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Lemma 3.1. \( 0 \cdot a = 0 \) for every number \( a \) . | Proof. Since \( 0 = 0 + 0 \), the distributive law gives\n\n\[ 0 \cdot a = \left( {0 + 0}\right) \cdot a = \left( {0 \cdot a}\right) + \left( {0 \cdot a}\right) . \]\n\nNow subtract \( 0 \cdot a \) from both sides (that is, use the additive cancellation law) to \( \operatorname{get}0 = 0 \cdot a \) . | Yes |
Lemma 3.2. If \( - a \) is that number which, when added to a, gives 0, then \( \left( {-1}\right) \left( {-a}\right) = a \) . | Proof. The distributive law and Lemma 3.1 give\n\n\[ 0 = 0 \cdot \left( {-a}\right) = \left( {-1 + 1}\right) \left( {-a}\right) = \left( {-1}\right) \left( {-a}\right) + \left( {-a}\right) ; \]\n\nnow add \( a \) to both sides (the additive cancellation law again) to get \( a = \left( {-1}\right) \left( {-a}\right) \) ... | Yes |
Corollary 3.3. \( \;\left( {-1}\right) a = - a \) for every number \( a \) . | Proof. By Lemma 3.2, \( \left( {-1}\right) \left( {-a}\right) = a \) . Multiplying both sides by -1 gives\n\n\[\n\left( {-1}\right) \left( {-1}\right) \left( {-a}\right) = \left( {-1}\right) a.\n\]\n\nBut Lemma 3.2 gives \( \left( {-1}\right) \left( {-1}\right) = 1 \), so that \( - a = \left( {-1}\right) a \) . | Yes |
Proposition 3.5. Lemma 3.1, Lemma 3.2, and Corollary 3.3 hold for every commutative ring. | Proof. Each of these results can be proved using only the defining axioms of a commutative ring. To illustrate, here is a very fussy proof of Lemma 3.1: if \( R \) is a commutative ring and \( a \in R \), then \( 0 \cdot a = 0 \) .\n\nSince \( 0 = 0 + 0 \), the distributive law gives\n\n\[ 0 \cdot a = \left( {0 + 0}\ri... | Yes |
Proposition 3.6 (Binomial Theorem). If \( a, b \in R \), where \( R \) is a commutative ring, then for all \( n \geq 0 \) ,\n\n\[{\left( a + b\right) }^{n} = \mathop{\sum }\limits_{{r = 0}}^{n}\left( \begin{array}{l} n \\ r \end{array}\right) {a}^{r}{b}^{n - r}\] | Proof. Adapt the proof of Proposition 1.15, the binomial theorem in \( \mathbb{Z} \). In particular, define \( {a}^{0} = 1 \) for every \( a \in R \), even for \( a = 0 \). | Yes |
Proposition 3.7. If \( R \) is a commutative ring in which \( 1 = 0 \), then \( R \) has only one element: \( R = \{ 0\} \) . One calls \( R \) the zero ring. | Proof. If \( r \in R \), then \( r = {1r} = {0r} = 0 \), by Proposition 3.5. \( \bullet \) | Yes |
Proposition 3.8. A commutative ring \( R \) is a domain if and only if it is not the zero ring and the product of any two nonzero elements of \( R \) is nonzero. | Proof. Assume that \( R \) is a domain, so that the cancellation law holds. Suppose, by way of contradiction, that there are nonzero elements \( a, b \in R \) with \( {ab} = 0 \) . Proposition 3.5 gives \( 0 \cdot b = 0 \), so that \( {ab} = 0 \cdot b \) . The cancellation law now gives \( a = 0 \) (for \( b \neq 0 \) ... | Yes |
A subring \( S \) of a commutative ring \( R \) is itself a commutative ring. | By hypothesis, \( 1 \in S \) and axiom (vii) in the definition of commutative ring on page 216 holds. We now show that \( S \) is closed under addition; that is, if \( s,{s}^{\prime } \in S \), then \( s + {s}^{\prime } \in S \) . Axiom (ii) in the definition of subring gives \( 0 = 1 - 1 \in S \) . Another application... | Yes |
Lemma 3.13. Let \( R \) be a commutative ring, and let \( a, b, c \) be elements of \( R \). (i) If \( a \mid b \) and \( b \mid c \), then \( a \mid c \). (ii) If \( a \mid b \) and \( a \mid c \), then \( a \) divides every number of the form \( {sb} + {tc} \), where \( s, t \in R \). | Proof. Exercises for the reader. - | No |
Proposition 3.15. Let \( R \) be a domain, and let \( a, b \in R \) be nonzero. Then \( a \mid b \) and \( b \mid a \) if and only if \( b = {ua} \) for some unit \( u \in R \) . | Proof. If \( a \mid b \) and \( b \mid a \), there are elements \( u, v \in R \) with \( b = {ua} \) and \( a = {vb} \) . Substituting, \( b = {ua} = {uvb} \) . Since \( b = {1b} \) and \( b \neq 0 \), the cancellation law in the domain \( R \) gives \( 1 = {uv} \), and so \( u \) is a unit.\n\nConversely, assume that ... | Yes |
Proposition 3.16. If \( a \) is an integer, then \( \left\lbrack a\right\rbrack \) is a unit in \( {\mathbb{I}}_{m} \) if and only if \( a \) and \( m \) are relatively prime. In fact, if \( {sa} + {tm} = 1 \), then \( {\left\lbrack a\right\rbrack }^{-1} = \left\lbrack s\right\rbrack \) . | Proof. If \( \left\lbrack a\right\rbrack \) is a unit in \( {\mathbb{I}}_{m} \), then there is \( \left\lbrack s\right\rbrack \in {\mathbb{I}}_{m} \) with \( \left\lbrack s\right\rbrack \left\lbrack a\right\rbrack = \left\lbrack 1\right\rbrack \) . Therefore, \( {sa} \equiv 1{\;\operatorname{mod}\;m} \), and so there i... | Yes |
Corollary 3.17. If \( p \) is a prime, then every nonzero \( \left\lbrack a\right\rbrack \) in \( {\mathbb{I}}_{p} \) is a unit. | Proof. If \( \left\lbrack a\right\rbrack \neq \left\lbrack 0\right\rbrack \), then \( a ≢ 0{\;\operatorname{mod}\;p} \), and hence \( p \nmid a \) . Therefore, \( a \) and \( p \) are relatively prime because \( p \) is prime. | Yes |
Proposition 3.18. Every field \( F \) is a domain. | Proof. Assume that \( {ab} = {ac} \), where \( a \neq 0 \) . Multiplying both sides by \( {a}^{-1} \) gives \( {a}^{-1}{ab} = {a}^{-1}{ac} \), and so \( b = c \) . | Yes |
Proposition 3.19. The commutative ring \( {\mathbb{I}}_{m} \) is a field if and only if \( m \) is prime. | Proof. If \( m \) is prime, then Corollary 3.17 shows that \( {\mathbb{I}}_{m} \) is a field.\n\nConversely, if \( m \) is composite, then Proposition 3.12 shows that \( {\mathbb{I}}_{m} \) is not a domain. By Proposition 3.18, \( {\mathbb{I}}_{m} \) is not a field. | Yes |
Lemma 3.20. If \( R \) is a domain and \( X = \{ \left( {a, b}\right) \in R \times R : b \neq 0\} \), then the relation \( \equiv \) on \( X \), defined by cross-multiplication:\n\n\[ \left( {a, b}\right) \equiv \left( {c, d}\right) \;\text{ if }\;{ad} = {bc}, \]\n\nis an equivalence relation. | Proof. Verifications of reflexivity and of symmetry are easy. For transitivity, assume that \( \left( {a, b}\right) \equiv \left( {c, d}\right) \) and \( \left( {c, d}\right) \equiv \left( {e, f}\right) \) . Now \( {ad} = {bc} \) gives \( {adf} = \) \( {bcf} \), and \( {cf} = {de} \) gives \( {bcf} = {bde} \) ; thus, \... | Yes |
Proposition 3.23. If \( R \) is a commutative ring and \( r,{s}_{i},{t}_{j} \in R \) for \( i \geq 0 \) and \( j \geq 0 \), then\n\n\[ \left( {{s}_{0} + {s}_{1}r + \cdots }\right) \left( {{t}_{0} + {t}_{1}r + \cdots }\right) = {a}_{0} + {a}_{1}r + \cdots + {a}_{k}{r}^{k} + \cdots ,\]\n\nwhere \( {a}_{k} = \mathop{\sum ... | Proof. Write \( \mathop{\sum }\limits_{i}{s}_{i}{r}^{i} = f\left( r\right) \) and \( \mathop{\sum }\limits_{j}{t}_{j}{r}^{j} = g\left( r\right) \) . Then\n\n\[ f\left( r\right) g\left( r\right) = \left( {{s}_{0} + {s}_{1}r + {s}_{2}{r}^{2} + \cdots }\right) g\left( r\right) \]\n\n\[ = {s}_{0}g\left( r\right) + {s}_{1}{... | Yes |
Lemma 3.24. Let \( R \) be a commutative ring and let \( \sigma ,\tau \in R\left\lbrack x\right\rbrack \) be nonzero polynomials.\n\n(i) Either \( {\sigma \tau } = 0 \) or \( \deg \left( {\sigma \tau }\right) \leq \deg \left( \sigma \right) + \deg \left( \tau \right) \).\n\n(ii) If \( R \) is a domain, then \( {\sigma ... | Proof.\n\n(i) Let \( \sigma = \left( {{s}_{0},{s}_{1},\ldots }\right) \) have degree \( m \), let \( \tau = \left( {{t}_{0},{t}_{1},\ldots }\right) \) have degree \( n \), and let \( {\sigma \tau } = \left( {{a}_{0},{a}_{1},\ldots }\right) \) . It suffices to prove that \( {a}_{k} = 0 \) for all \( k > m + n \) . By de... | Yes |
(i) If \( \sigma = \left( {{s}_{0},{s}_{1},\ldots ,{s}_{j},\ldots }\right) \), then\n\n\[ \n{x\sigma } = \left( {0,{s}_{0},{s}_{1},\ldots ,{s}_{j},\ldots }\right) ; \n\]\n\nthat is, multiplying by \( x \) shifts each coefficient one step to the right. | (i) Write \( x = \left( {{t}_{0},{t}_{1},\ldots ,{t}_{i},\ldots }\right) \), where \( {t}_{1} = 1 \) and all other \( {t}_{i} = 0 \), and let \( {x\sigma } = \left( {{a}_{0},{a}_{1},\ldots ,{a}_{k},\ldots }\right) \) . Now \( {a}_{0} = {t}_{0}{s}_{0} = 0 \) because \( {t}_{0} = 0 \) . If \( k \geq 1 \), then the only n... | Yes |
Proposition 3.27. If \( \sigma = \left( {{s}_{0},{s}_{1},\ldots ,{s}_{n},0,0,\ldots }\right) \), then\n\n\[ \sigma = {s}_{0} + {s}_{1}x + {s}_{2}{x}^{2} + \cdots + {s}_{n}{x}^{n}, \]\n\nwhere each element \( s \in R \) is identified with the polynomial \( \left( {s,0,0,\ldots }\right) \) . | Proof.\n\n\[ \sigma = \left( {{s}_{0},{s}_{1},\ldots ,{s}_{n},0,0,\ldots }\right) \]\n\n\[ = \left( {{s}_{0},0,0,\ldots }\right) + \left( {0,{s}_{1},0,\ldots }\right) + \cdots + \left( {0,0,\ldots ,{s}_{n},0,\ldots }\right) \]\n\n\[ = {s}_{0}\left( {1,0,0,\ldots }\right) + {s}_{1}\left( {0,1,0,\ldots }\right) + \cdots ... | Yes |
Corollary 3.28. Polynomials \( f\left( x\right) = {s}_{0} + {s}_{1}x + {s}_{2}{x}^{2} + \cdots + {s}_{n}{x}^{n} \) and \( g\left( x\right) = {t}_{0} + {t}_{1}x + {t}_{2}{x}^{2} + \cdots + {t}_{m}{x}^{m} \) are equal if and only if \( {s}_{i} = {t}_{i} \) for all \( i \in \mathbb{N} \) . | Proof. We have merely restated the definition of equality of polynomials in terms of the familiar notation. - | No |
Proposition 3.29. The elements of \( F\left( x\right) \) have the form \( f\left( x\right) /g\left( x\right) \), where \( f\left( x\right) \) , \( g\left( x\right) \in F\left\lbrack x\right\rbrack \) and \( g\left( x\right) \neq 0 \) . | Proof. By Theorem 3.21, every element in the fraction field \( F\left( x\right) \) has the form \( f\left( x\right) g{\left( x\right) }^{-1} \) . | Yes |
Proposition 3.30. If \( p \) is a prime, then the field of rational functions \( {\mathbb{F}}_{p}\left( x\right) \) is an infinite field whose prime field is \( {\mathbb{F}}_{p} \) . | Proof. By Proposition 3.25, \( {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) is a domain. Its fraction field \( {\mathbb{F}}_{p}\left( x\right) \) is a field containing \( {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) as a subring, while \( {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) contains \( {\mathbb{F}}_{p} \... | No |
Corollary 3.34. If \( R \) is a commutative ring and \( s \in R \), then the evaluation map \( {e}_{s} : R\left\lbrack x\right\rbrack \rightarrow R \) is a homomorphism. | Proof. If we set \( R = S \) and \( \varphi = {1}_{R} \) in Proposition 3.33, then \( \widetilde{\varphi } = {e}_{s} \) . \( \; \bullet \) | No |
Lemma 3.35. If \( f : A \rightarrow R \) is a ring homomorphism, then, for all \( a \in A \) ,\n\n(i) \( f\left( {a}^{n}\right) = f{\left( a\right) }^{n} \) for all \( n \geq 0 \) ;\n\n(ii) if \( a \) is a unit, then \( f\left( a\right) \) is a unit and \( f\left( {a}^{-1}\right) = f{\left( a\right) }^{-1} \) ;\n\n(iii... | Proof.\n\n(i) If \( n = 0 \), then \( f\left( {a}^{0}\right) = 1 = {\left( f\left( a\right) \right) }^{0} \) ; this follows from our convention that \( {r}^{0} = 1 \) for any ring element \( r \), together with the property \( f\left( 1\right) = 1 \) satisfied by every ring homomorphism. The statement for positive \( n... | Yes |
Corollary 3.36. If \( f : A \rightarrow R \) is a ring homomorphism, then\n\n\[ f\left( {U\left( A\right) }\right) \subseteq U\left( R\right) \]\n\nwhere \( U\left( A\right) \) is the group of units of \( A \) ; if \( f \) is an isomorphism, then there is a group isomorphism\n\n\[ U\left( A\right) \cong U\left( R\right... | Proof. The first statement is just a rephrasing of part (ii) of Lemma 3.35: if \( a \) is a unit in \( A \), then \( f\left( a\right) \) is a unit in \( R \) .\n\nIf \( f \) is an isomorphism, then its inverse \( {f}^{-1} : R \rightarrow A \) is also a ring homomorphism, by Exercise 3.41(i) on page 248; hence, if \( r ... | Yes |
Proposition 3.38. If \( f : A \rightarrow R \) is a ring homomorphism, where \( R \) is a nonzero ring, then \( \operatorname{im}f \) is a subring of \( R \) and \( \ker f \) is a proper subset of \( A \) satisfying the conditions:\n\n(i) \( 0 \in \ker f \) ;\n\n(ii) \( x, y \in \ker f \) implies \( x + y \in \ker f \)... | Proof. If \( r,{r}^{\prime } \in \operatorname{im}f \), then \( r = f\left( a\right) \) and \( {r}^{\prime } = f\left( {a}^{\prime }\right) \) for some \( a,{a}^{\prime } \in A \) . Hence, \( r - {r}^{\prime } = f\left( a\right) - f\left( {a}^{\prime }\right) = f\left( {a - {a}^{\prime }}\right) \in \operatorname{im}f ... | Yes |
Theorem 3.40. Every ideal in \( \mathbb{Z} \) is a principal ideal. | Proof. This is just a restatement of Corollary 1.34. - | No |
Proposition 3.41. If \( R \) is a commutative ring and \( a = {ub} \) for some unit \( u \in R \) , then \( \left( a\right) = \left( b\right) \) . Conversely, if \( R \) is a domain, then \( \left( a\right) = \left( b\right) \) implies \( a = {ub} \) for some unit \( u \in R \) . | Proof. Suppose that \( a = {ub} \) for some unit \( u \in R \) . If \( x \in \left( a\right) \), then \( x = {ra} = \) \( {rub} \in \left( b\right) \) for some \( r \in R \), so that \( \left( a\right) \subseteq \left( b\right) \) . For the reverse inclusion, if \( y \in \left( b\right) \), then \( y = {sb} \) for some... | Yes |
Proposition 3.43. A nonzero commutative ring \( R \) is a field if and only if its only ideals are \( \{ 0\} \) and \( R \) itself. | Proof. Assume that \( R \) is a field. If \( I \neq \{ 0\} \), it contains some nonzero element, and every nonzero element in a field is a unit. Therefore, \( I = R \), by Example 3.42.\n\nConversely, assume that \( R \) is a commutative ring whose only ideals are \( \{ 0\} \) and \( R \) itself. If \( a \in R \) and \... | Yes |
Proposition 3.44. A ring homomorphism \( f : A \rightarrow R \) is an injection if and only if \( \ker f = \{ 0\} \) . | Proof. If \( f \) is an injection, then \( a \neq 0 \) implies \( f\left( a\right) \neq f\left( 0\right) = 0 \), and so \( a \notin \) \( \ker f \) . Therefore, \( \ker f = \{ 0\} \) . Conversely, if \( \ker f = \{ 0\} \) and \( f\left( a\right) = f\left( {a}^{\prime }\right) \) , then \( 0 = f\left( a\right) - f\left(... | Yes |
Corollary 3.45. If \( f : k \rightarrow R \) is a ring homomorphism, where \( R \) is not the zero ring, and if \( k \) is a field, then \( f \) is an injection. | Proof. By the proposition, it suffices to prove that \( \ker f = \{ 0\} \) . But \( \ker f \) is a proper ideal in \( k \), by Proposition 3.38, and Proposition 3.43 shows that \( k \) has only two ideals: \( k \) and \( \{ 0\} \) . Now \( \ker f \neq k \), because \( f\left( 1\right) = 1 \neq 0(R \) is not the zero ri... | Yes |
Theorem 3.46 (Division Algorithm). Let \( R \) be a commutative ring, let \( f\left( x\right) \) , \( g\left( x\right) \in R\left\lbrack x\right\rbrack \), and let the leading coefficient of \( f\left( x\right) \) be a unit in \( R \) .\n\n(i) There are polynomials \( q\left( x\right), r\left( x\right) \in R\left\lbrac... | Proof.\n\n(i) We prove the existence of \( q\left( x\right), r\left( x\right) \in R\left\lbrack x\right\rbrack \) as in the statement. If \( f \mid g \) , then \( g = {qf} \) for some \( q \) ; define the remainder \( r = 0 \), and we are done. If \( f \nmid g \), then consider all (necessarily nonzero) polynomials of ... | Yes |
For every positive integer \( n \), the cyclotomic polynomial \( {\Phi }_{n}\left( x\right) \) is a monic polynomial all of whose coefficients are integers. | The proof is by induction on \( n \geq 1 \) . The base step holds, for \( {\Phi }_{1}\left( x\right) = \) \( x - 1 \) . For the inductive step, we assume that \( {\Phi }_{d}\left( x\right) \) is a monic polynomial with integer coefficients. From the equation \( {x}^{n} - 1 = \mathop{\prod }\limits_{d}{\Phi }_{d}\left( ... | Yes |
Lemma 3.48. Let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \), where \( k \) is a field, and let \( a \in k \) . Then there is \( q\left( x\right) \in k\left\lbrack x\right\rbrack \) with\n\n\[ f\left( x\right) = q\left( x\right) \left( {x - a}\right) + f\left( a\right) . \] | Proof. Use the division algorithm to obtain\n\n\[ f\left( x\right) = q\left( x\right) \left( {x - a}\right) + r \]\n\nthe remainder \( r \) is a constant because \( x - a \) has degree 1 . By Corollary 3.34, \( {e}_{a} : k\left\lbrack x\right\rbrack \rightarrow k \), evaluation at \( a \), is a ring homomorphism:\n\n\[... | Yes |
Proposition 3.49. If \( f\left( x\right) \in k\left\lbrack x\right\rbrack \), where \( k \) is a field, then \( a \in k \) is a root of \( f\left( x\right) \) in \( k \) if and only if \( x - a \) divides \( f\left( x\right) \) in \( k\left\lbrack x\right\rbrack \) . | Proof. If \( a \) is a root of \( f\left( x\right) \) in \( k \), then \( f\left( a\right) = 0 \) and the lemma gives \( f\left( x\right) = \) \( q\left( x\right) \left( {x - a}\right) \) . Conversely, if \( f\left( x\right) = q\left( x\right) \left( {x - a}\right) \), then evaluating at \( a \) gives \( f\left( a\righ... | Yes |
Theorem 3.50. Let \( k \) be a field and let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) .\n\n(i) If \( f\left( x\right) \) has degree \( n \), then \( f\left( x\right) \) has at most \( n \) roots in \( k \) . | Proof.\n\n(i) We prove the statement by induction on \( n \geq 0 \) . If \( n = 0 \), then \( f\left( x\right) \) is a nonzero constant, and so the number of its roots in \( k \) is zero. Now let \( n > 0 \) . If \( f\left( x\right) \) has no roots in \( k \), then we are done, for \( 0 \leq n \) . Otherwise, we may as... | Yes |
Corollary 3.52. Let \( k \) be an infinite field and let \( f\left( x\right) \) and \( g\left( x\right) \) be polynomials in \( k\left\lbrack x\right\rbrack \) . If \( f\left( x\right) \) and \( g\left( x\right) \) determine the same polynomial function, i.e., if \( {f}^{ \circ }\left( a\right) = {g}^{ \circ }\left( a\... | Proof. If \( f\left( x\right) \neq g\left( x\right) \), then the polynomial \( f\left( x\right) - g\left( x\right) \) is nonzero, so that it has some degree, say, \( n \) . Now every element of \( k \) is a root of \( f\left( x\right) - g\left( x\right) \) ; since \( k \) is infinite, this polynomial of degree \( n \) ... | Yes |
Corollary 3.53. Let \( k \) be any (possibly finite) field, let \( f\left( x\right), g\left( x\right) \in k\left\lbrack x\right\rbrack \), and let \( n = \max \{ \deg \left( f\right) ,\deg \left( g\right) \} \) . If there are \( n + 1 \) distinct elements \( a \in k \) with \( f\left( a\right) = g\left( a\right) \), th... | Proof. If \( f\left( x\right) \neq g\left( x\right) \), then \( h\left( x\right) = f\left( x\right) - g\left( x\right) \neq 0 \), and\n\n\[ \deg \left( h\right) \leq \max \{ \deg \left( f\right) ,\deg \left( g\right) \} = n. \]\n\nBy hypothesis, there are \( n + 1 \) elements \( a \in k \) with \( h\left( a\right) = f\... | Yes |
Corollary 3.54 (Lagrange Interpolation). Let \( k \) be a field, and let \( {u}_{0},\ldots ,{u}_{n} \) be distinct elements of \( k \) . Given any list \( {y}_{0},\ldots ,{y}_{n} \) in \( k \), there exists a unique \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) of degree \( \leq n \) with \( f\left( {u}_{i}\r... | Proof. The polynomial \( f\left( x\right) \) displayed in the formula has degree at most \( n \) and \( f\left( {u}_{i}\right) = {y}_{i} \) for all \( i \) . Uniqueness follows from Corollary 3.53. - | Yes |
Lemma 3.55. Let \( f\left( x\right) \) be a nonzero polynomial in \( k\left\lbrack x\right\rbrack \), where \( k \) is a field. If \( h\left( x\right) = {a}_{n}{x}^{n} + \cdots + {a}_{0} \in k\left\lbrack x\right\rbrack \) is a divisor of \( f\left( x\right) \), then \( {a}_{n}^{-1}h\left( x\right) \) is a monic diviso... | Proof. Since \( k \) is a field, \( {a}_{n} \in k \) nonzero implies \( {a}_{n}^{-1} \in k \) . If \( f\left( x\right) = c\left( x\right) h\left( x\right) \) , then \( f\left( x\right) = \left\lbrack {{a}_{n}c\left( x\right) }\right\rbrack \left\lbrack {{a}_{n}^{-1}h\left( x\right) }\right\rbrack \) . Of course, \( \de... | Yes |
Proposition 3.56. If \( k \) is a field, then every pair \( f\left( x\right), g\left( x\right) \in k\left\lbrack x\right\rbrack \) has a gcd. | Proof. Since the theorem is obvious if both \( f\left( x\right) \) and \( g\left( x\right) \) are 0, we may assume that \( f\left( x\right) \neq 0 \) . If \( h\left( x\right) \) is a divisor of \( f\left( x\right) \), then \( \deg \left( h\right) \leq \deg \left( f\right) \) ; a fortiori, \( \deg \left( f\right) \) is ... | No |
Theorem 3.57. If \( k \) is a field and \( f\left( x\right), g\left( x\right) \in k\left\lbrack x\right\rbrack \), then their gcd is a linear combination of \( f\left( x\right) \) and \( g\left( x\right) \). | Proof. We may assume that at least one of \( f \) and \( g \) is not zero (for the gcd is 0 otherwise). Consider the set \( I \) of all the linear combinations:\n\n\[ I = \{ s\left( x\right) f\left( x\right) + t\left( x\right) g\left( x\right) : s\left( x\right), t\left( x\right) \in k\left\lbrack x\right\rbrack \} .\n... | Yes |
Theorem 3.59. If \( k \) is a field, then every ideal \( I \) in \( k\left\lbrack x\right\rbrack \) is a principal ideal. Moreover, if \( I \neq \{ 0\} \), there is a monic polynomial that generates \( I \) . | Proof. If \( I \) consists of 0 alone, take \( d = 0 \) . If there are nonzero polynomials in \( I \), the least integer axiom allows us to choose a polynomial \( d\left( x\right) \in I \) of least degree. As in Lemma 3.55, we may assume that \( d\left( x\right) \) is monic.\n\nWe claim that every \( f \) in \( I \) is... | Yes |
Proposition 3.63. Assume that \( k \) is a field and that \( f\left( x\right), g\left( x\right) \in k\left\lbrack x\right\rbrack \) are nonzero.\n\n(i) \( \left\lbrack {f\left( x\right), g\left( x\right) }\right\rbrack \) is the monic generator of \( \left( {f\left( x\right) }\right) \cap \left( {g\left( x\right) }\rig... | Proof.\n\n(i) Since \( f\left( x\right) \neq 0 \) and \( g\left( x\right) \neq 0 \), we have \( \left( f\right) \cap \left( g\right) \neq 0 \), because \( 0 \neq {fg} \in \) \( \left( f\right) \cap \left( g\right) \) . By Theorem 3.59, \( \left( f\right) \cap \left( g\right) = \left( m\right) \), where \( m \) is the m... | Yes |
Proposition 3.64. If \( k \) is a field, then a nonconstant polynomial \( p\left( x\right) \in k\left\lbrack x\right\rbrack \) is irreducible in \( k\left\lbrack x\right\rbrack \) if and only if \( p\left( x\right) \) has no factorization in \( k\left\lbrack x\right\rbrack \) of the form \( p\left( x\right) = f\left( x... | Proof. If \( p\left( x\right) \) is irreducible, it must be nonconstant. If \( p\left( x\right) = f\left( x\right) g\left( x\right) \) in \( k\left\lbrack x\right\rbrack \), where both factors have degree smaller than \( \deg \left( p\right) \), then neither \( \deg \left( f\right) \) nor \( \deg \left( g\right) \) is ... | Yes |
Proposition 3.65. Let \( k \) be a field and let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) be a quadratic or cubic polynomial. Then \( f\left( x\right) \) is irreducible in \( k\left\lbrack x\right\rbrack \) if and only if \( f\left( x\right) \) does not have a root in \( k \). | Proof. If \( f\left( x\right) \) has a root \( a \) in \( k \), then Proposition 3.49 shows that \( f\left( x\right) \) has an honest factorization, and so it is not irreducible.\n\nConversely, assume that \( f\left( x\right) \) is not irreducible, i.e., there is a factorization \( f\left( x\right) = g\left( x\right) h... | Yes |
If \( k \) is a field, then every nonconstant \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) has a factorization\n\n\[ f\left( x\right) = a{p}_{1}\left( x\right) \cdots {p}_{t}\left( x\right) \]\n\nwhere \( a \) is a nonzero constant and the \( {p}_{i}\left( x\right) \) are monic irreducible polynomials. | We prove the proposition for a polynomial \( f\left( x\right) \) by (the second form of) induction on \( \deg \left( f\right) \geq 1 \) . If \( \deg \left( f\right) = 1 \), then \( f\left( x\right) = {ax} + c = a\left( {x + {a}^{-1}c}\right) \) ; as every linear polynomial, \( x + {a}^{-1}c \) is irreducible, and so it... | Yes |
Lemma 3.67. Let \( k \) be a field, let \( p\left( x\right), f\left( x\right) \in k\left\lbrack x\right\rbrack \), and let \( d\left( x\right) = \left( {p, f}\right) \) be their \( \gcd \) . If \( p\left( x\right) \) is a monic irreducible, then\n\n\[ d\left( x\right) = \left\{ \begin{array}{ll} 1 & \text{ if }p\left( ... | Proof. The only monic divisors of \( p\left( x\right) \) are 1 and \( p\left( x\right) \) . If \( p\left( x\right) \mid f\left( x\right) \), then \( d\left( x\right) = p\left( x\right) \), for \( p\left( x\right) \) is monic. If \( p\left( x\right) \nmid f\left( x\right) \), then the only monic common divisor is 1, and... | Yes |
Theorem 3.68 (Euclid’s Lemma). Let \( k \) be a field and let \( f\left( x\right), g\left( x\right) \in \) \( k\left\lbrack x\right\rbrack \) . If \( p\left( x\right) \) is an irreducible polynomial in \( k\left\lbrack x\right\rbrack \), and if \( p\left( x\right) \mid f\left( x\right) g\left( x\right) \), then \( p\le... | Proof. If \( p \mid f \), we are done. If \( p \nmid f \), then the lemma says that \( \gcd \left( {p, f}\right) = 1 \) . There are thus polynomials \( s\left( x\right) \) and \( t\left( x\right) \) with \( 1 = {sp} + {tf} \), and so\n\n\[ g = {spg} + {tfg}. \]\n\nSince \( p \mid {fg} \), it follows that \( p \mid g \)... | Yes |
Corollary 3.69. Let \( f\left( x\right), g\left( x\right), h\left( x\right) \in k\left\lbrack x\right\rbrack \), where \( k \) is a field, and let \( h\left( x\right) \) and \( f\left( x\right) \) be relatively prime. If \( h\left( x\right) \mid f\left( x\right) g\left( x\right) \), then \( h\left( x\right) \mid g\left... | Proof. By hypothesis, \( {fg} = {hq} \) for some \( q\left( x\right) \in k\left\lbrack x\right\rbrack \) . There are polynomials \( s \) and \( t \) with \( 1 = {sf} + {th} \), and so \( g = {sfg} + {thg} = {shq} + {thg} = h\left( {{sq} + {tg}}\right) \) ; that is, \( h \mid g \) . | Yes |
Proposition 3.70. If \( k \) is a field, every nonzero \( f\left( x\right) /g\left( x\right) \in k\left( x\right) \) can be put in lowest terms. | Proof. If \( d = \left( {f, g}\right) \), then \( f = d{f}^{\prime } \) and \( g = d{g}^{\prime } \) in \( k\left\lbrack x\right\rbrack \) . Moreover, \( {f}^{\prime } \) and \( {g}^{\prime } \) are relatively prime, for if \( h \) were a nonconstant common divisor of \( {f}^{\prime } \) and \( {g}^{\prime } \) , then ... | Yes |
Theorem 3.71 (Euclidean Algorithm). If \( k \) is a field and \( f\left( x\right), g\left( x\right) \in k\left\lbrack x\right\rbrack \) , then there is an algorithm computing the \( \gcd \left( {f\left( x\right), g\left( x\right) }\right) \), and there is an algorithm finding a pair of polynomials \( s\left( x\right) \... | Proof. The proof is just a repetition of the proof of the euclidean algorithm in \( \mathbb{Z} \) : iterated application of the division algorithm.\n\n\[ g = {q}_{0}f + {r}_{1}\;\deg \left( {r}_{1}\right) < \deg \left( f\right) \]\n\n\[ f = {q}_{1}{r}_{1} + {r}_{2}\;\deg \left( {r}_{2}\right) < \deg \left( {r}_{1}\righ... | Yes |
Find the gcd in \( \mathbb{Q}\left\lbrack x\right\rbrack \) of\n\n\[ f\left( x\right) = {x}^{3} - {x}^{2} - x + 1\;\text{ and }\;g\left( x\right) = {x}^{3} + 4{x}^{2} + x - 6. \] | Note that \( f\left( x\right), g\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \), and \( \mathbb{Z} \) is not a field. As we proceed, rational numbers may enter, for \( \mathbb{Q} \) is the smallest field containing \( \mathbb{Z} \). Here are the equations.\n\n\[ g = 1 \cdot f + \left( {5{x}^{2} + {2x} - 7}... | Yes |
Find the gcd in \( {\\mathbb{F}}_{5}\\left\\lbrack x\\right\\rbrack \) of\n\n\\[ f\\left( x\\right) = {x}^{3} - {x}^{2} - x + 1\\;\\text{ and }\\;g\\left( x\\right) = {x}^{3} + 4{x}^{2} + x - 6. \\] | The terms in the euclidean algorithm simplify considerably.\n\n\\[ g = 1 \\cdot f + \\left( {{2x} + 3}\\right) \\]\n\n\\[ f = \\left( {3{x}^{2} + 2}\\right) \\left( {{2x} + 3}\\right) . \\]\n\nThe gcd is \( x - 1 \) (which is \( {2x} + 3 \) made monic). | Yes |
Corollary 3.75. Let \( k \) be a subfield of a field \( K \), so that \( k\left\lbrack x\right\rbrack \) is a subring of \( K\left\lbrack x\right\rbrack \) . If \( f\left( x\right), g\left( x\right) \in k\left\lbrack x\right\rbrack \), then their \( \gcd \) in \( k\left\lbrack x\right\rbrack \) is equal to their \( \gc... | Proof. The division algorithm in \( K\left\lbrack x\right\rbrack \) gives\n\n\[ g\left( x\right) = Q\left( x\right) f\left( x\right) + R\left( x\right) ,\]\n\nwhere \( Q\left( x\right), R\left( x\right) \in K\left\lbrack x\right\rbrack \) and either \( R\left( x\right) = 0 \) or \( \deg \left( R\right) < \deg \left( f\... | Yes |
Proposition 3.78. Every euclidean ring \( R \) is a PID. In particular, the ring \( \mathbb{Z}\left\lbrack i\right\rbrack \) of Gaussian integers is a PID. | Proof. The reader can adapt the proof of Proposition 3.59: if \( I \) is a nonzero ideal in \( R \), then \( I = \left( d\right) \), where \( d \) is an element in \( I \) of least degree. \( \; \bullet \) | No |
Proposition 3.80. If \( R \) is a euclidean ring that is not a field, then \( R \) has a universal side divisor. | Proof. Define\n\n\[ S = \{ \partial \left( v\right) : v \neq 0\text{ and }v\text{ is not a unit }\} \]\n\nwhere \( \partial \) is the degree function on \( R \) . Since \( R \) is not a field, there is some \( v \in {R}^{ \times } \) which is not a unit, and so \( S \) is a nonempty subset of the natural numbers. By th... | Yes |
Proposition 3.81. Let \( R \) be a PID.\n\n(i) Each \( \alpha ,\beta \in R \) has a gcd, \( \delta \), which is a linear combination of \( \alpha \) and \( \beta \) : there are \( \sigma ,\tau \in R \) such that\n\n\[ \delta = {\sigma \alpha } + {\tau \beta } \]\n\n(ii) If an irreducible element \( \pi \in R \) divides... | Proof.\n\n(i) We may assume that at least one of \( \alpha \) and \( \beta \) is not zero (otherwise, the gcd is 0 and the result is obvious). Consider the set \( I \) of all the linear combinations:\n\n\[ I = \{ {\sigma \alpha } + {\tau \beta } : \sigma ,\tau \text{ in }R\} .\n\nNow \( \alpha \) and \( \beta \) are in... | Yes |
Lemma 3.82. If \( p \) is a prime and \( p \equiv 1{\;\operatorname{mod}\;4} \), then there is an integer \( m \) with\n\n\[{m}^{2} \equiv - 1{\;\operatorname{mod}\;p}.\] | Proof. If \( G = {\left( {\mathbb{F}}_{p}\right) }^{ \times } \) is the multiplicative group of nonzero elements in \( {\mathbb{F}}_{p} \) , then \( \left| G\right| = p - 1 \equiv 0{\;\operatorname{mod}\;4} \) ; that is, \( 4 \) is a divisor of \( \left| G\right| \) . By Proposition 2.122, \( G \) contains a subgroup \... | Yes |
Theorem 3.83 (Fermat’s Two-Squares Theorem). \( {}^{14} \) An odd prime \( p \) is a sum of two squares,\n\n\[ p = {a}^{2} + {b}^{2} \]\n\nwhere \( a \) and \( b \) are integers, if and only if \( p \equiv 1{\;\operatorname{mod}\;4} \) . | Proof. For any integer \( a \) we have \( a \equiv r{\;\operatorname{mod}\;4} \), where \( r = 0,1,2 \) or 3, and so \( {a}^{2} \equiv {r}^{2}{\;\operatorname{mod}\;4} \) . But, \( {\;\operatorname{mod}\;4} \),\n\n\[ {0}^{2} \equiv 0,\;{1}^{2} \equiv 1,\;{2}^{2} = 4 \equiv 0,\;\text{ and }\;{3}^{2} = 9 \equiv 1, \]\n\n... | Yes |
Theorem 3.84 (Unique Factorization). If \( k \) is a field, then every polynomial \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) of degree \( \geq 1 \) is a product of a nonzero constant and monic irreducibles. Moreover, if\n\n\[ f\left( x\right) = a{p}_{1}\left( x\right) \cdots {p}_{m}\left( x\right) \;\text{... | Proof. The existence of a factorization of a polynomial \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) into irre-ducibles was proved in Proposition 3.66, and so we need only prove the uniqueness assertion.\n\nAn equation \( f\left( x\right) = a{p}_{1}\left( x\right) \cdots {p}_{m}\left( x\right) \) gives \( a ... | Yes |
Proposition 3.86. Let \( k \) be a field and let \( g\left( x\right) = a{p}_{1}^{{e}_{1}}\cdots {p}_{n}^{{e}_{n}} \in k\left\lbrack x\right\rbrack \) and let \( h\left( x\right) = b{p}_{1}^{{f}_{1}}\cdots {p}_{n}^{{f}_{n}} \in k\left\lbrack x\right\rbrack \), where \( a, b \in k \), the \( {p}_{i} \) are distinct monic... | Proof. The proof is a just an adaptation of the proof of Proposition 1.52. | No |
Lemma 3.87. Let \( k \) be a field, and let \( b\left( x\right) \in k\left\lbrack x\right\rbrack \) have \( \deg \left( b\right) \geq 1 \) . Each nonzero \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) has an expression\n\n\[ f\left( x\right) = {d}_{m}\left( x\right) b{\left( x\right) }^{m} + \cdots + {d}_{j}\l... | Proof. By the division algorithm, there are \( g\left( x\right) ,{d}_{0}\left( x\right) \in k\left\lbrack x\right\rbrack \) with\n\n\[ f\left( x\right) = g\left( x\right) b\left( x\right) + {d}_{0}\left( x\right) ,\]\n\nwhere either \( {d}_{0}\left( x\right) = 0 \) or \( \deg \left( {d}_{0}\right) < \deg \left( b\right... | Yes |
Lemma 3.88. Let \( k \) be a field, let \( f\left( x\right) /g\left( x\right) \in k\left( x\right) \), and suppose that \( g\left( x\right) = \) \( {q}_{1}\left( x\right) \cdots {q}_{m}\left( x\right) \), where \( {q}_{1}\left( x\right) ,\ldots ,{q}_{m}\left( x\right) \in k\left\lbrack x\right\rbrack \) are pairwise re... | Proof. The proof is by induction on \( m \geq 1 \) . The base step \( m = 1 \) is clearly true. Since \( {q}_{1} \) and \( {q}_{2}\ldots {q}_{m} \) are relatively prime, there are polynomials \( s \) and \( t \) with \( 1 = s{q}_{1} + t{q}_{2}\cdots {q}_{m} \) . Therefore,\n\n\[ \frac{f}{g} = \left( {s{q}_{1} + t{q}_{2... | Yes |
Theorem 3.89 (Partial Fractions). Let \( k \) be a field, and let the factorization into irreducibles of a monic polynomial \( g\left( x\right) \in k\left\lbrack x\right\rbrack \) be\n\n\[ g\left( x\right) = {p}_{1}{\left( x\right) }^{{e}_{1}}\cdots {p}_{m}{\left( x\right) }^{{e}_{m}}. \]\n\nIf \( f\left( x\right) /g\l... | Proof. Clearly, the polynomials \( {p}_{1}{\left( x\right) }^{{e}_{1}},{p}_{2}{\left( x\right) }^{{e}_{2}},\ldots ,{p}_{m}{\left( x\right) }^{{e}_{m}} \) are pairwise relatively prime. By Lemma 3.88, there are \( {a}_{i}\left( x\right) \in k\left\lbrack x\right\rbrack \) with\n\n\[ \frac{f\left( x\right) }{g\left( x\ri... | Yes |
Theorem 3.90. Let \( f\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} \in \mathbb{Z}\left\lbrack x\right\rbrack \subseteq \mathbb{Q}\left\lbrack x\right\rbrack \) . Every rational root \( r \) of \( f\left( x\right) \) has the form \( r = b/c \), where \( b \mid {a}_{0} \) and \( c \mid {a}_{n} \) . | Proof. We may assume that \( r = b/c \) is in lowest terms, that is, \( \left( {b, c}\right) = 1 \) . Substituting \( r \) into \( f\left( x\right) \) gives\n\n\[ 0 = f\left( {b/c}\right) = {a}_{0} + {a}_{1}b/c + \cdots + {a}_{n}{b}^{n}/{c}^{n}, \]\n\nand multiplying through by \( {c}^{n} \) gives\n\n\[ 0 = {a}_{0}{c}^... | Yes |
Corollary 3.91. A rational number \( z \) that is an algebraic integer must lie in \( \mathbb{Z} \) . More precisely, if \( f\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \subseteq \mathbb{Q}\left\lbrack x\right\rbrack \) is a monic polynomial, then every rational root of \( f\left( x\right) \) is an intege... | Proof. If \( f\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} \) is monic, then \( {a}_{n} = 1 \), and Theorem 3.90 applies at once. | Yes |
Lemma 3.93 (Gauss’s Lemma). If \( f\left( x\right), g\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) are both primitive, then their product \( f\left( x\right) g\left( x\right) \) is also primitive. | Proof. Let \( f\left( x\right) = \sum {a}_{i}{x}^{i}, g\left( x\right) = \sum {b}_{j}{x}^{j} \), and \( f\left( x\right) g\left( x\right) = \sum {c}_{k}{x}^{k} \) . If \( f\left( x\right) g\left( x\right) \) is not primitive, then there is a prime \( p \) that divides every \( {c}_{k} \) . Since \( f\left( x\right) \) ... | Yes |
Lemma 3.94. Every nonzero \( f\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) has a unique factorization\n\n\[ f\left( x\right) = c\left( f\right) {f}^{\# }\left( x\right) \]\n\nwhere \( c\left( f\right) \in \mathbb{Q} \) is positive and \( {f}^{\# }\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack... | Proof. There are integers \( {a}_{i} \) and \( {b}_{i} \) with\n\n\[ f\left( x\right) = \left( {{a}_{0}/{b}_{0}}\right) + \left( {{a}_{1}/{b}_{1}}\right) x + \cdots + \left( {{a}_{n}/{b}_{n}}\right) {x}^{n} \in \mathbb{Q}\left\lbrack x\right\rbrack . \]\n\nDefine \( B = {b}_{0}{b}_{1}\ldots {b}_{n} \), so that \( g\lef... | Yes |
Corollary 3.95. If \( f\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \), then \( c\left( f\right) \in \mathbb{Z} \) . | Proof. If \( d \) is the gcd of the coefficients of \( f\left( x\right) \), then \( \left( {1/d}\right) f\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) is primitive. Since \( d\left\lbrack {\left( {1/d}\right) f\left( x\right) }\right\rbrack \) is a factorization of \( f\left( x\right) \) as the product o... | Yes |
Corollary 3.96. If \( f\left( x\right) \in \mathbb{Q}\left\lbrack x\right\rbrack \) factors as \( f\left( x\right) = g\left( x\right) h\left( x\right) \), then\n\n\[ c\left( f\right) = c\left( g\right) c\left( h\right) \;\text{ and }\;{f}^{\# }\left( x\right) = {g}^{\# }\left( x\right) {h}^{\# }\left( x\right) . \] | Proof. We have\n\n\[ f\left( x\right) = g\left( x\right) h\left( x\right) \]\n\n\[ c\left( f\right) {f}^{\# }\left( x\right) = \left\lbrack {c\left( g\right) {g}^{\# }\left( x\right) }\right\rbrack \left\lbrack {c\left( h\right) {h}^{\# }\left( x\right) }\right\rbrack \]\n\n\[ = c\left( g\right) c\left( h\right) {g}^{\... | Yes |
Theorem 3.97 (Gauss). Let \( f\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) . If\n\n\[ f\left( x\right) = G\left( x\right) H\left( x\right) \text{ in }\mathbb{Q}\left\lbrack x\right\rbrack ,\]\n\nthen there is a factorization\n\n\[ f\left( x\right) = g\left( x\right) h\left( x\right) \text{ in }\mathbb{Z... | Proof. By Corollary 3.96, there is a factorization\n\n\[ f\left( x\right) = c\left( G\right) c\left( H\right) {G}^{\# }\left( x\right) {H}^{\# }\left( x\right) \text{ in }\mathbb{Q}\left\lbrack x\right\rbrack ,\]\n\nwhere \( {G}^{\# }\left( x\right) ,{H}^{\# }\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \)... | Yes |
Theorem 3.98. Let \( f\left( x\right) = {a}_{0} + {a}_{1}x + {a}_{2}{x}^{2} + \cdots + {x}^{n} \in \mathbb{Z}\left\lbrack x\right\rbrack \) be monic, and let \( p \) be a prime. If \( {f}^{ * }\left( x\right) = \left\lbrack {a}_{0}\right\rbrack + \left\lbrack {a}_{1}\right\rbrack x + \left\lbrack {a}_{2}\right\rbrack {... | Proof. By Exercise 3.44 on page 248, the natural map \( \varphi : \mathbb{Z} \rightarrow {\mathbb{F}}_{p} \) defines a homomorphism \( {\varphi }^{ * } : \mathbb{Z}\left\lbrack x\right\rbrack \rightarrow {\mathbb{F}}_{p}\left\lbrack x\right\rbrack \) by\n\n\[{\varphi }^{ * }\left( {{b}_{0} + {b}_{1}x + {b}_{2}{x}^{2} +... | No |
Lemma 3.102. Let \( g\left( x\right) \in \mathbb{Z}\left\lbrack x\right\rbrack \) . If there is \( c \in \mathbb{Z} \) with \( g\left( {x + c}\right) \) irreducible in \( \mathbb{Z}\left\lbrack x\right\rbrack \), then \( g\left( x\right) \) is irreducible in \( \mathbb{Q}\left\lbrack x\right\rbrack \) . | Proof. By Proposition 3.33, the function \( \varphi : \mathbb{Z}\left\lbrack x\right\rbrack \rightarrow \mathbb{Z}\left\lbrack x\right\rbrack \), given by\n\n\[ f\left( x\right) \mapsto f\left( {x + c}\right) \]\n\nis an isomorphism. If \( g\left( x\right) = s\left( x\right) t\left( x\right) \), then \( g\left( {x + c}... | Yes |
Corollary 3.104 (Gauss). For every prime \( p \), the pth cyclotomic polynomial \( {\Phi }_{p}\left( x\right) \) is irreducible in \( \mathbb{Q}\left\lbrack x\right\rbrack \) . | Proof. Since \( {\Phi }_{p}\left( x\right) = \left( {{x}^{p} - 1}\right) /\left( {x - 1}\right) \), we have\n\n\[ \n{\Phi }_{p}\left( {x + 1}\right) = \left\lbrack {{\left( x + 1\right) }^{p} - 1}\right\rbrack /x \n\]\n\n\[ \n= {x}^{p - 1} + \left( \begin{array}{l} p \\ 1 \end{array}\right) {x}^{p - 2} + \left( \begin{... | Yes |
Lemma 3.105. If \( R \) is a commutative ring and \( I \) is an ideal in \( R \), then congruence mod \( I \) is an equivalence relation on \( R \) . | Proof.\n\n(i) Reflexivity: if \( a \in R \), then \( a - a = 0 \in I \) ; hence, \( a \equiv a{\;\operatorname{mod}\;I} \).\n\n(ii) Symmetry: if \( a \equiv b{\;\operatorname{mod}\;I} \), then \( a - b \in I \). Since \( - 1 \in R \), we have \( b - a = \left( {-1}\right) \left( {a - b}\right) \in I \), and so \( b \eq... | Yes |
Lemma 3.106. The functions\n\n\\[ \n\\alpha : \\left( {R/I}\\right) \\times \\left( {R/I}\\right) \\rightarrow R/I\\text{, given by}\\left( {\\left\\lbrack a\\right\\rbrack ,\\left\\lbrack b\\right\\rbrack }\\right) \\mapsto \\left\\lbrack {a + b}\\right\\rbrack \\text{,}\n\\]\n\nand\n\n\\[ \n\\mu : \\left( {R/I}\\righ... | Proof. Let us prove that \\( \\alpha \\) and \\( \\mu \\) are well-defined. Recall Lemma 2.19: if \\( \\equiv \\) is an equivalence relation on a set \\( X \\), then \\( \\left\\lbrack a\\right\\rbrack = \\left\\lbrack {a}^{\\prime }\\right\\rbrack \\) if and only if \\( a \\equiv {a}^{\\prime } \\) ; here, \\( \\left\... | Yes |
Lemma 3.108. If \( R \) is a commutative ring and \( I \) is an ideal, then for each \( a \in R \), the congruence class \( \left\lbrack a\right\rbrack \) in \( R/I \) is a coset:\n\n\[ \left\lbrack a\right\rbrack = a + I \] | Proof. If \( b \in \left\lbrack a\right\rbrack \), then \( b - a \in I \) . Hence, \( b = a + \left( {b - a}\right) \in a + I \), and so \( \left\lbrack a\right\rbrack \subseteq a + I \) . For the reverse inclusion, if \( c \in a + I \), then \( c = a + i \) for some \( i \in I \), and so \( c - a \in I \) . Hence, \( ... | Yes |
Theorem 3.110 (First Isomorphism Theorem). If \( \varphi : R \rightarrow S \) is a homomorphism of commutative rings, then \( \ker \varphi \) is an ideal in \( R,\operatorname{im}\varphi \) is a subring of \( S \), and there is an isomorphism\n\n\[ \widetilde{\varphi } : R/\ker \varphi \rightarrow \operatorname{im}\var... | Proof. Let \( I = \ker \varphi \) . We have already seen, in Proposition 3.38, that \( I \) is an ideal in \( R \) and that \( \operatorname{im}\varphi \) is a subring of \( A \) .\n\n\( \widetilde{\varphi } \) is well-defined.\n\nIf \( a + I = b + I \), then \( a - b \in I = \ker \varphi \), so that \( \varphi \left( ... | Yes |
Proposition 3.111. If \( k \) is a field, then its prime field is isomorphic to \( \mathbb{Q} \) or to \( {\mathbb{F}}_{p} \) for some prime \( p \) . | Proof. Denote the one in \( k \) by \( \varepsilon \), and consider the homomorphism \( \chi : \mathbb{Z} \rightarrow k \) defined by \( \chi \left( n\right) = {n\varepsilon } \) . Since every ideal in \( \mathbb{Z} \) is principal, there is an integer \( m \geq 0 \) with \( \ker \chi = \left( m\right) \) . If \( m = 0... | Yes |
Theorem 3.113. If \( k \) is a field and \( I = \left( {p\left( x\right) }\right) \), where \( p\left( x\right) \in k\left\lbrack x\right\rbrack \) is nonconstant, then the following statements are equivalent.\n\n(i) \( k\left\lbrack x\right\rbrack /I \) is a field.\n\n(ii) \( k\left\lbrack x\right\rbrack /I \) is a do... | Proof.\n\n(i) \( \Rightarrow \) (ii). Every field is a domain.\n\n(ii) \( \Rightarrow \) (iii). If \( p\left( x\right) \) is not irreducible, then there is a factorization \( p\left( x\right) = \) \( g\left( x\right) h\left( x\right) \) in \( k\left\lbrack x\right\rbrack \) with \( \deg \left( g\right) < \deg \left( p\... | Yes |
Theorem 3.115. Let \( k \) be a field, let \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) be a nonzero polynomial of degree \( n \geq 1 \), let \( I = \left( {f\left( x\right) }\right) \), and let \( K = k\left\lbrack x\right\rbrack /I \) . Then every element in \( K \) has a unique expression of the form\n\n\... | Proof. Every element of \( K \) has the form \( g\left( x\right) + I \), where \( g\left( x\right) \in k\left\lbrack x\right\rbrack \) . By the division algorithm, there are polynomials \( q\left( x\right), r\left( x\right) \in k\left\lbrack x\right\rbrack \) with \( g\left( x\right) = \) \( q\left( x\right) f\left( x\... | Yes |
Corollary 3.117. Let \( k \) be a field and \( p\left( x\right) \in k\left\lbrack x\right\rbrack \) be an irreducible polynomial. If \( K = k\left\lbrack x\right\rbrack /I \), where \( I = \left( {p\left( x\right) }\right) \), and if \( \alpha \in K \), then there exists a unique monic irreducible polynomial \( h\left(... | Proof. By Theorem 3.115, \( \alpha = {b}_{0} + {b}_{1}z + \cdots + {b}_{n - 1}{z}^{n - 1} \), where \( z = x + I \) , all \( {b}_{i} \in k \), and \( n = \deg \left( p\right) \) . Thus, \( \alpha \) is a root of \( {b}_{0} + {b}_{1}x + \cdots + {b}_{n - 1}{x}^{n - 1} \in \) \( k\left\lbrack x\right\rbrack \), and Propo... | Yes |
Theorem 3.118 (Kronecker). If \( k \) is a field and \( f\left( x\right) \in k\left\lbrack x\right\rbrack \) is nonconstant, then there exists a field \( K \), containing \( k \) as a subfield, with \( f\left( x\right) \) splitting in \( K\left\lbrack x\right\rbrack \) ; that is, \( f\left( x\right) \) is a product of ... | Proof. We prove the theorem by induction on \( \deg \left( f\right) \), and we modify the statement a bit to enable us to prove the inductive step more easily: if \( E \) is any field containing \( k \) as a subfield (so that \( f\left( x\right) \in k\left\lbrack x\right\rbrack \subseteq E\left\lbrack x\right\rbrack \)... | Yes |
Proposition 3.119. If \( E \) is a finite field, then \( \left| E\right| = {p}^{n} \) for some prime \( p \) and some \( n \geq 1 \) . | Proof. If \( k \) is the prime field of \( E \), then Proposition 3.111 says that \( k \cong \mathbb{Q} \) or \( k \cong {\mathbb{F}}_{p} \) for some prime \( p \) ; since \( \mathbb{Q} \) is infinite, we have \( k \) of characteristic \( p \) . Therefore, \( {pa} = 0 \) for all \( a \in E \) ; that is, as an additive ... | Yes |
Lemma 3.120. Let \( E \) be a finite field, and let \( k \) be its prime field.\n\n(i) There is a prime \( p \) with \( k \cong {\mathbb{F}}_{p} \).\n\n(ii) There is an integer \( M > 0 \) so that every nonzero \( z \in E \) is a root of \( {x}^{M} - 1 \).\n\n(iii) If \( S \) is a subfield of \( E \) and \( z \in E \),... | Proof.\n\n(i) Since \( E \) is finite, it cannot contain a copy of \( \mathbb{Q} \), and so Proposition 3.111 says that its prime field \( k \cong {\mathbb{F}}_{p} \) for some prime \( p \).\n\n(ii) Let \( z \in E \) be nonzero. Since \( E \) is finite, there must be repetitions on the list \( 1, z,{z}^{2},\ldots \) Le... | Yes |
Proposition 3.121. If \( E \) is a finite field, then \( \left| E\right| = {p}^{n} \) for some prime \( p \) and some \( n \geq 1 \) . | Proof. Since \( E \) is finite, Lemma 3.120(i) says that its prime field \( k \cong {\mathbb{F}}_{p} \) for some prime \( p \) . If \( k = E \), we are done. Otherwise, there is some element \( {z}_{1} \in E \) with \( {z}_{1} \notin k \) . By Corollary 3.120(iii), \( \left| {k\left( {z}_{1}\right) }\right| = {p}^{m} \... | Yes |
Every finite field \( E \) has exactly \( {p}^{n} \) elements for some prime \( p \) and some \( n \geq 1 \) . | Proof. Since \( E \) is finite, its prime field \( k \cong {\mathbb{F}}_{p} \) for some prime \( p \) . There is a primitive element \( \pi \in E \), by Theorem 3.122, and so \( \left| {k\left( \pi \right) }\right| = {p}^{n} \) for some \( n \geq 1 \), by Lemma 3.120(iii). But \( k\left( \pi \right) = E \), because the... | Yes |
Theorem 3.124 (Galois). If \( p \) is a prime and \( n \) is a positive integer, then there exists a field that has exactly \( {p}^{n} \) elements. | Proof. Write \( q = {p}^{n} \), and consider the polynomial\n\n\[ g\left( x\right) = {x}^{q} - x \in {\mathbb{F}}_{p}\left\lbrack x\right\rbrack .\n\]\n\nBy Kronecker’s theorem, there is a field \( E \) containing \( {\mathbb{F}}_{p} \) such that \( g\left( x\right) \) is a product of linear factors in \( E\left\lbrack... | No |
There are exactly two \( 2 \times 2 \) Latin squares having entries 0 and 1 : | \[ A = \left\lbrack \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right\rbrack \;\text{ and }\;B = \left\lbrack \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right\rbrack \; \blacktriangleleft \] | Yes |
Lemma 3.131. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \) be a Latin square whose entries lie in a set \( X \) with \( n \) elements. If \( x \mapsto {x}^{\prime } \) is a permutation of \( X \), then \( {A}^{\prime } = \left\lbrack {\left( {a}_{ij}\right) }^{\prime }\right\rbrack \) is a Latin square; that is, if ... | Proof. As the \( i \) th row \( \left( {{a}_{i1},\ldots ,{a}_{in}}\right) \) of \( A \) is a permutation of \( X \), so is the \( i \) th row \( \left( {{\left( {a}_{i1}\right) }^{\prime },\ldots ,{\left( {a}_{in}\right) }^{\prime }}\right) \) of \( {A}^{\prime } \) (the composite of two permutations is again a permuta... | Yes |
Corollary 3.133. For every prime power \( {p}^{e} > 2 \), there exists a pair of orthogonal \( {p}^{e} \times {p}^{e} \) Latin squares. | Proof. By Galois’ theorem, there exists a finite field \( k \) with \( \left| k\right| = {p}^{e} \) . In order to have an orthogonal pair of Latin squares, we need \( \left| {k}^{ \times }\right| \geq 2 \) ; that is, \( {p}^{e} - 1 \geq 2 \) , hence \( {p}^{e} > 2 \) . | No |
Theorem 3.134 (Euler). If \( n ≢ 2{\;\operatorname{mod}\;4} \), then there exists an orthogonal pair of \( n \times n \) Latin squares. | Proof. We merely state the main steps of the proof. One shows first that if \( A \) and \( B \) are Latin squares, then \( A \otimes B \) is a Latin square. Second, one proves that if \( A \) and \( {A}^{\prime } \) are orthogonal \( k \times k \) Latin squares, and if \( B \) and \( {B}^{\prime } \) are orthogonal \( ... | No |
Proposition 3.135. If \( A \) is an \( n \times n \) magic square, then its magic number is\n\n\[ \sigma = \frac{1}{2}n\left( {{n}^{2} - 1}\right) \] | Proof. If \( {\rho }_{i} \) denotes the sum of the entries in the \( i \) th row of \( A \), then \( {\rho }_{i} = \sigma \) for all \( i \), and so\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{\rho }_{i} = {n\sigma } \]\n\nBut this last number is the sum of all the entries in \( A \) ; that is,\n\n\[ {n\sigma } = 1 + 2 +... | Yes |
Proposition 3.136. If \( A = \left\lbrack {a}_{ij}\right\rbrack \) and \( B = \left\lbrack {b}_{ij}\right\rbrack \) are orthogonal Latin squares with entries in \( 0,1,\ldots, n - 1 \), then the matrix \( M = \left\lbrack {{a}_{ij}n + {b}_{ij}}\right\rbrack \) is an \( n \times n \) magic square. | Proof. Since \( A \) and \( B \) are orthogonal, the entries \( \left( {{a}_{ij},{b}_{ij}}\right) \) of their Hadamard product \( A \circ B \) are all distinct. It follows from Proposition 1.44, which says that the \( n \) -adic digits of a non-negative number are unique, that every number from 0 through \( {n}^{2} - 1... | Yes |
Proposition 3.139. If \( n \in \mathbb{N} \) is an odd integer that is not a multiple of 3, then there exists an \( n \times n \) diabolic square. | Proof. Let \( A = \left\lbrack {a}_{ij}\right\rbrack \) and \( B = \left\lbrack {b}_{ij}\right\rbrack \) be diagonal orthogonal \( n \times n \) Latin squares, which exist, by Lemma 3.138. By Proposition 3.136, the matrix \( M = \left\lbrack {{a}_{ij}n + {b}_{ij}}\right\rbrack \) is a magic square with magic number \( ... | Yes |
Lemma 3.140. If \( {A}_{1},{A}_{2},\ldots ,{A}_{t} \) is an orthogonal set of \( n \times n \) Latin squares, then \( t \leq n - 1 \) . | Proof. There is no loss in generality in assuming that each \( {A}_{v} \) has entries lying in \( X = \{ 0,1,\ldots, n - 1\} \) . Permute the entries of \( {A}_{1} \) so that its first row is \( 0,1,\ldots, n - 1 \) in this order. By Lemma 3.131, this new matrix \( {A}_{1}^{\prime } \) is a Latin square which is orthog... | Yes |
Theorem 3.141. If \( q = {p}^{e} \), then there exists a complete orthogonal set of \( q - 1 \) \( q \times q \) Latin squares. | Proof. If \( k \) is a finite field with \( q \) elements, then there are \( q - 1 \) elements \( a \in {k}^{ \times } \) , and so there are \( q - 1 \) Latin squares \( {L}_{a} \), each pair of which is orthogonal, by Theorem 3.132. - | Yes |
Proposition 4.2. Every subspace \( U \) of a vector space \( V \) over a field \( k \) is itself a vector space over \( k \) . | Proof. By hypothesis, \( U \) is closed under scalar multiplication: if \( u \in U \) and \( c \in k \), then \( {cu} \in U \) . Axioms (i) through (iv) in the definition of vector space hold for all scalars and for all vectors in \( V \) ; in particular, they hold for all vectors in \( U \) . For example, axiom (iii) ... | Yes |
Proposition 4.7. If \( X = {v}_{1},\ldots ,{v}_{m} \) is a list in a vector space \( V \), then \( \langle X\rangle \) is a subspace of \( V \) containing the subset \( \left\{ {{v}_{1},\ldots ,{v}_{m}}\right\} \) . | Proof. Let us write \( L = \left\langle {{v}_{1},\ldots ,{v}_{m}}\right\rangle \) . Now \( 0 \in L \), for\n\n\[ 0 = 0{v}_{1} + \cdots + 0{v}_{m} \]\n\nIf \( u = {a}_{1}{v}_{1} + \cdots + {a}_{m}{v}_{m} \) and \( v = {b}_{1}{v}_{1} + \cdots + {b}_{m}{v}_{m} \in L \), then\n\n\[ u + v = {a}_{1}{v}_{1} + \cdots + {a}_{n}... | Yes |
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