \( m \geq 2 \) if and only if \( \alpha \) is an \( \left( {m - 1}\right) \) -tuple root of \( {f}^{\prime } \) . As a consequence, \( \alph... | Proof If \( f\left( x\right) = {\left( x - \alpha \right) }^{m}g\left( x\right) \), then \( {f}^{\prime }\left( x\right) = {\left( x - \alpha \right) }^{m - 1}\left( {{mg}\left( x\right) + \left( {x - \alpha }\right) {g}^{\prime }\left( x\right) }\right) \) . For \( g\left( \alpha \right) \neq 0 \) and since \( m \neq ... | Yes |
Example 3.4 (Irreducible Polynomials) Let \( f \in \mathbb{k}\left\lbrack x\right\rbrack \) be irreducible. Then \( f \) is coprime to all nonzero polynomials of smaller degree. | Therefore, an irreducible polynomial \( f \) is separable as soon as \( {f}^{\prime } \neq 0 \) . Since for char \( \mathbb{k} = 0 \) and \( \deg f > 0 \) we always have \( {f}^{\prime } \neq 0 \), every irreducible polynomial over a field of characteristic zero is separable. If char \( \mathbb{k} = p > 0 \), then \( {... | No |
Proposition 3.8 For a field \( \mathbb{k} \) and nonconstant polynomial \( f \in \mathbb{k}\left\lbrack x\right\rbrack \), the residue class ring \( \mathbb{k}\left\lbrack x\right\rbrack /\left( f\right) \) is a field if and only if \( f \) is irreducible in \( \mathbb{k}\left\lbrack x\right\rbrack \) . | Proof If \( f = {gh} \), where \( \deg g,\deg h < \deg f \), then both classes \( {\left\lbrack g\right\rbrack }_{p},{\left\lbrack h\right\rbrack }_{p} \) are nonzero but have zero product in \( \mathbb{k}\left\lbrack x\right\rbrack /\left( f\right) \) . This prevents the latter from being a field. If \( f \) is irredu... | Yes |
Proposition 3.9 (Chinese Remainder Theorem) Let \( \mathbb{k} \) be an arbitrary field and \( f \in \mathbb{k}\left\lbrack x\right\rbrack \) a product of \( m \) mutually coprime polynomials:\n\n\[ f = {f}_{1}{f}_{2}\cdots {f}_{m}\text{, where}\forall i \neq j,\;\operatorname{GCD}\left( {{f}_{i},{f}_{j}}\right) = 1\tex... | Proof (of Proposition 3.9) It remains to verify that \( \varphi \) is surjective, i.e., that for every collection of residue classes \( \left\lbrack {r}_{i}\right\rbrack \in \mathbb{k}\left\lbrack x\right\rbrack /\left( {f}_{i}\right) \), there exists a polynomial \( g \in \mathbb{k}\left\lbrack x\right\rbrack \) such ... | Yes |
For a field \( \mathbb{k} \) and polynomial \( f \in \mathbb{k}\left\lbrack x\right\rbrack \) of positive degree, there exists a field \( \mathbb{F} \supset \mathbb{k} \) such that in \( \mathbb{F}\left\lbrack x\right\rbrack \), the polynomial \( f \) can be factored completely into a product of \( \deg f \) linear fac... | Proof By induction on \( n = \deg f \) . For \( n = 1 \), the statement is trivial. Assume that it holds for all fields \( \mathbb{k} \) and all polynomials of degree less than \( n \) . Consider a field \( \mathbb{k} \) , and let \( f \in \mathbb{k}\left\lbrack x\right\rbrack \) have degree \( n \) . If \( f = {gh} \)... | Yes |
Lemma 3.2 The radial vectors of the points \( z \in {\mathbb{R}}^{2} \) form a field with respect to the usual addition of vectors \( {}^{11} \) and multiplication defined by the rule that lengths are multiplied, arguments are added, i.e., by the formulas\n\n\[ \left| {{z}_{1}{z}_{2}}\right| \triangleq \left| {z}_{1}\r... | Proof (of Lemma 3.2) We have seen in Example 2.4 on p. 22 that vectors form an additive abelian group. Multiplication (3.19) is clearly commutative and associative. The unit direction vector of the \( {OX} \) -axis, which has length 1 and argument 0, is the neutral element for multiplication. The inverse of a nonzero v... | Yes |
Let us take \( z = \cos \varphi + i\sin \varphi \) and compute \( {z}^{n} = \) \( \cos \left( {n\varphi }\right) + i\sin \left( {n\varphi }\right) \) by expanding \( {\left( \cos \varphi + i\sin \varphi \right) }^{n} \) via the binomial formula (1.7) on p. 6. | \[ \cos \left( {n\varphi }\right) + i\sin \left( {n\varphi }\right) = {\left( \cos \varphi + i\sin \varphi \right) }^{n} \] \[ = {\cos }^{n}\varphi + i\left( \begin{array}{l} n \\ 1 \end{array}\right) {\cos }^{n - 1}\varphi \sin \varphi - \left( \begin{array}{l} n \\ 2 \end{array}\right) {\cos }^{n - 2}\varphi {\sin }^... | Yes |
Example 3.6 (The Equation \( {z}^{n} = a \) ) The complex roots of the equation \( {z}^{n} = a \) , where \( a = \left| a\right| \cdot \left( {\cos \alpha + i\sin \alpha }\right) \neq 0 \), are the numbers | \[ {z}_{k} = \sqrt[n]{\left| a\right| } \cdot \left( {\cos \frac{\alpha + {2\pi k}}{n} + i \cdot \sin \frac{\alpha + {2\pi k}}{n}}\right) ,\;0 \leq k \leq n - 1. \] | Yes |
Theorem 3.2 Every finite subgroup \( A \) of the multiplicative group \( {\mathbb{k}}^{ * } \) of a field \( \mathbb{k} \) is cyclic. | Proof Let \( m = \mathop{\max }\limits_{{b \in A}}\operatorname{ord}\left( b\right) \) . If the order of every element in \( A \) divides \( m \), then all of them are roots of the polynomial \( {x}^{m} - 1 = 0 \) . Since this polynomial has at most \( m \) roots in the field \( \mathbb{k} \) by Corollary 3.2 on p. 50,... | Yes |
Lemma 3.3 For finite fields \( \mathbb{k} \subset \mathbb{F} \), there exists \( n \in \mathbb{N} \) such that \( \left| \mathbb{F}\right| = {\left| \mathbb{k}\right| }^{n} \) . | Proof We use induction \( {}^{20} \) on the difference \( \left| \mathbb{F}\right| - \left| \mathbb{k}\right| \) . The case \( \left| \mathbb{F}\right| = \left| \mathbb{k}\right| \) is trivial. If there exists \( \zeta \in \mathbb{F} \smallsetminus \mathbb{k} \), then \( \zeta \) is algebraic over \( \mathbb{k} \), bec... | Yes |
Theorem 3.3 For every \( n \in \mathbb{N} \) and prime \( p \in \mathbb{N} \), there exists a finite field \( {\mathbb{F}}_{q} \) of characteristic \( p \) and cardinality \( q = {p}^{n} \) . | Proof Consider the polynomial \( f\left( x\right) = {x}^{q} - x \in {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) . By Theorem 3.1, there exists a field \( \mathbb{F} \supset {\mathbb{F}}_{p} \) such that \( f \) acquires \( n \) roots in \( \mathbb{F} \) . Since \( {f}^{\prime } \equiv 1 \), all these roots are disti... | "Yes" |
Theorem 3.4 Two finite fields are isomorphic if and only if they have equal cardinalities. | Proof Let \( \mathbb{F} \) be a field with \( \left| \mathbb{F}\right| = q \) and char \( \mathbb{F} = p \) . Then \( q = {p}^{n} \) by Corollary 3.4. It is enough to show that \( \mathbb{F} \) is isomorphic to the field \( {\mathbb{F}}_{q} \) constructed in the proof of Theorem 3.3. The multiplicative group \( {\mathb... | Yes |
Show that for every prime \( p \in \mathbb{N} \), the Legendre-Jacobi symbol \( {}^{31}\left( \frac{n}{p}\right) \) is a multiplicative character \( {}^{32} \) of \( n \) and evaluate \( \mathop{\sum }\limits_{{n = 1}}^{{p - 1}}\left( \frac{n}{p}\right) \) . Then compare the sign of \( \left( \frac{m}{p}\right) \) with... | Then take \( m = q \) in this product, factorize each fraction as in Problem 3.24 on p. 68, and prove that for every prime \( q \in \mathbb{N} \) ,\n\n\[ \left( \frac{p}{q}\right) \cdot \left( \frac{q}{p}\right) = {\left( -1\right) }^{\frac{p - 1}{2}\frac{q - 1}{2}}.\] | No |
Lemma 4.1 The equality \( a/r = b/t \) in \( K{S}^{-1} \) holds if and only if there exists \( s \in S \) such that ats \( = \) brs in \( K \) . | Proof Let us provisionally write \( \left( {a, r}\right) \approx \left( {b, t}\right) \) if at \( s = {brs} \) for some \( s \in S \) . This relation is contained in \( { \sim }_{S} \), because it can be achieved in two steps by means of the relations generating \( { \sim }_{S} \) as follows: \( \left( {a, r}\right) \s... | Yes |
Lemma 4.2 Addition and multiplication of fractions are well defined by the rules\n\n\\[ \n\\frac{a}{r} + \\frac{b}{s} \\triangleq \\frac{{as} + {br}}{rs},\\;\\frac{a}{r} \\cdot \\frac{b}{s} \\triangleq \\frac{ab}{rs}, \n\\]\n\n(4.1)\n\nand they provide \\( K{S}^{-1} \\) with the structure of a commutative ring with uni... | Proof The consistency of the definitions (4.1) means that the results of the operations are unchanged after replacement of \\( \\frac{a}{r} \\) and \\( \\frac{b}{s} \\) by \\( \\frac{au}{ru} \\) and \\( \\frac{bw}{sw} \\) respectively, where \\( u, w \\in S \\) . This is obvious:\n\n\\[ \n\\frac{au}{ru} + \\frac{bw}{sw... | No |
Theorem 4.1 The map \( {\iota }_{S} : K \rightarrow K{S}^{-1}, a \mapsto a/1 \), is a ring homomorphism with kernel \( \ker {\iota }_{S} = \{ a \in K \mid \exists s \in S : {as} = 0\} \) . Every element of \( {\iota }_{S}\left( S\right) \) is invertible in \( K{S}^{-1} \) . For every ring homomorphism \( \varphi : K \r... | Proof It is clear that \( {\iota }_{S} \) respects both ring operations. Given \( s \in S \), the inverse to \( {\iota }_{S}\left( s\right) = s/1 \) is \( 1/s \) . The fraction \( {\iota }_{S}\left( a\right) = a/1 \) equals \( 0/1 \) if or only if there exists \( s \in S \) such that \( a \cdot 1 \cdot s = 0 \cdot 1 \c... | Yes |
Example 4.2 (Laurent Series) For a field \( \mathbb{k} \), the ring of formal power series \( \mathbb{k}\llbracket x\rrbracket \) is an integral domain. Its field of fractions \( {Q}_{\mathbb{k}\left\lbrack x\right\rbrack } \) is described as follows. Every power series \( q\left( x\right) \in \mathbb{k}\llbracket x\rr... | We conclude that the field of fractions \( {Q}_{\mathbb{k}\llbracket x\rrbracket } \) coincides with the localization of \( \mathbb{k}\llbracket x\rrbracket \) in the multiplicative system of powers \( {x}^{m} \) . The latter consists of formal power series with integer exponents bounded from below:\n\n\[ f\left( x\rig... | Yes |
Proposition 4.1 Assume that the denominator of the simplified representation \( f/g \) is factorized as \( g = {g}_{1} \cdot {g}_{2}\cdots {g}_{m} \), where \( \operatorname{GCD}\left( {{g}_{i}{g}_{j}}\right) = 1 \) for all \( i \neq j \) and all \( {g}_{i} \) are monic. Then the fraction \( f/g \) is uniquely expanded... | Proof Write \( {G}_{i} = g/{g}_{v} \) for the product of all \( {g}_{v} \) except \( {g}_{i} \) . Then (4.4) is equivalent to \[ f = {hg} + {f}_{1}{G}_{1} + {f}_{2}{G}_{2} + \cdots + {f}_{m}{G}_{m}, \] where \( \deg \left( {{f}_{1}{G}_{1} + {f}_{2}{G}_{2} + \cdots + {f}_{m}{G}_{m}}\right) < \deg g \), because of the ab... | Yes |
Proposition 4.2 Every simplified fraction \( f/{g}^{m} \), where \( \deg f < \deg \left( {g}^{m}\right) \) and \( g \) is monic, is uniquely expanded in \( \mathbb{k}\left( x\right) \) as a sum of simplified fractions\n\n\[ \frac{f}{{g}^{m}} = \frac{{f}_{1}}{g} + \frac{{f}_{2}}{{g}^{2}} + \cdots + \frac{{f}_{m}}{{g}^{m... | Proof The expansion (4.5) is equivalent to the expansion\n\n\[ f = {f}_{1}{g}^{m - 1} + {f}_{2}{g}^{m - 2} + \cdots + {f}_{m - 1}g + {f}_{m}, \]\n\n(4.6)\n\nwhich is nothing but the representation of \( f \) in \( g \) -adic notation, where \( {f}_{m} \) is the remainder on division of \( f \) by \( g,{f}_{m - 1} \) is... | Yes |
Let us compute the antiderivative \( {}^{4} \) and 2016th derivative of \( 1/\left( {1 + {x}^{2}}\right) \) . | The partial fraction expansion of \( 1/\left( {1 + {x}^{2}}\right) \) in \( \mathbb{C}\left( x\right) \) looks like\n\n\[ \frac{1}{1 + {x}^{2}} = \frac{\alpha }{1 + {ix}} + \frac{\beta }{1 - {ix}}\text{, where }\alpha ,\beta \in \mathbb{C}. \]\n\nSubstituting \( x = \pm i \) in the equality \( 1 = \alpha \left( {1 - {i... | Yes |
Let us find the \( k \) th element of the Fibonacci sequence \( {z}_{k} \) defined recursively as\n\n\[ \n{z}_{0} = 0,\;{z}_{1} = 1,\;{z}_{k} = {z}_{k - 1} + {z}_{k - 2}\;\text{ for }\;k \geq 2.\n\] | This sequence satisfies a linear recurrence equation of second order: \( {z}_{k} - {z}_{k - 1} - \) \( {z}_{k - 2} = 0 \) . The equality (4.16) becomes \( x\left( {1 - x - {x}^{2}}\right) = {b}_{0} + {b}_{1} + \cdots \) and gives \( {b}_{0} = 0,{b}_{1} = 1 \) . Thus, \( {z}_{k} \) equals the \( k \) th coefficient in t... | Yes |
Proposition 4.3 Let the polynomial (4.8) of the linear order-n recurrence equation \( {z}_{k} + {a}_{1}{z}_{k - 1} + {a}_{2}{z}_{k - 2} + \cdots + {a}_{n}{z}_{k - n} = 0 \), where \( {a}_{i} \in \mathbb{k} \), be completely factorized in \( \mathbb{k}\left\lbrack t\right\rbrack \) as \( {t}^{n} + {a}_{1}{t}^{n - 1} + \... | Proof The generating series \( \sum {z}_{k}{x}^{k} \in \mathbb{k}\llbracket x\rrbracket \) of any solution is a power series expansion for a sum of partial fractions of the form \( \beta \cdot {\left( 1 - \alpha x\right) }^{-m} \), where \( \alpha \) is a root of the polynomial (4.8), the integer \( m \) is in the rang... | Yes |
Lemma 4.3 For all \( u, w \in U \), the equalities \( u = w,{u}^{\prime } = {w}^{\prime },\log \left( u\right) = \log \left( w\right) \) , \( {u}^{\prime }/u = {w}^{\prime }/w \) are equivalent. | Proof The first equality implies all the others. For two power series \( u, w \) with equal constant terms, the first two equalities are equivalent by the differentiation formula (4.17). Replacing \( u, w \) by \( \log u,\log w \), we get the equivalence of the last two equalities. It remains to deduce the first equali... | Yes |
Theorem 4.2 The maps (4.22) and (4.20) taking the exponential and logarithm of power series \n\n(4.23)\n\nare isomorphisms of abelian groups each the inverse of the other. That is, the equalities\n\n\\[ \n\\log {e}^{... | Proof Differentiation of both sides immediately verifies the equality \\( \\log {e}^{\\tau } = \\tau \\) . After that, taking the logarithms of both sides verifies the equality \\( {e}^{\\log u} = u \\) . Therefore, the maps (4.23) are bijections that are inverses to each other. The power series \\( \\log \\left( {{u}_... | Yes |
Example 4.5 (Binomial with Rational Exponent) If the exponent \( \alpha \) is equal to \( n \in \) \( \mathbb{N} \), then the numerator (4.27) acquires zero factors for all \( k > n \), and the binomial expansion (4.28) becomes finite: | \[ {\left( 1 + x\right) }^{n} = 1 + {nx} + \frac{n\left( {n - 1}\right) }{2}{x}^{2} + \cdots + {x}^{n} = \mathop{\sum }\limits_{{k = 0}}^{n}\left( \begin{array}{l} n \\ k \end{array}\right) \cdot {x}^{k}. \] | Yes |
Let us use the square root expansion (4.29) to deduce an explicit formula for the Catalan numbers, which appear in many combinatorial problems. The product of \( \\left( {n + 1}\\right) \) quantities\n\n\\[ \n{a}_{0}{a}_{1}{a}_{2}\\cdots {a}_{n}\\;\\text{(an}n\\text{-fold product)} \n\\]\n\n(4.30)\n\ncan be computed in... | For \( n \\geq 2 \), the set of all admissible arrangements of \( n \) pairs of parentheses splits into \( n \) disjoint classes in accordance with the position of the penultimate pair of parentheses:\n\n\\[ \n\\left( {{a}_{0}\\left( {{a}_{2}\\ldots {a}_{n}}\\right) }\\right) ,\\left( {\\left( {{a}_{0}{a}_{1}}\\right) ... | Yes |
Example 4.7 (Shift Operators) The Appell polynomials of exponent \( {e}^{x} = 1 + x + \) \( \frac{1}{2}{x}^{2} + \frac{1}{6}{x}^{3} + \cdots \) are | \[ \mathop{\sum }\limits_{{k \geq 0}}\frac{1}{k!}{D}^{k}{t}^{m} = \mathop{\sum }\limits_{{k \geq 0}}\frac{m\left( {m - 1}\right) \cdots \left( {m - k + 1}\right) }{k!}{t}^{m - k} \] \[ = \mathop{\sum }\limits_{{k = 0}}^{m}\left( \begin{array}{l} m \\ k \end{array}\right) {t}^{m - k} = {\left( t + 1\right) }^{m}. \] Hen... | Yes |
We are looking for polynomials \( {S}_{m}\left( t\right) \in \) \( \mathbb{Q}\left\lbrack t\right\rbrack \), numbered by integers \( m \geq 0 \), such that\n\n\[ \n{S}_{m}\left( n\right) = {0}^{m} + {1}^{m} + {2}^{m} + {3}^{m} + \cdots + {n}^{m} = \mathop{\sum }\limits_{{k = 0}}^{n}{k}^{m} \n\]\n\nfor all integers \( n... | To analyze the general case, let us consider the difference operator \( \nabla = 1 - {e}^{-D} \) : \( \varphi \left( t\right) \mapsto \varphi \left( t\right) - \varphi \left( {t - 1}\right) \) . If a required polynomial \( {S}_{m}\left( t\right) \) exists, then\n\n\[ \n\nabla {S}_{m}\left( t\right) = {t}^{m} \n\]\n\nbe... | Yes |
For an algebraically closed field \( \mathbb{k} \) of zero characteristic, the field of Puiseux series in \( x \) with coefficients in \( \mathbb{k} \) is algebraically closed too. | We give two proofs of Theorem 4.3. The first, short and conceptual, goes back to van der Waerden and Hensel. The second, which allows us to expand implicit algebraic functions in Puiseux series effectively, was the original discovery of Newton. | No |
Lemma 4.5 Let \( \mathbb{k} \) be an algebraically closed field of zero characteristic. Then for every polynomial\n\n\[ F\left( {t, x}\right) = {a}_{n}\left( t\right) {x}^{n} + {a}_{n - 1}\left( t\right) {x}^{n - 1} + \cdots + {a}_{0}\left( x\right) \in \mathbb{k}\left( \left( t\right) \right) \left\lbrack x\right\rbra... | Proof Without loss of generality, we may assume that the coefficients of \( F \) lie in \( \mathbb{k}\llbracket t\rrbracket \), that the leading coefficient is \( {a}_{n} = 1 \), and that the next coefficient is \( {a}_{n - 1} = 0 \) . The first is achieved via multiplication of \( F \) by an appropriate power of \( t ... | Yes |
Example 4.9 (Counterexample to Theorem 4.3 in Positive Characteristic) The proof of Lemma 4.5 essentially uses the assumption char \( \mathbb{k} = 0 \), without which both Lemma 4.5 and Theorem 4.3 fail. To demonstrate this, put \( \mathbb{k} = {\mathbb{F}}_{p} \) and consider the equation \( {x}^{p} - x = {t}^{-1} \) ... | Since \( {c}^{p} = c \) for all \( c \in {\mathbb{F}}_{p} \), the substitution of \( x = x\left( t\right) \) into \( t{x}^{p} - {tx} = 1 \) leads to\n\n\[ - {c}_{1}{t}^{{\lambda }_{1} + 1} + {c}_{2}{t}^{{\lambda }_{2} + 1} - {c}_{1}{t}^{p{\lambda }_{1} + 1} + {c}_{3}{t}^{{\lambda }_{3} + 1} - {c}_{2}{t}^{p{\lambda }_{2... | Yes |
Proposition 4.5 Every output series \( x\left( t\right) \) of Newton’s method applied to the polynomial \( F\left( {t, x}\right) \) is a Puiseux series that satisfies the equation \( {}^{15}F\left( {t, x\left( t\right) }\right) = 0 \) in \( \mathbb{k}\llbracket t\rrbracket \) . | Proof Let us show that the exponents of an output series \( x\left( t\right) \) have a common denominator. It is enough to verify that all \( {\varepsilon }_{i} \) but a finite number are integers. As we\n\n\( {}^{15} \) Note that this gives another proof of Lemma 4.5. have just seen, the denominator of \( {\varepsilon... | No |
Problem 4.4 Let \( g\left( x\right) = \prod \left( {x - {\alpha }_{i}}\right) \), where all \( {\alpha }_{i} \) are distinct. For a polynomial \( f \in \mathbb{k}\left\lbrack x\right\rbrack \) such that \( \deg f < \deg g \), prove the following partial fraction expansion \( {}^{16} \) : | \[ f\left( x\right) /g\left( x\right) = \sum \frac{f\left( {\alpha }_{i}\right) /{g}^{\prime }\left( {\alpha }_{i}\right) }{\left( x - {\alpha }_{i}\right) }\text{, where }{g}^{\prime } = \frac{d}{dx}g. \] | Yes |
Proposition 5.1 A commutative ring \( K \) with unit is a field if and only if there are no nontrivial ideals in \( K \) . | Proof An ideal in a field is trivial by Exercise 5.1. Conversely, if a nonzero ideal coincides with \( K \), then \( \left( b\right) = K \) for every \( b \neq 0 \) . Hence, \( 1 = {sb} \) for some \( s \in K \) . Thus, all \( b \neq 0 \) are invertible. | No |
Lemma 5.1 The following properties of a commutative ring \( K \) are equivalent:\n\n(1) Every subset \( M \subset K \) contains some finite collection of elements \( {a}_{1},{a}_{2},\ldots ,{a}_{k} \in \) \( M \) such that \( \left( M\right) = \left( {{a}_{1},{a}_{2},\ldots ,{a}_{k}}\right) \) .\n\n(2) Every ideal \( I... | Proof Clearly, \( \left( 1\right) \Rightarrow \left( 2\right) \) . To deduce (3) from (2), write \( I = \bigcup {I}_{v} \) for the union of all ideals in the chain. Then \( I \) is an ideal as well. By (2), \( I \) is generated by some finite set of its elements. All these elements belong to some \( {I}_{n} \) . Theref... | Yes |
Theorem 5.1 (Hilbert's Basis Theorem) For every Noetherian commutative ring \( K \), the polynomial ring \( K\left\lbrack x\right\rbrack \) is Noetherian as well. | Proof Consider an arbitrary ideal \( I \subset K\left\lbrack x\right\rbrack \) and write \( {L}_{d} \subset K \) for the set of leading coefficients of all polynomials of degree \( \leq d \) in \( I \) including the zero polynomial. Also write \( {L}_{\infty } = { \cup }_{d}{L}_{d} \) for the set of all leading coeffic... | Yes |
Corollary 5.2 Every infinite system of polynomial equations with coefficients in a Noetherian ring \( K \) is equivalent to some finite subsystem. | Proof Since \( K\left\lbrack {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right\rbrack \) is Noetherian, among the right-hand sides of a polynomial equation system\n\n\[ \n{f}_{v}\left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) = 0 \n\]\n\nthere is some finite collection \( {f}_{1},{f}_{2},\ldots ,{f}_{m} \) that generates the sam... | Yes |
Example 5.1 (Non-Noetherian Rings) Consider a countably infinite set of variables \( {x}_{i} \) numbered by \( i \in \mathbb{N} \) and define the polynomial ring \( \mathbb{Q}\left\lbrack {{x}_{1},{x}_{2},{x}_{3},\ldots }\right\rbrack \) in these variables to be the set of all finite sums of finite monomials \( {x}_{{v... | Thus, the ideal \( I \) is not finitely generated, and the ring \( \mathbb{Q}\left\lbrack {{x}_{1},{x}_{2},{x}_{3},\ldots }\right\rbrack \) is not Noetherian. The less artificial rings \( \mathcal{C}\left( \mathbb{R}\right) \) and \( {\mathcal{C}}^{\infty }\left( \mathbb{R}\right) \) of all continuous and all infinitel... | Yes |
Example 5.2 (Image of a Ring Homomorphism) It follows from Sect. 2.6.4 on p. 33 that the image of a ring homomorphism \( \varphi : {K}_{1} \rightarrow {K}_{2} \) is isomorphic to the quotient ring \( {K}_{1}/\ker \left( \varphi \right) \) . | The isomorphism sends an element \( b = \varphi \left( a\right) \in \operatorname{im}\varphi \) to the coset \( {\left\lbrack a\right\rbrack }_{\ker \varphi } = {\varphi }^{-1}\left( b\right) \) . Therefore, every homomorphism of rings \( \varphi : {K}_{1} \rightarrow {K}_{2} \) can be decomposed as the quotient epimor... | Yes |
Corollary 5.3 Every finitely generated commutative algebra A over a Noetherian ring \( K \) is Noetherian, and all polynomial relations between generators of A follow from some finite set of those relations. | Proof The result follows immediately from Exercise 5.9 and Corollary 5.1. | No |
Theorem 5.2 Every Euclidean domain \( K \) is a principal ideal domain. | Proof Given a nonzero ideal \( I \subset K \), write \( d \in I \) for some nonzero element of lowest degree. Clearly, \( \left( d\right) \subset I \) . Each \( a \in I \) can be written as \( a = {dq} + r \), where either \( r = 0 \) or \( v\left( r\right) < v\left( d\right) \) . The latter is impossible by the choice... | Yes |
Proposition 5.2 The following properties of an element \( p \) in a principal ideal domain \( K \) are equivalent:\n\n(1) The quotient ring \( K/\left( p\right) \) is a field.\n\n(2) The quotient ring \( K/\left( p\right) \) has no zero divisors.\n\n(3) The element \( p \) is irreducible. | Proof The implication (1) \( \Rightarrow \) (2) holds trivially for every commutative ring \( {}^{11}\mathrm{\;K} \) . The implication \( \left( 2\right) \Rightarrow \left( 3\right) \) holds for every integral domain \( {}^{12}K \) . Indeed, if \( p = {ab} \) in \( K \), then \( \left\lbrack a\right\rbrack \left\lbrack... | Yes |
Proposition 5.3 If \( K \) is Noetherian, then every noninvertible element \( a \in K \) is a finite product of irreducible elements. | Proof If \( a \) is irreducible, there is nothing to do. If not, we write \( a \) as a product of two noninvertible elements and repeat the process on each factor. If this procedure stops after a finite number of iterations, we get the required irreducible factorization. If not, we can form an infinite sequence of elem... | Yes |
Example 5.5 (Non-unique Factorization Domain) The ring \( \mathbb{Z}\left\lbrack \sqrt{5}\right\rbrack \overset{\text{ def }}{ = }\mathbb{Z}\left\lbrack x\right\rbrack /\left( {{x}^{2} - 5}\right) \) consists of elements \( a + b\sqrt{5} \), where \( a, b \in \mathbb{Z} \) and \( \sqrt{5}\overset{\text{ def }}{ = }\lef... | In analogy with the complex numbers, let us call the elements \( \vartheta = a + b\sqrt{5} \) and \( \bar{\vartheta } = a - b\sqrt{5} \) conjugates and introduce the norm \( \parallel \vartheta \parallel \triangleq \vartheta \cdot \bar{\vartheta } = {a}^{2} - 5{b}^{2} \in \mathbb{Z} \) .\n\nThe above exercise implies t... | Yes |
Proposition 5.4 A Noetherian integral domain \( K \) is a unique factorization domain if and only if every irreducible element in \( K \) is prime. | Proof Let \( K \) be a unique factorization domain and \( q \in K \) irreducible. If \( q \) divides some product \( {ab} \), then an irreducible factorization of \( {ab} \) contains a factor associated with \( q \) . On the other hand, an irreducible factorization of \( {ab} \) is the product of irreducible factorizat... | Yes |
Corollary 5.5 Every principal ideal domain is a unique factorization domain. | Proof This follows at once from Proposition 5.2 on p. 111 | No |
Lemma 5.2 For all \( f, g \in K\left\lbrack x\right\rbrack \), the equality \( \operatorname{cont}\left( {fg}\right) = \operatorname{cont}\left( f\right) \cdot \operatorname{cont}\left( g\right) \) holds. | Proof Since \( K \) is a unique factorization domain, it is enough to check that for every irreducible \( q \in K \), the following statement holds: \( q \) divides \( \operatorname{cont}\left( {fg}\right) \) if and only if \( q \) divides \( \operatorname{cont}\left( f\right) \cdot \operatorname{cont}\left( g\right) \... | Yes |
Lemma 5.3 (Simplified Form of a Polynomial) Every polynomial \( f\left( x\right) \in {Q}_{K}\left\lbrack x\right\rbrack \) can be written as\n\n\[ f\left( x\right) = \frac{a}{b} \cdot {f}_{\text{red }}\left( x\right) \]\n\nwhere \( {f}_{\text{red }} \in K\left\lbrack x\right\rbrack, a, b \in K \), and \( \operatorname{... | Proof Factor the lowest common denominator from the coefficients of \( f \) . Then factor out the greatest common divisor from the numerators. We get a number \( c \in {Q}_{K} \) multiplied by a polynomial of content 1 with coefficients in \( K \) . Denote this polynomial by \( {f}_{\text{red }} \in K\left\lbrack x\rig... | Yes |
Lemma 5.4 (Gauss’s Lemma) Every irreducible polynomial \( f \in K\left\lbrack x\right\rbrack \) remains irreducible in \( {Q}_{K}\left\lbrack x\right\rbrack \) . | Proof Since \( f \) is irreducible, \( \operatorname{cont}\left( f\right) = 1 \) . Let \( f\left( x\right) = g\left( x\right) \cdot h\left( x\right) \) in \( {Q}_{K}\left\lbrack x\right\rbrack \) . Write \( g \) and \( h \) in the simplified form (5.11) and simplify the resulting constant factor. Then we get the equali... | Yes |
Theorem 5.3 If \( K \) is a unique factorization domain, then the polynomial ring \( K\left\lbrack x\right\rbrack \) is a unique factorization domain as well. | Proof Since the principal ideal domain \( {Q}_{K}\left\lbrack x\right\rbrack \) is a unique factorization domain, every \( f \in K\left\lbrack x\right\rbrack \) is factorized within \( {Q}_{K}\left\lbrack x\right\rbrack \) into a finite product of irreducible polynomials \( {f}_{v} \in {Q}_{K}\left\lbrack x\right\rbrac... | Yes |
Let us show that the polynomial \( f\left( x\right) = {x}^{5} + {x}^{2} + 1 \) is irreducible in \( \mathbb{Z}\left\lbrack x\right\rbrack \) | Since \( f \) has no integer roots, a nontrivial factorization \( f = {gh} \) in \( \mathbb{Z}\left\lbrack x\right\rbrack \) is possible only for \( \deg \left( g\right) = 2 \) and \( \deg \left( h\right) = 3 \) . Reduction of the coefficients modulo 2 leads to the polynomial \( {\left\lbrack f\right\rbrack }_{2} = {x}... | Yes |
Assume that \( f \in \mathbb{Z}\left\lbrack x\right\rbrack \) is monic and every coefficient of \( f \) except the leading coefficient is divisible by a prime \( p \in \mathbb{N} \) . Let \( f\left( x\right) = g\left( x\right) h\left( x\right) \) in \( \mathbb{Z}\left\lbrack x\right\rbrack \) . Since reduction modulo \... | This observation is known as Eisenstein's criterion. | Yes |
Example 5.9 (Cyclotomic Polynomial \( {\Phi }_{p} \) ) Eisenstein’s criterion allows us to see easily that for every prime \( p \in \mathbb{N} \), the cyclotomic polynomial \( {}^{19} \n\n\[ \n{\Phi }_{p}\left( x\right) = {x}^{p - 1} + {x}^{p - 2} + \cdots + x + 1 = \frac{{x}^{p} - 1}{x - 1} \n\] \n\nis irreducible in ... | Just pass to the new variable \( t = x - 1 \) . Then \n\n\[ \nf\left( t\right) = {\Phi }_{p}\left( {t + 1}\right) = \frac{{\left( t + 1\right) }^{p} - 1}{t} = {t}^{p - 1} + \left( \begin{array}{l} p \\ 1 \end{array}\right) {t}^{p - 2} + \cdots + \left( \begin{matrix} p \\ p - 1 \end{matrix}\right) \n\] \n\nsatisfies Ei... | Yes |
Example 6.4 (Coordinate Plane) The simplest example of a nonzero vector space different from \( \mathbb{k} \) is the coordinate plane \( {\mathbb{k}}^{2} = \mathbb{k} \times \mathbb{k} \) . By definition, it consists of ordered pairs of numbers arranged in columns of height two:\n\n\[ v = \left( \begin{array}{l} {x}_{1... | Addition and multiplication by scalars are defined componentwise:\n\n\[ \lambda \left( \begin{array}{l} {a}_{1} \\ {a}_{2} \end{array}\right) + \mu \left( \begin{array}{l} {b}_{1} \\ {b}_{2} \end{array}\right) \overset{\text{ def }}{ = }\left( \begin{array}{l} \lambda {a}_{1} + \mu {b}_{1} \\ \lambda {a}_{2} + \mu {b}_... | Yes |
Any two nonproportional vectors \( a, b \in {\mathbb{k}}^{2} \) form a basis of the coordinate plane in the sense that every vector \( v \in {\mathbb{k}}^{2} \) admits a unique representation\n\n\[ v = x \cdot a + y \cdot b\text{, where}x, y \in \mathbb{k}\text{.} \] | Indeed, given such an expression, then in virtue of the relations \( \det \left( {a, a}\right) = \) \( \det \left( {b, b}\right) = 0 \), the evaluation of liner functions (6.10) on the both sides of (6.12) leads to\n\n\[ \det \left( {v, b}\right) = \det \left( {x \cdot a + y \cdot b, b}\right) = x \cdot \det \left( {a,... | Yes |
A coordinate module \( {K}^{n} \) over an arbitrary commutative ring \( K \) with unit is defined in the same way as the coordinate plane from Example 6.4. By definition, a vector of \( {K}^{n} \) is an ordered collection of \( n \) elements \( {}^{6} \) of \( K \) arranged in either columns or rows \( {}^{7} \) of siz... | \[ \left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) ,\;{x}_{i} \in K. \]\n\nAddition of vectors and multiplication of vectors by scalars are defined componentwise:\n\n\[ \left( {{x}_{1},{x}_{2},\ldots ,{x}_{n}}\right) + \left( {{y}_{1},{y}_{2},\ldots ,{y}_{n}}\right) \overset{\text{ def }}{ = }\left( {{x}_{1} + {y}_{1},... | Yes |
Example 6.7 (Matrices) An \( m \times n \) matrix over \( K \) is a rectangular array with \( m \) rows and \( n \) columns,\n\n\[ A = \\left( {a}_{ij}\\right) = \\left( \\begin{matrix} {a}_{11} & {a}_{12} & \\ldots & {a}_{1n} \\\\ {a}_{21} & {a}_{22} & \\ldots & {a}_{2n} \\\\ \\vdots & \\vdots & \\vdots & \\vdots \\\\... | In particular, the coordinate module \( {K}^{n} \) can be thought of either as a module of one-row matrices \( {\\operatorname{Mat}}_{1 \\times n}\\left( K\\right) \) or as a module of one-column matrices \( {\\operatorname{Mat}}_{n \\times 1}\\left( K\\right) \) . We write \( {E}_{ij} \) for a matrix that has 1 in the... | Yes |
For a set \( X \), the ring \( {K}^{X} \) of all functions \( f : X \rightarrow K \) can be considered a \( K \) -module with respect to the standard pointwise addition and multiplication by constants | \[ {f}_{1} + {f}_{2} : x \mapsto {f}_{1}\left( x\right) + {f}_{2}\left( x\right) \;\text{ and }\;{\lambda f} : x \mapsto {\lambda f}\left( x\right) . \] For the finite set \( X = \{ 1,2,\ldots, n\} \), there exists an isomorphism of \( K \) -modules \( {K}^{X} \simeq {K}^{n} \) sending a function \( f : X \rightarrow K... | Yes |
Lemma 6.1 Assume that a \( K \) -module \( V \) is generated by a set \( E \subset V \) . Then \( E \) is a basis of \( V \) if and only if \( E \) is linearly independent. | Proof If \( \sum {\lambda }_{i}{e}_{i} = 0 \) for some \( {e}_{i} \in E \) and some \( {\lambda }_{i} \in K \) not all equal to zero, then the zero vector has two different linear expansions \( 0 = \sum {x}_{i}{e}_{i} = \sum 0 \cdot {e}_{i} \) . Conversely, if \( \sum {x}_{i}{e}_{i} = \sum {y}_{i}{e}_{i} \), where \( {... | No |
Lemma 6.2 (Exchange Lemma) Assume that the vectors \( {w}_{1},{w}_{2},\ldots ,{w}_{m} \) generate the vector space \( V \) and that the vectors \( {u}_{1},{u}_{2},\ldots ,{u}_{k} \in V \) are linearly independent. Then \( m \geq k \), and after appropriate renumbering of the vectors \( {w}_{i} \) and replacing the firs... | Proof Let \( {u}_{1} = {x}_{1}{w}_{1} + {x}_{2}{w}_{2} + \cdots + {x}_{m}{w}_{m} \) . Since the vectors \( {u}_{i} \) are linearly independent, we have \( {u}_{1} \neq 0 \), and therefore, not all the \( {x}_{i} \) equal zero. Renumber the \( {w}_{i} \) in order to have \( {x}_{1} \neq 0 \) . Then\n\n\[ \n{w}_{1} = \fr... | Yes |
In every vector space \( V \) over a field \( \mathbb{k} \), every generating set of vectors contains a basis of \( V \), and every linearly independent set of vectors can be extended to a basis of \( V \). Moreover, all bases of \( V \) have the same cardinality. | Proof We prove the theorem under the additional assumption that \( V \) is finitely generated. In the general case, the proof is similar but uses some transfinite arguments. We sketch those arguments in Sect. 6.2.4 below.\n\nBy Lemma 6.1, the bases of \( V \) are exactly the linearly independent generating sets of vect... | No |
In an n-dimensional vector space \( V \), the following properties of a collection of \( n \) vectors are equivalent:\n\n(1) They are linearly independent.\n\n(2) They span \( V \) .\n\n(3) They form a basis of \( V \) . | Proof Let \( {e}_{1},{e}_{2},\ldots ,{e}_{n} \) be a basis in \( V \) and \( {v}_{1},{v}_{2},\ldots ,{v}_{n} \) the given vectors. If they are linearly independent, then by Lemma 6.2, replacement of all vectors \( {e}_{i} \) by vectors \( {v}_{i} \) leads to a generating collection of vectors. Hence, \( \left( 1\right)... | Yes |
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