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Theorem 9.12 (Trotter) There exists a (unique) process \( {\left( {L}_{t}^{a}\left( B\right) \right) }_{a \in \mathbb{R}, t \geq 0} \), whose sample paths are continuous functions of the pair \( \left( {a, t}\right) \), such that, for every fixed \( a \in \mathbb{R},{\left( {L}_{t}^{a}\left( B\right) \right) }_{t \geq ... | Proof The first assertion follows by applying Theorem 9.4 and Corollary 9.7 to \( X = B \), noting that \( \langle B, B{\rangle }_{t} = t \) . We have already seen that the inclusion (9.17) holds with probability one if \( a \) is fixed, hence simultaneously for all rationals, a.s. A continuity argument allows us to ge... | Yes |
Proposition 9.13 (i) Let \( a \in \mathbb{R} \smallsetminus \{ 0\} \) and \( {T}_{a} \mathrel{\text{:=}} \inf \left\{ {t \geq 0 : {B}_{t} = a}\right\} \) . Then \( {L}_{{T}_{a}}^{0}\left( B\right) \) has an exponential distribution with mean \( 2\left| a\right| \) . | (i) By simple scaling and symmetry arguments, it is enough to take \( a = 1 \) . We then observe that \( {L}_{\infty }^{0}\left( B\right) = \infty \) a.s. Indeed, the scaling argument of the preceding proof shows that \( {L}_{\infty }^{0}\left( B\right) \) has the same distribution as \( \lambda {L}_{\infty }^{0}\left(... | Yes |
Theorem 9.14 (Lévy) The two processes \( {\left( {S}_{t},{S}_{t} - {B}_{t}\right) }_{t \geq 0} \) and \( {\left( {L}_{t}^{0}\left( B\right) ,\left| {B}_{t}\right| \right) }_{t \geq 0} \) have the same distribution. | Proof By Tanaka’s formula, for every \( t \geq 0 \) ,\n\n\[ \left| {B}_{t}\right| = - {\beta }_{t} + {L}_{t}^{0}\left( B\right) \]\n\n(9.18)\n\nwhere\n\n\[ {\beta }_{t} = - {\int }_{0}^{t}\operatorname{sgn}\left( {B}_{s}\right) \mathrm{d}{B}_{s} \]\n\nSince \( \langle \beta ,\beta {\rangle }_{t} = t \), Theorem 5.12 en... | Yes |
Proposition 9.15 We have a.s.\n\n\[ \n\\left\\{ {t \\geq 0 : {B}_{t} = 0}\\right\\} = \\left\\{ {{\\tau }_{s} : s \\geq 0}\\right\\} \\cup \\left\\{ {{\\tau }_{s - } : s \\in D}\\right\\} \n\]\n\nwhere \( D \) is the countable set of jump times of \( {\\left( {\\tau }_{s}\\right) }_{s \\geq 0} \) . | Proof We know from (9.17) that a.s.\n\n\[ \n\\operatorname{supp}\\left( {{\\mathrm{d}}_{t}{L}_{t}^{0}\\left( B\\right) }\\right) \\subset \\left\\{ {t \\geq 0 : {B}_{t} = 0}\\right\\} .\n\]\n\nIt follows that any time \( t \) of the form \( t = {\\tau }_{s} \) or \( t = {\\tau }_{s - } \) must belong to the zero set of... | Yes |
Lemma 1. The homomorphic image \( \bar{G} \) of a group \( G \) is a group. If \( G \) is commutative, so is \( G \) . | PROOF. We first prove the associative law in \( \bar{G} \) . Let \( \bar{a},\bar{b},\bar{c} \) be arbitrary elements of \( \bar{G} \) ; they are images of certain elements \( a, b, c \) of \( G \) , since \( T \) maps \( G \) onto \( G \) . We have \( \left( {ab}\right) c = a\left( {bc}\right) \) . We have \( \left\lbr... | Yes |
Lemma 1. (Lemma of Gauss) If \( f\left( x\right), g\left( x\right) \in {R}^{r}X\rbrack \), then \( c\left( {fg}\right) = \) \( c\left( f\right) c\left( g\right) \) . In particular, the product of two primitive polynomials is primitive. | PROOF. If \( c = c\left( f\right), d = c\left( g\right) \), then \( f\left( x\right) = c{f}_{1}\left( x\right), g\left( x\right) = d{g}_{1}\left( x\right) \), and \( {f}_{1} \) and \( {g}_{1} \) are primitive. Since \( {fg} = \left( {cd}\right) {f}_{1}{g}_{1} \), we need only prove that \( {f}_{1}{g}_{1} \) is primitiv... | Yes |
Lemma 2. Let \( k \subset L \subset \Delta \subset K \) be successive finite algebraic extensions of \( k \), where \( K \) is a normal extension of \( k \) . If \( \Delta \) possesses \( {nL} \) -isomorphisms into \( K \), then every \( k \) -isomorphism of \( L \) into \( K \) has exactly \( n \) extensions which are... | PROOF. Let \( G \) be the group of all \( k \) -automorphisms of \( K \) and let \( G\left( L\right) \) (respectively, \( G\left( \Delta \right) \) ) be the subgroup of \( G \) consisting of those auto-morphisms of \( K \) which leave fixed every element of \( L \) (respectively, of \( \Delta \) ). It is clear that \( ... | Yes |
Corollary 3. There exist transcendence bases of \( K/k \) . | We apply Corollary 2 for the case \( S = K \) . | No |
Lemma 2. Let \( {x}_{1},{x}_{2},\cdots ,{x}_{n},{x}_{n + 1} \) be elements of an extension field \( K \) of a field \( k \) and assume that these \( n + 1 \) elements \( {x}_{i} \) are algebraically dependent over \( k \) but that the n elements \( {x}_{1},{x}_{2},\cdots ,{x}_{n} \) are algebraically independent over \... | PROOF. Since the \( n + 1 \) elements \( {x}_{i} \) are algebraically dependent over \( k \), the set \( A \) contains polynomials different from zero. Let \( f\left( {{X}_{1},{X}_{2},\cdots ,{X}_{n},{X}_{n + 1}}\right) \) be a non-zero polynomial in \( A \), of smallest possible degree \( q \) in \( {X}_{n + 1} \), an... | Yes |
Corollary 1. Let \( k, K \) and \( S \) be fields such that \( k \subset K \subset S \) . If \( X \) is a (finite or infinite) set of algebraically independent elements of \( S \) over \( K \), then the subfields \( K \) and \( k\left( X\right) \) of \( S \) are linearly disjoint over \( k \) . | It is obvious that \( K \) and \( k\left( X\right) \) are linearly disjoint over \( k \) if and only if \( K \) and \( k\left\lbrack X\right\rbrack \) are linearly disjoint over \( k \) . Now, the set of all monomials \( {x}_{1}{}^{i}{}_{1}{x}_{2}{}^{i}{}_{2}\cdots {x}_{n}{}^{i}{}_{n},{x}_{\alpha } \in X \), is a basis... | Yes |
Corollary 2. If a field \( K \) is a purely transcendental extension of a field \( k \), then \( K \) and \( {k}^{{p}^{-1}} \) are linearly disjoint over \( k \) . | For, if \( K = k\left( X\right) \), where \( X \) is a suitable transcendence basis of \( K/k \) , then the elements of \( X \), being algebraically independent over \( k \), are also algebraically independent over \( {k}^{{p}^{-1}} \) . Therefore, by the preceding corollary, \( K \) and \( {k}^{{p}^{-1}} \) are linear... | Yes |
Lemma 2. Let \( K \) be an algebraic extension of \( k \) and let \( \{ z\} \) be a transcendence basis of \( k\left( x\right) /k \) . If we have \( {\left\lbrack {k}^{\prime }\left( x\right) : {k}^{\prime }\left( z\right) \right\rbrack }_{i} = {\left\lbrack k\left( x\right) : k\left( z\right) \right\rbrack }_{i} \) fo... | PROOF. For any field \( {k}^{\prime } \) between \( k \) and \( K \) we denote by \( L\left( {k}^{\prime }\right) \) the maximal separable extension of \( {k}^{\prime }\left( z\right) \) in \( {k}^{\prime }\left( x\right) \) . Let \( \left\{ {{\xi }_{1},{\xi }_{2},\cdots ,{\xi }_{m}}\right\} \) be a basis of \( k\left(... | Yes |
Corollary 1. If \( \overrightarrow{k} \) is the algebraically closed field given by the algebraic closure of \( k \) in \( S \), then\n\n(13)\n\n\[{\left\lbrack k\left( x\right) : k\left( z\right) \right\rbrack }_{i} = {\left\lbrack k\left( x\right) : k\right\rbrack }_{i}{\left\lbrack \bar{k}\left( x\right) : \bar{k}\l... | For \( \overrightarrow{k} \) being a perfect field, the order of inseparability of \( \overrightarrow{k}\left( x\right) \) over \( \overrightarrow{k} \) is equal to 1 . | No |
Corollary 1. Let \( K \) be a field and let \( F = K\\left( x\\right) \) be a simple transcendental extension of \( K \) . If \( D \) is a derivation of \( K \) with values in some field \( L \) containing \( F \), and if \( u \) is any element of \( L \), then there exists one and only one derivation \( {D}^{\\prime }... | In fact, 0 is the only polynomial \( f \) in \( K\\left\\lbrack X\\right\\rbrack \) such that \( f\\left( x\\right) = 0 \) . | No |
Corollary 1. Any two tensor products of \( A \) and \( B \) are equivalent. | This has been established in the second part of the above proof. | No |
Corollary 2. If the zero-ideal in \( R \otimes {R}^{\prime } \) is primary (or equivalently: if every zero-divisor in \( R \otimes {R}^{\prime } \) is nilpotent), then any two free joins of \( R/k \) and \( {R}^{\prime }/k \) are equivalent. | For in that case the radical of \( \left( 0\right) \) is a prime ideal \( \mathfrak{p} \) containing all the zero divisors of \( R \otimes {R}^{\prime } \), and any other prime ideal in \( R \otimes {R}^{\prime } \) must contain \( \mathfrak{p} \) . | No |
Lemma 3. Given a ring \( R \), an ideal \( \mathfrak{m} \) of \( R \) and a family \( \left\{ {\mathfrak{a}}_{\lambda }\right\} \) of ideals of \( R \) which are closed (with respect to \( \mathfrak{m} \) ), the intersection \( \mathop{\bigcap }\limits_{\lambda }{\mathfrak{a}}_{\lambda } \) is closed. | This follows from the obvious inclusion \( \left( {\mathop{\bigcap }\limits_{\lambda }{\mathfrak{a}}_{\lambda }}\right) + {\mathfrak{m}}^{n} \subset \mathop{\bigcap }\limits_{\lambda }\left( {{\mathfrak{a}}_{\lambda } + {\mathfrak{m}}^{n}}\right) \nand from the associativity of intersections. | No |
Corollary 2. We have \( {\mathfrak{a}}^{e} \neq {R}_{M} \) if and only if \( \mathfrak{a} \cap M = \varnothing \) . | We notice that \( {\mathfrak{a}}^{e} = {R}_{M} \) is equivalent to \( 1 \in {\mathfrak{a}}^{ec} \), and we use (a). | No |
Lemma 1. If a module \( M \) over a principal ideal ring \( R \) has a basis of \( n \) elements, then every submodule \( N \) of \( M \) has a basis of \( n \) elements. | PROOF. If \( n = 1 \), we have \( M = {Rx} \) and then clearly \( N = \mathfrak{A}x \), where \( \mathfrak{A} \) is an ideal in \( \mathrm{R} \) . Since \( R \) is a PIR, we have \( \mathfrak{A} = {Rt} \), whence \( N = {Ry} \) , where \( y = {tx} \), and this establishes the lemma in the case \( n = 1 \) . In the gene... | Yes |
Theorem 2 (Transitivity of Integral Dependence). Let \( A \) be a ring, \( B \) an overring of \( A \) integral over \( A \), and \( C \) an overring of \( B \) integral over \( B \) . Then \( C \) is integral over \( A \) . | PROOF. Let \( x \) be an element of \( C \), and let\n\n\[ \n{x}^{n} + {b}_{n - 1}{x}^{n - 1} + \cdots + {b}_{0} = 0\left( {{b}_{i} \in B}\right)\n\]\n\nbe an equation of integral dependence for \( x \) over \( B \) . Then the ring \( {B}^{\prime } = A\left\lbrack {{b}_{0},\cdots ,{b}_{n - 1}}\right\rbrack \) is a fini... | Yes |
Corollary 1. The assumptions being the same as in Theorem 7, let us furthermore assume that the ring \( A \) is noetherian. Then \( {A}^{\prime } \) is a finite \( A \) - module and is a noetherian ring. | In fact, \( {A}^{\prime } \) is a submodule of the finite \( A \) -module \( \mathop{\sum }\limits_{i}A{x}_{i} \), and is therefore a finite \( A \) -module. Thus \( {A}^{\prime } \) satisfies the a.c.c. as an \( A \) -module (III,§ 10, Theorem 18), and a fortiori satisfies the a.c.c. as an \( {A}^{\prime } \) -module ... | Yes |
Lemma 2. Let \( R \) be a Dedekind domain, \( K \) its quotient field, \( \mathfrak{p} \) a proper prime ideal in \( R, L \) a finite algebraic extension of \( K \), and \( {R}^{\prime } \) a noetherian overring of \( R \) contained in \( L \) . Let \( {\mathfrak{p}}^{e} = {\mathfrak{Q}}_{1} \cap \cdots \cap {\mathfrak... | PROOF OF THE LEMMA. We first notice that since \( \mathfrak{p} \) is maxima \( {}^{1} \), we have \( {\mathfrak{P}}_{i} \cap R = {\mathfrak{p}}^{e} \cap R = \mathfrak{p} \) ; thus \( R/\mathfrak{p} \) may be identified with subfields of \( {R}^{\prime }/{\mathfrak{P}}_{i} \) and of \( {R}^{\prime }/{\mathfrak{p}}^{e} \... | Yes |
Corollary 1. If \( \mathfrak{o} \) is an integral domain, not a field, and if \( K \) is a field containing \( \mathfrak{o} \) as subring, then there exist non-trivial places \( \mathcal{P} \) of \( K \) such that \( {K}_{\mathcal{P}} > 0 \) . | For \( \mathfrak{o} \) contains ideals different from \( \left( 0\right) \) and \( \mathfrak{o} \) . | No |
Corollary 2. A field \( K \) possesses only trivial places if and only if \( K \) is an absolutely algebraic field, of characteristic \( p \neq 0 \) (i.e., if and only if \( K \) is an algebraic extension of the prime field of characteristic \( p \neq 0 \) ). | For, the absolutely algebraic fields, of characteristic \( p \neq 0 \), are the only fields with the property that all their subrings are fields, whereas the valuation ring of a non-trivial place is not a field. | Yes |
THEOREM 7. Let \( \mathfrak{o} \) be a subring of a field \( K,\mathfrak{p} \) and \( \mathfrak{q} \) two prime ideals in \( \mathfrak{v} \) such that \( \mathfrak{p} \subset \mathfrak{q} \) . Suppose that \( \mathcal{P} \) is a place of \( K \) with center \( \mathfrak{p} \) in \( \mathfrak{o} \) . Then there exists a... | PROOF. Without loss of generality we may assume that \( {K}_{\mathcal{P}}/{\mathfrak{M}}_{\mathcal{P}} \) is the residue field of \( \mathcal{P} \) . Consider now the subring \( o/\mathfrak{p} \) of the residue field \( {K}_{\mathcal{P}}/{\mathfrak{M}}_{\mathcal{P}} \) of \( \mathcal{P} \), the prime ideal \( \mathfrak... | Yes |
Property 1. A place \( \mathcal{P} \) of \( k\left( V\right) /k \) is trivial if and only if its center on \( V \) is a general point of \( V \) over \( k \) . | The proof is straightforward and may be left to the reader. | No |
Property 2. If \( Q \) is the center on \( V \) of a place \( \mathcal{P} \) of \( k\left( V\right) /k \) then \( \dim Q/k \leqq \dim \mathcal{P}/k \leqq \dim V \), and \( \mathcal{P} \) is trivial if and only if \( \dim \mathcal{P}/k = \) \( \dim V \) . | Obvious. | No |
Property 3. Let \( \mathcal{P} \) and \( \mathcal{Q} \) be places of \( k\left( V\right) /k \) and let \( P \) and \( Q \) be their respective centers on \( V \) . If \( \mathcal{P}\overset{k}{ \rightarrow }\mathcal{Q} \) then also \( P\overset{k}{ \rightarrow }Q \) . | Obvious. | No |
Property 4. Let \( P \) and \( Q \) be points of \( V \) such that \( P\overset{k}{ \rightarrow }Q \) . Suppose that \( \mathcal{P} \) is a place of \( k\left( V\right) /k \) which admits \( P \) as center on \( V \) . Then there exists a place 2 of \( k\left( V\right) /k \) which is a specialization of \( \mathcal{P} ... | This is the analogue of Theorem \( 7,§5 \), and the proof is the same. | No |
Property 5. If \( Q \) is the center on \( V \) of a place \( \mathcal{P} \) of \( k\left( V\right) /k \) then \( \mathfrak{v}\left( {Q;V}\right) \subset k\left( V\right) \), and \( \mathfrak{m}\left( {Q;V}\right) = {\mathfrak{M}}_{\mathcal{P}} \cap \mathfrak{v}\left( {Q;V}\right) \) . Conversely, if these two conditio... | Obvious. | No |
Property 6. If \( Q \) is a point of \( V \) then the integral closure of \( \mathfrak{o}\left( {Q;V}\right) \) is the intersection of all the valuation rings which belong to places \( \mathcal{P} \) of \( k\left( V\right) /k \) having center \( Q \) on \( V \) . | This is a particular case of Theorem \( 8,§5 \) . | No |
For any place \( \mathcal{P} \) of \( K \) there exist extensions \( \mathcal{P} \star \) in \( {K}^{ \star } \) such that \( {\dim }_{K}\mathcal{P} \star \) is any preassigned cardinal number \( \geqq 0 \) and \( \leqq \) transcendence degree of \( {K}^{ \star }/K \) . | Let \( \left\{ {y}_{j}\right\} \) be a transcendence basis of \( {K}^{ \star }/K \) and let \( \left\{ {u}_{j}\right\} \) be a set of indeterminates over \( \Delta \), in \( \left( {1,1}\right) \) correspondence with the set \( \left\{ {y}_{j}\right\} \) . Let \( f \) be the (uniquely determined) homomorphism of the po... | Yes |
Corollary 2. If \( {\mathfrak{P}}^{ \star } \) is any maximal ideal in \( {K}_{\mathcal{P}}^{ \star } \), then the quotient ring of \( K \), with respect to \( {\mathfrak{P}}^{ \star } \) is the valuation ring of a place \( {\mathcal{P}}^{ \star } \) of \( {K}^{ \star } \) which is an extension of \( \mathcal{P} \) . | For, by Theorem \( 4,§4 \), there exists a place \( \mathcal{P} \star \) of \( {K}^{ \star } \) such that \( {K}_{{\mathcal{P}}^{ \star }}^{ \star } \supset {K}_{\mathcal{P}}^{ \star } \) and \( {\mathfrak{M}}_{{\mathcal{P}}^{ \star }} \supset {\mathfrak{P}}^{ \star } \) . Since \( {K}_{\mathcal{P}}^{ \star } \) is int... | Yes |
Corollary 3. Let \( K \) be the quotient field of an integrally closed noetherian domain \( R \) . If \( w \) is any element of \( K, w \neq 0 \), then (1) there is only a finite number of prime ideals \( \mathfrak{p} \) in the set \( S \) such that \( {v}_{\mathfrak{p}}\left( w\right) \neq 0 \) ; (2) \( w \) belongs t... | If \( w \in R \), then \( {Rw} = {\mathfrak{p}}_{1}^{\left( {n}_{1}\right) } \cap {\mathfrak{p}}_{2}^{\left( {n}_{2}\right) } \cap \cdots \cap {\mathfrak{p}}_{s}^{\left( {n}_{s}\right) } \), where \( s \geqq 0 \), the \( {\mathfrak{p}}_{i} \) are minimal prime ideals in \( R,{n}_{i} \geqq 1 \) and \( s = 0 \) if and on... | Yes |
Lemma 3. If \( F\left( {{X}_{1},{X}_{2},\cdots ,{X}_{n}}\right) \) is a non-zero power series in \( k\left\lbrack \left\lbrack {{X}_{1},{X}_{2},\cdots ,{X}_{n}}\right\rbrack \right\rbrack \left( {k, a\text{field}}\right) \), then there exists an automorphism \( \varphi \) of \( k\left\lbrack \left\lbrack {{X}_{1},{X}_{... | PROOF. We assume first that \( k \) is an infinite field. Let \( {f}_{q} \) be the initial form of \( F \) . Since \( k \) is infinite we can find elements \( {a}_{1},{a}_{2},\cdots \) , \( {a}_{n - 1} \) in \( k \) such that \( {f}_{q}\left( {{a}_{1},{a}_{2},\cdots ,{a}_{n - 1},1}\right) \neq 0 \) . Then we may use th... | Yes |
If \( \mathfrak{p} \) is any prime ideal in \( k\left\lbrack {{X}_{1},{X}_{2},\cdots ,{X}_{n}}\right\rbrack \), then \( \mathfrak{p} \) is the ideal of its own variety \( \mathcal{V}\left( \mathfrak{p}\right) \), and hence \( \mathcal{V}\left( \mathfrak{p}\right) \) is irreducible and \( \mathfrak{p} \in \mathbf{I} \) ... | For, \( \sqrt{\mathfrak{p}} = \mathfrak{p} \), whence \( \mathfrak{p} = \mathcal{I}\left( {\mathcal{V}\left( \mathfrak{p}\right) }\right) \in \mathbf{I} \) . The irreducibility of \( \mathcal{V}\left( \mathfrak{p}\right) \) follows from Theorem 12. | No |
Corollary 4. Let \( {}^{h}R = k\left\lbrack {{y}_{0},{y}_{1},\cdots ,{y}_{n}}\right\rbrack \) be a homogeneous finite integral domain and let \( \mathfrak{P} \) and \( {\mathfrak{P}}^{\prime } \) be prime ideals in \( {}^{h}R \), of dimension \( s + 1 \) and \( {s}^{\prime } + 1 \) respectively, such that \( \mathfrak{... | Assuming that \( {y}_{0} \neq 0 \) we set \( {x}_{i} = {y}_{i}/{y}_{0}, i = 1,2,\cdots, n \), and we consider the integral domain \( R = k\left\lbrack {{x}_{1},{x}_{2},\cdots ,{x}_{n}}\right\rbrack \) . We apply the results proved in \( §5 \) in regard to the relationship between homogeneous ideals in \( {}^{h}R \) and... | Yes |
Lemma 2. Let \( \mathfrak{A} \) be an ideal in \( R = k\left\lbrack {{X}_{1},\cdots ,{X}_{n}}\right\rbrack \), different from \( R \), and let \( \left\{ {{z}_{1},\cdots ,{z}_{d}}\right\} \) be a finite set of algebraically independent elements of \( R/\mathfrak{A} \) over \( k \) such that \( R/\mathfrak{A} \) is inte... | PROOF. Let \( \mathfrak{p} \) be any prime ideal of \( R \) containing \( \mathfrak{A} \), and let \( \overline{\mathfrak{p}} = \mathfrak{p}/\mathfrak{A} \) . Then \( R/\mathfrak{p} \) is integral over \( k\left\lbrack z\right\rbrack /\left( {\overline{\mathfrak{p}} \cap k\left\lbrack z\right\rbrack }\right) \), whence... | Yes |
Corollary 1. \( A,\mathfrak{m} \) and \( E \) being as in Theorem 7, suppose that \( {G}_{\mathfrak{m}}\left( E\right) \) is a finite \( {G}_{\mathfrak{m}}\left( A\right) \) -module. Then \( E \) is a finite \( A \) -module. | We apply Theorem 7 to the case \( F = E \) . | No |
Corollary 5. Let \( A \) be a Zariski ring, \( E \) a finite \( A \) -module, \( {E}^{\prime } \) a submodule of \( E \) and \( z \) an element of \( E \) . Then \( \widehat{A}\left( {{E}^{\prime } : {Az}}\right) = \widehat{A}{E}^{\prime } : {Az} \) . | We recall that \( {E}^{\prime } : {Az} \) is the ideal of all elements \( a \) in \( A \) such that \( {az} \in {E}^{\prime } \) . We apply Corollary 3 to the case \( E = A, F = E,{F}^{\prime } = {E}^{\prime } \), and take for \( f \) the mapping \( a \rightarrow {az} \) . | No |
Corollary 2. With the same assumptions on \( A \) and \( \mathfrak{m} \) as in Theorem 12, assume furthermore that the closed ideal \( \mathfrak{a} \) admits an irredundant primary representation \( \mathfrak{a} = {\mathfrak{Q}}_{1} \cap {\mathfrak{Q}}_{2} \cap \cdots \cap {\mathfrak{Q}}_{h} \) such that none of the pr... | As in Theorem 12, let \( \widehat{A}\mathfrak{a} = {\mathfrak{q}}_{1}{}^{ \star } \cap {\mathfrak{q}}_{2}{}^{ \star } \cap \cdots \cap {\mathfrak{q}}_{n}{}^{ \star } \) be an irredundant primary prepresentation of \( \widehat{A}\mathfrak{a} \) and let \( {\mathfrak{q}}_{i} = {\mathfrak{q}}_{i} \star \cap A,{\mathfrak{p... | Yes |
Corollary 1. Let \( A \) be a local ring, \( K \) a subfield of \( A \), and \( \left\{ {{x}_{1},\cdots ,{x}_{d}}\right\} \) a system of parameters of \( A \) . Then the elements \( {x}_{1},\cdots ,{x}_{d} \) are algebraically independent over \( K \) . | Let \( G\left( {{X}_{1},\cdots ,{X}_{d}}\right) \) be a non-zero polynomial over \( K \) such that \( G\left( {{x}_{1},\cdots ,{x}_{d}}\right) = 0 \) . Denote by \( F\left( {{X}_{1},\cdots ,{X}_{d}}\right) \) the lowest degree form of \( G \), and by \( s \) the degree of \( F \) . From \( G\left( {{x}_{1},\cdots ,{x}_... | Yes |
Corollary 1. The hypothesis and notations being as in Theorem 24, suppose furthermore that all the local rings \( {B}_{{\mathfrak{p}}_{i}} \) have the same dimension as A. Then \[ \left\lbrack {B : A}\right\rbrack e\left( \mathfrak{q}\right) = \mathop{\sum }\limits_{i}\left\lbrack {B/{\mathfrak{p}}_{i} : A/\mathfrak{m}... | In fact, all the polynomials \( {\bar{P}}_{{\mathfrak{q}}_{i}}\left( n\right) = {\bar{P}}_{\mathfrak{q}{B}_{{\mathfrak{p}}_{i}}}\left( n\right) \) have then the same degree \( d = \dim \left( A\right) \) . | No |
Proposition 1. Let \( \mathfrak{o} \) be a noetherian domain and let \( {\mathfrak{o}}^{\prime } = \mathfrak{o}\left\lbrack t\right\rbrack \) be a domain which contains \( \mathfrak{o} \) and is a simple ring extension of \( \mathfrak{o} \) . Let \( {\mathfrak{p}}^{\prime } \) be a prime ideal in \( {\mathfrak{o}}^{\pr... | PROOF. We first make a remark which will be useful in the proof of either part of the proposition. Let \( {q}^{\prime } \) be a prime ideal in \( {o}^{\prime } \) such that \( {\mathfrak{p}}^{\prime } > {\mathfrak{q}}^{\prime } \) and assume that \( {\mathfrak{p}}^{\prime } \cap \mathfrak{o} = {\mathfrak{q}}^{\prime } ... | Yes |
Proposition 3. Let \( \mathfrak{o} \) be a noetherian integral domain and let \( {T}_{1},{T}_{2},\cdots ,{T}_{n} \) be transcendentals which are algebraically independent over \( \mathfrak{o} \) . If for any \( n \) the domain \( \mathfrak{o}\left\lbrack {{T}_{1},{T}_{2},\cdots ,{T}_{n}}\right\rbrack \) satisfies the c... | PROOF. Let \( {\mathfrak{o}}^{\prime },\mathfrak{p},{\mathfrak{p}}^{\prime } \) have the same meaning as in Proposition 2 and let \( {\mathfrak{O}}^{\prime } = \mathfrak{o}\left\lbrack {{T}_{1},{T}_{2},\cdots ,{T}_{n}}\right\rbrack \) . We have \( {\mathfrak{o}}^{\prime } = {\mathfrak{O}}^{\prime }/{\mathfrak{M}}^{\pri... | Yes |
Lemma 1. Let \( K \) be a field, \( {K}_{0} \) a subfield of \( K, v \) a valuation of \( K \) and \( {v}_{0} \) the restriction of \( v \) to \( {K}_{0} \) . If \( \Delta \) and \( {\Delta }_{0} \) are the residue fields of \( v \) and \( {v}_{0} \) respectively, and if \( \operatorname{tr.d.}K/{K}_{0} \) is finite, t... | PROOF. Let tr.d. \( K/{K}_{0} = g \) and tr.d. \( \Delta /{\Delta }_{0} = h \), so that \( h \leqq g \) . Fix \( h \) elements \( {x}_{1},{x}_{2},\cdots ,{x}_{h} \) in \( K \) such that their \( v \) -residues \( {\bar{x}}_{i} \) are algebraically independent over \( {\Delta }_{0} \) . Let \( {K}^{\prime } = {K}_{0}\le... | Yes |
Corollary 1. If the assumption \( \mathrm{r} \) . \( \operatorname{rank}v + {\dim }_{0}v = \dim \left( 0\right) + \) tr.d. \( {K}^{\prime }/K \) of Proposition 3 is replaced by the stronger assumption \( \operatorname{rank}v + \) \( {\dim }_{\mathfrak{o}}v = \dim \left( \mathfrak{o}\right) + \operatorname{tr}.\mathrm{d... | This follows from Proposition 3 and from the remark made just before the statement of Lemma 2. | No |
Proposition 1. If a \( v \) -ideal \( \mathfrak{A} \) (associated with a given valuation \( v \) ) in \( R \) admits an irredundant primary decomposition \( \mathfrak{A} = {\mathfrak{q}}_{1} \cap {\mathfrak{q}}_{2} \cap \cdots \cap {\mathfrak{q}}_{h} \) , then the prime ideals \( {\mathfrak{p}}_{i} = \sqrt{{\mathfrak{q... | PROOF. If \( \mathfrak{p} \) is a prime ideal of \( \mathfrak{A} \) there exists an element \( c \) in \( R \) such that \( c \notin \mathfrak{A} \) and \( \mathfrak{A} : \left( c\right) \) is primary for \( \mathfrak{p} \) (Vol. I, Ch. IV, \( §5 \), Theorem 6). By Lemma 1, \( \mathfrak{A} : \left( c\right) \) is a \( ... | Yes |
Proposition 2. Let \( r \) be the rank of \( v \), let \( \mathfrak{M} > {\mathfrak{M}}_{1} > \cdots > {\mathfrak{M}}_{r - 1} > \left( 0\right) \) be the prime ideals of \( {R}_{v} \) and let \( \mathfrak{p} = {\mathfrak{p}}_{0} > {\mathfrak{p}}_{1} > \cdots > {\mathfrak{p}}_{h - 1}\left( { > \left( 0\right) }\right) \... | PROOF. The proposition is obvious if \( r = 1 \) (see Lemma 3). We shall therefore use induction with respect to \( r \) .\n\nLet \( v = {v}_{1} \circ \bar{v} \), where \( {v}_{1} \) is of rank \( r - 1 \) and \( \bar{v} \) is of rank 1 . Then \( {\mathfrak{M}}_{1},{\mathfrak{M}}_{2},\cdots ,{\mathfrak{M}}_{r - 1} \) a... | Yes |
Proposition 1. The operation \( M \rightarrow {M}^{\prime } \) satisfies the following conditions:\n\n(a) \( {\mathfrak{o}}^{\prime } = \overline{\mathfrak{d}} \).\n\n(b) \( {M}^{\prime } \supset M \).\n\n(c) If \( M \supset N \) then \( {M}^{\prime } \supset {N}^{\prime } \).\n\n(d) \( {\left( {M}^{\prime }\right) }^{... | PROOF. Property (a) follows from VI,§ 4, Theorem 6, while (b) and (c) are self-evident. From (b) and (c) follows \( {\left( {M}^{\prime }\right) }^{\prime } \supset {M}^{\prime } \), but on the other hand, we have for any \( v \) in \( S : {\left( {M}^{\prime }\right) }^{\prime } \subset {R}_{v}{M}^{\prime } \subset {R... | Yes |
Proposition 1. Assuming that \( \bar{g} \neq {z}_{1} \) we set \( {\tau }^{\prime } = {t}_{2}/{t}_{1} \) (where \( \left( {{t}_{1},{t}_{2}}\right) \) is a fixed pair of regular parameters of \( \mathfrak{o} \) ) and \( {R}^{\prime } = \mathfrak{o}\left\lbrack {\tau }^{\prime }\right\rbrack + \) The set of elements \( F... | PROOF. Let \( \bar{G}\left( z\right) = \bar{g}\left( {1, z}\right) \) and let \( \alpha \) be a root of the irreducible polynomial \( \bar{G}\left( z\right) \) in some extension field of \( k \) (note that since \( {z}_{1} \neq \) \( \bar{g}\left( {{z}_{1},{z}_{2}}\right) ,\bar{G}\left( z\right) \) has positive degree)... | Yes |
Proposition 2. Let \( \mathfrak{A} \) be an ideal in \( \mathfrak{o} \), of order \( r \), and let \( {\bar{g}}^{\sigma } \) be the highest power of \( \bar{g} \) which divides the characteristic form \( c\left( \mathfrak{A}\right) \) of \( \mathfrak{A} \) . Then:\n\n(a) The order \( {r}^{\prime } \) of the transform \... | PROOF. We fix an element \( x \) in \( \mathfrak{A} \) such that the initial form \( \widetilde{x} \) is of degree \( r \) and is exactly divisible by \( {\bar{g}}^{\sigma } \) . Then \( {x}^{\prime } = x/{t}_{1}{}^{r} \in {\mathfrak{A}}^{\prime } \) . Let \( \bar{x} = {\bar{g}}^{\sigma }\psi \), where \( \bar{\psi } \... | Yes |
Proposition 4. Let \( \mathfrak{A} \) be an ideal in \( \mathfrak{o} \), primary for \( \mathfrak{m} \), and let \( r \) be the order of \( \mathfrak{A} \) . We assume that \( {\mathfrak{v}}^{\prime }\mathfrak{A} \cap \mathfrak{v} = \mathfrak{A} \), where \( {\mathfrak{o}}^{\prime } \) is the semi-local ring defined in... | PROOF. We set \( {\mathfrak{C}}^{\prime } = {\mathfrak{o}}^{\prime }\mathfrak{A},{\mathfrak{C}}_{i}^{\prime } = {\mathfrak{o}}_{i}^{\prime }\mathfrak{A} = {\mathfrak{o}}_{i}^{\prime }{\mathfrak{C}}^{\prime } \) . From the theory of quotient rings we know (Vol. I, Ch. IV, \( §{11} \), Theorem 19) that \( {\mathfrak{C}}_... | Yes |
Corollary 3. With the assumptions and notations as in Theorem 1, the decomposition (22) of \( \mathfrak{A} \) is the only decomposition of \( \mathfrak{A} \) into contracted ideals \( {\mathfrak{A}}_{i} \) satisfying (23) and such that \( c\left( {\mathfrak{A}}_{i}\right) \) is a power of \( {\widetilde{g}}_{i} \) . | Let \( \mathfrak{A} = {\mathfrak{A}}_{1} \cap {\mathfrak{A}}_{2} \cap \cdots \cap {\mathfrak{A}}_{m} \) be another such decomposition. Then \( c\left( {\widetilde{\mathfrak{A}}}_{i}\right) \) divides \( c\left( \mathfrak{A}\right) \), and thus the degree \( {\sigma }_{i} \) of \( c\left( {\widetilde{\mathfrak{A}}}_{i}\... | Yes |
Lemma 2. Let \( A \) be a local ring, \( \left\{ {{a}_{1},\cdots ,{a}_{n}}\right\} \) a prime sequence in \( A \) , and \( j \rightarrow i\left( j\right) \) a permutation of the indexing set \( \{ 1,2,\cdots, n\} \) . Then \( \left\{ {{a}_{i\left( 1\right) },{a}_{i\left( 2\right) },\cdots ,{a}_{i\left( n\right) }}\righ... | By elementary properties of permutations, it is sufficient to prove that, for every \( j,\left\{ {{a}_{1},\cdots ,{a}_{j - 1},{a}_{j + 1},{a}_{j},{a}_{j + 2},\cdots ,{a}_{n}}\right\} \) is a prime sequence. The property that \( {a}_{i} \) is prime to the ideal generated by the elements \( {a}_{k} \) which precede it in... | Yes |
Corollary 2. Let \( A \) be a Macaulay ring. For a finite subset \( S \) of \( A \) to be a prime sequence, it is necessary and sufficient that it be a subset of some system of parameters. | In fact, if \( S \) is a prime sequence, it is contained in a maximal prime sequence, i.e., in a system of parameters. The converse follows from Corollary 1 ((a) implies (c)), since any subset of a prime sequence is a prime sequence (Lemma 2). | Yes |
Corollary 3. Let \( A \) be a Macaulay ring. For every prime ideal \( \mathfrak{p} \) in \( A \), we have \( h\left( \mathfrak{p}\right) + \dim \left( {A/\mathfrak{p}}\right) = \dim \left( A\right) \) . | In fact, among the prime sequences which are contained in \( \mathfrak{p} \), we consider a maximal one, say \( \left\{ {{a}_{1},\cdots ,{a}_{j}}\right\} \) . Let \( \left\{ {{\mathfrak{p}}^{\prime }{}_{i}}\right\} \) be the set of associated prime ideals of \( \mathfrak{a} = A{a}_{1} + \cdots + A{a}_{j} \) . We have \... | Yes |
Corollary 6. Let \( A \) be a local ring, \( \widetilde{A} \) its completion. For \( A \) to be a Macaulay ring, it is necessary and sufficient that \( \widehat{A} \) be a Macaulay ring. | Let \( {a}_{1},\cdots ,{a}_{i} \) be elements of \( A \) . By Corollary 5 to VIII,§ 4, Theorem 11, and since \( \mathfrak{b}\widehat{A} \cap A = \mathfrak{b} \) for every ideal \( \mathfrak{b} \) in \( A \) (VIII, \( §2 \) , Corollary 2 to Theorem 5), the relations \( \left( {A{a}_{1} + \cdots + A{a}_{j - 1}}\right) : ... | Yes |
Theorem 1. Any group is isomorphic to a transformation group. | Proof. The transformation group that we shall define will act in the set \( \mathfrak{G} \) of the given group. With each element \( a \) of the group & we associate the mapping\n\n\[ x \rightarrow {xa} \]\n\nof the set \( \mathcal{G} \) into itself. We denote this mapping as \( {a}_{r} \) and call it the right multipl... | Yes |
Theorem 4. Let 3 be cyclic of order \( r\left( { < \infty }\right) \) . Then the order of any subgroup of 3 is a divisor of \( r \) and, if \( t \) is any positive divisor of \( r,3 \) possesses one and only one subgroup of order \( t \) . | It is customary to denote the number of positive divisors of an integer \( r \) by \( d\left( r\right) \) . Thus 3 possesses \( d\left( r\right) \) subgroups. | No |
Theorem 6. The image \( {\mathfrak{G}}_{\eta } \) of a homomorphism of \( \mathfrak{G} \) into \( {\mathfrak{G}}^{\prime } \) is a subgroup of \( {\mathfrak{G}}^{\prime } \) . | Proof. Since \( \left( {x\eta }\right) \left( {y\eta }\right) = \left( {xy}\right) \eta ,\& \eta \) is closed under the composition in \( {\mathcal{B}}^{\prime } \) . Also \( \left( {1\eta }\right) \left( {1\eta }\right) = {1\eta } \) so that \( {1\eta } \) is the identity \( {1}^{\prime } \) of \( {\mathcal{O}}^{\prim... | Yes |
Theorem 7. If \( \eta \) is a homomorphism of \( \mathfrak{G} \) into \( {\mathfrak{G}}^{\prime } \), the inverse image \( \mathcal{Q} = {\eta }^{-1}\left( {1}^{\prime }\right) \) of the identity of \( {\mathcal{B}}^{\prime } \) is an invariant subgroup of \( \mathfrak{G} \) . | Proof. We know that \( 1\mathrm{e}\mathfrak{K} \) . If \( {k}_{1},{k}_{2}\mathrm{e}\mathfrak{K} \), then \( \left( {{k}_{1}{k}_{2}}\right) \eta = \) \( \left( {{k}_{1}\eta }\right) \left( {{k}_{2}\eta }\right) = {1}^{\prime }{1}^{\prime } = {1}^{\prime } \) . Hence \( {k}_{1}{k}_{2} \) e \( \mathfrak{K} \) . Also if \(... | Yes |
Theorem 8. Let \( \eta \) be a homomorphism of \( \mathfrak{G} \) into \( {\mathfrak{G}}^{\prime } \) and let \( \mathfrak{H} \) be an invariant subgroup of \( \mathfrak{G} \) contained in \( \mathfrak{K} = {\eta }^{-1}\left( {1}^{\prime }\right) \) . Then the rule \( a\mathfrak{H} \rightarrow {a\eta } \) is a homomorp... | Suppose now that we particularize our considerations to the case in which \( \eta \) is a homomorphism of \( \mathfrak{G} \) onto \( {\mathfrak{G}}^{\prime } \) . If \( \mathfrak{K} \) is the kernel, then we see that the induced mapping \( \bar{\eta } \) of \( \overline{\mathcal{B}} = \mathcal{G}/\mathcal{R} \) onto \(... | No |
Theorem 1. If \( \mathfrak{R} \) is a commutative ring with an identity, a matrix (a) \( \varepsilon {\Re }_{n} \) is a unit if and only if its determinant is a unit in \( \Re \) . | To prove the necessity we require the fundamental multiplication rule\n\n(11)\n\[ \det \left( a\right) \left( b\right) = \det \left( a\right) \det \left( b\right) \text{.} \]\n\nIf \( \left( a\right) \left( b\right) = 1 \), then this gives det \( \left( a\right) \) det \( \left( b\right) = 1 \) . Hence det \( \left( a\... | No |
Theorem 2. The order of the group \( M \) of units of \( I/\left( m\right) \) is the number of positive integers that are less than \( m \) and are relatively prime to \( m\left( {\left( {a, m}\right) = 1}\right) \) . | This number is denoted as \( \phi \left( m\right) \) and the function of \( m \) thus determined is called Euler \( \phi \) -function (totient). We know that, if \( \mathfrak{G} \) is a finite group of order \( n \), then \( {a}^{n} = 1 \) for every \( {a\varepsilon }\& \) Applying this to \( M \) we see that, if \( \l... | Yes |
Theorem 4. If \( \eta \) is a homomorphism of \( \mathfrak{A} \) into \( {\mathfrak{A}}^{\prime } \), the image set \( \mathfrak{A}\eta \) is a subring of \( {\mathfrak{A}}^{\prime } \) . | Proof. Since \( \eta \) is a homomorphism of the additive group of \( \mathfrak{A},\mathfrak{A}\eta \) is a subgroup of the additive group of \( {\mathfrak{A}}^{\prime } \) . Since \( \left( {a\eta }\right) \left( {b\eta }\right) = \) \( {\left( ab\right) }_{\eta },{\mathfrak{A}}_{\eta } \) is closed under multiplicati... | Yes |
Theorem 5. The kernel of a homomorphism of a ring \( \mathfrak{A} \) is an ideal in \( \mathfrak{A} \) . | Proof. Let \( \mathcal{R} = {\eta }^{-1}\left( 0\right) \) . We know that \( \mathcal{R} \) is a subgroup of the additive group of \( \mathfrak{A} \) . Now let \( b \) e \( \mathfrak{K} \) and let \( a \) be arbitrary in \( \mathfrak{A} \) . Then \( \left( {ab}\right) \eta = \left( {a\eta }\right) \left( {b\eta }\right... | Yes |
Theorem 6. Let \( \eta \) be a homomorphism of the ring \( \mathfrak{A} \) into the ring \( {\mathfrak{A}}^{\prime } \) with kernel \( \mathfrak{K} \) and let \( \mathfrak{B} \) be an ideal of \( \mathfrak{A} \) contained in \( \mathfrak{K} \) . Then the correspondence \( \bar{\eta } : a + \mathfrak{B} \rightarrow {a\e... | If \( {\mathfrak{A}}^{\prime } = \mathfrak{A}\eta \) and \( \mathfrak{B} = \mathfrak{K} \), then \( \bar{\eta } \) is an isomorphism of \( \overline{\mathfrak{A}} \) onto \( {\mathfrak{A}}^{\prime } \) . This, together with an earlier result, gives the Fundamental theorem of homomorphism of rings. The difference ring \... | Yes |
Theorem 8. A ring \( \mathfrak{A} \) with an identity \( 1 \neq 0 \) is a division ring if and only if it has no proper left (right )ideals. | Proof. Suppose first that \( \mathfrak{A} \) is a division ring. Then, if \( \mathfrak{B} \) is a left ideal in \( \mathfrak{A} \neq 0 \) , \( \mathfrak{B} \) contains an element \( b \neq 0 \) . Then \( 1 = \) \( {b}^{-1}{b\varepsilon }\mathfrak{B} \) and every \( x = {x1} \) is in \( \mathfrak{B} \) . Hence \( \mathf... | Yes |
Theorem 9. Let \( \mathfrak{G} \) be an arbitrary commutative group (written additively) and let \( \mathfrak{E} \) be the totality of endomorphisms of \( \mathfrak{G} \) . Then \( \mathfrak{E} \) is closed relative to the addition composition defined by \( a\left( {\eta + \rho }\right) \) \( = {a\eta } + {a\rho } \) a... | We call \( \mathfrak{E} \) the ring of endomorphisms of \( \mathfrak{G} \) . More generally we shall be interested in considering subrings of rings \( \mathfrak{E} \) . Such a subring will be called a ring of endomorphisms and we shall see in the next section that these rings play the same role in ring theory that tran... | No |
Theorem 11. If \( \mathfrak{A} \) is a ring with an identity, then any mapping in \( \mathfrak{A} \) , + that commutes with all the left (right) multiplications is a right (left) multiplication. | The proof of this theorem is identical with that of the corresponding group result given on p. 30. | No |
Theorem 1. Any ring can be imbedded in a ring with an identity. | We note also that the ring of integers is imbedded in the ring \( \mathfrak{B} \) since the mapping \( m \rightarrow \left( {m,0}\right) \) is an isomorphism of \( I \) onto a subring \( {I}^{\prime } \) of \( \mathfrak{B} \) . We now simplify our notation by writing \( m \) for \( \left( {m,0}\right) \) and \( a \) fo... | No |
Theorem 2. Any commutative integral domain \( \left( { \neq 0}\right) \) can be imbedded in a field. | We shall now note that \( \mathfrak{F} \) is a minimal field containing the image \( \mathfrak{A} \) of \( \mathfrak{A} \) . This is clear since any \( a/b \) of \( \mathfrak{F} \) can be written in the form \( a/b = \left( {{ab}/b}\right) \left( {b/{b}^{2}}\right) = \left( {{ab}/b}\right) {\left( {b}^{2}/b\right) }^{-... | No |
Theorem 6. Every ideal in \( \mathfrak{F}\left\lbrack x\right\rbrack \) , \( \mathfrak{F} \) a field, is a principal ideal. | Proof. Let \( \mathfrak{B} \) be an ideal in \( \mathfrak{F}\left\lbrack x\right\rbrack \) . If \( \mathfrak{B} = 0 \), the ideal consisting of 0 alone, then \( \mathfrak{B} = \left( 0\right) \), the principal ideal generated by 0 . Assume therefore that \( \mathfrak{B} \neq 0 \) . Let \( g\left( x\right) \) be a non-z... | Yes |
Corollary 1. If \( \mathfrak{F} \) is a field, any polynomial ring \( \mathfrak{F}\left\lbrack u\right\rbrack \cong \) \( \mathfrak{F}\left\lbrack x\right\rbrack /\left( {g\left( x\right) }\right) \) where either \( g\left( x\right) = 0 \) or \( g\left( x\right) \) is a polynomial of positive degree. | The possibility that \( g\left( x\right) \) is a non-zero polynomial of 0 degree is excluded since it implies that \( \left( {g\left( x\right) }\right) = \mathfrak{F}\left\lbrack x\right\rbrack \) . | No |
Theorem 8. Let \( {\mathfrak{A}}_{i}, i = 1,2 \), be a ring with an identity and let \( {\mathfrak{A}}_{i}\left\lbrack {{x}_{1i},{x}_{2i},\cdots ,{x}_{ri}}\right\rbrack \) be a ring of polynomials in the algebraically independent elements \( {x}_{ji} \) . Then any homomorphism (isomorphism) of \( {\mathfrak{A}}_{1} \) ... | The case \( r = 1 \) of this theorem has been proved in the preceding section. The extension to arbitrary \( r \) is immediate by induction. The details of the argument will be left to the reader. | No |
Theorem 9. Every symmetric polynomial is expressible as a polynomial in the elementary symmetric polynomials \( {p}_{i} \) defined in (25). The elementary symmetric polynomials \( {p}_{1},{p}_{2},\cdots ,{p}_{r} \) are algebraically independent over \( \mathfrak{A} \) . Every \( {x}_{i} \) is algebraic over \( \mathfra... | The last statement of the theorem is clear since\n\n\[ F\left( {x}_{i}\right) = {x}_{i}^{r} - {p}_{1}{x}_{i}{}^{r - 1} + \cdots + {\left( -1\right) }^{r}{p}_{r} = 0. \] | No |
Theorem 10. If \( \mathfrak{F} \) is an infinite field and \( f\left( {{x}_{1},{x}_{2},\cdots ,{x}_{r}}\right) \) is a polynomial \( \neq 0 \) in the polynomial domain \( \mathfrak{F}\left\lbrack {{x}_{1},{x}_{2},\cdots ,{x}_{r}}\right\rbrack ,{x}_{i} \) algebraically independent, then there exist elements \( {c}_{1},{... | Proof. The case \( r = 1 \) has been proved above. Hence we assume that the theorem holds for \( r - 1{x}^{\prime }\mathrm{s} \) . We write\n\n\[ f\left( {{x}_{1},{x}_{2},\cdots ,{x}_{r}}\right) = {B}_{0} + {B}_{1}{x}_{r} + {B}_{2}{x}_{r}^{2} + \cdots + {B}_{n}{x}_{r}^{n} \]\n\nwhere \( {B}_{i} \) e \( \mathfrak{F}\lef... | Yes |
Lemma 1. If \( C \) holds in \( \mathfrak{S} \), then any finite number of elements of \( \mathfrak{S} \) have a g.c.d. | Let \( a, b, c\mathrm{e}\mathfrak{S} \) and set \( r = \left( {a,\left( {b, c}\right) }\right) \) . Then \( r \mid a \) and \( r \mid \left( {b, c}\right) \) so that \( r \mid b \) and \( r \mid c \) . Also if \( s \mid a, b, c \) then \( s \mid a \) and \( s \mid \left( {b, c}\right) \) so that \( s \mid \left( {a,\le... | Yes |
Lemma 3. \( c\left( {a, b}\right) \sim \left( {{ca},{cb}}\right) \) . | Proof. Write \( d = \left( {a, b}\right) \) and \( e = \left( {{ca},{cb}}\right) \) . Then \( {cd} \mid {ca} \) and \( {cd} \mid {cb} \) . Hence \( {cd} \mid e \) . On the other hand, \( {ca} = {ex} \) and \( {cb} = {ey} \) and if \( e = {cdu} \), then\n\n\[ \n{ca} = {cdux},\;{cb} = {cduy}.\n\]\n\nHence \( a = {dux} \)... | Yes |
Lemma 4. If \( \left( {a, b}\right) \sim 1 \) and \( \left( {a, c}\right) \sim 1 \) then \( \left( {a,{bc}}\right) \sim 1 \). | Proof. If \( \left( {a, b}\right) \sim 1 \), then \( \left( {{ac},{bc}}\right) \sim c \) . Hence \( 1 \sim \left( {a, c}\right) \sim \) \( \left( {a,\left( {{ac},{bc}}\right) }\right) \sim \left( {\left( {a,{ac}}\right) ,{bc}}\right) \sim \left( {a,{bc}}\right) . | No |
Theorem 1. If \( \otimes \) is a commutative semi-group with identity and cancellation law and \( \mathfrak{S} \) satisfies \( A \) and \( C \), then \( \mathfrak{S} \) is Gaussian. | We have seen in the Introduction that the semi-group of positive integers and the domain of integers have the greatest common divisor property \( \mathrm{C} \) . Also it is clear by consideration of absolute values that \( A \) holds in these systems. Hence we see that they are Gaussian. | No |
Theorem 2. Every principal ideal domain is Gaussian. | We have seen that, if \( \mathfrak{F} \) is a field, then \( \mathfrak{F}\left\lbrack x\right\rbrack, x \) transcendental, is a principal ideal domain (Chapter III,§ 6). Hence \( \mathfrak{F}\left\lbrack x\right\rbrack \) is Gaussian. | No |
Theorem 3. Every Euclidean domain is a principal ideal domain. | Proof. Let \( \mathfrak{B} \) be any ideal in the Euclidean domain \( \mathfrak{A} \) . If \( \mathfrak{B} = 0 \), then \( \mathfrak{B} = \left( 0\right) \) . Now let \( \mathfrak{B} \neq 0 \) . Then \( \mathfrak{B} \) contains elements for which \( \delta > 0 \) and since the \( \delta \) ’s are non-negative integers ... | Yes |
Lemma 1. If \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) are primitive in \( \mathfrak{A}\left\lbrack x\right\rbrack \) and are associates in \( \mathfrak{F}\left\lbrack x\right\rbrack \), then \( {f}_{1}\left( x\right) \) and \( {f}_{2}\left( x\right) \) are associates in \( \mathfrak{A}\left\lbrack x... | Proof. Since the units of \( \mathfrak{F}\left\lbrack x\right\rbrack \) are the non-zero elements of \( \mathfrak{F} \), we have \( {f}_{1}\left( x\right) = \alpha {f}_{2}\left( x\right) ,\alpha \neq 0 \) in \( \mathfrak{F} \) . Write \( \alpha = {d}_{2}{d}_{1}{}^{-1},{d}_{i} \) in A. Then \( {d}_{1}{f}_{1}\left( x\rig... | Yes |
Lemma 2 (Gauss). The product of primitive polynomials is primitive. | Proof. Let \( f\left( x\right) = {a}_{0} + {a}_{1}x + \cdots + {a}_{n}{x}^{n} \) and \( g\left( x\right) = {b}_{0} + {b}_{1}x \) \( + \cdots + {b}_{m}{x}^{m} \) be primitive and suppose that \( f\left( x\right) g\left( x\right) = {c}_{0} + {c}_{1}x + \) \( \cdots + {c}_{n + m}{x}^{m + n} \) is not primitive. Then there... | Yes |
Lemma 3. If \( f\left( x\right) \) is an irreducible polynomial of degree \( > 0 \) in \( \mathfrak{A}\left\lbrack x\right\rbrack, f\left( x\right) \) is irreducible in \( \mathfrak{F}\left\lbrack x\right\rbrack \) . | Proof. Since \( f\left( x\right) \) is irreducible, it is primitive. Now let \( f\left( x\right) \) be any primitive polynomial in \( \mathfrak{A}\left\lbrack x\right\rbrack \) and suppose that, in \( \mathfrak{F}\left\lbrack x\right\rbrack \) , \( f\left( x\right) = {\phi }_{1}\left( x\right) {\phi }_{2}\left( x\right... | Yes |
Theorem 2. Let \( \eta \) be an \( M \) -homomorphism of \( \mathfrak{G} \) onto \( {\mathfrak{G}}^{\prime } \) with kernel \( \mathfrak{K} \) and let \( \{ \mathfrak{H}\} \) be the collection of \( M \) -subgroups of \( \mathfrak{G} \) that contain \( \mathfrak{K} \) . Then the mapping \( \mathfrak{H} \rightarrow \mat... | Proof. We have seen that \( \mathfrak{H} \rightarrow \mathfrak{H}\eta \) is a mapping of \( \{ \mathfrak{H}\} \) onto the set of \( M \) -subgroups of \( {\mathfrak{G}}^{\prime } \) . Also if \( {\mathfrak{H}}_{1} \) and \( {\mathfrak{H}}_{2}\varepsilon \{ \mathfrak{H}\} \) and \( {\mathfrak{H}}_{1}\eta = {\mathfrak{H}... | Yes |
Theorem 4. A necessary and sufficient condition that an M-group \( \mathfrak{G} \) have a composition series is that \( \mathfrak{G} \) satisfies the two chain conditions. | Sufficiency. We shall show first that if \( \mathfrak{H} \neq 1 \) is a term of a normal series, then \( \mathfrak{H} \) contains a maximal invariant \( M \) -subgroup. Thus, either \( {\mathfrak{H}}_{1} = 1 \) is maximal invariant or there exists a proper invariant \( M \) -subgroup \( {\mathfrak{H}}_{2} \) of \( \mat... | Yes |
Theorem 6. If \( \mathfrak{G} \) is a finite cyclic group of order \( n = {p}_{1}{}^{{e}_{1}}{p}_{2}{}^{{e}_{2}}\cdots \) \( {p}_{s}{}^{{e}_{i}},{p}_{i} \) prime, \( {p}_{i} \neq {p}_{j} \) if \( i \neq j \), then \( \mathfrak{G} \) is a direct product of cyclic groups of orders \( {p}_{i}^{{e}_{i}}, i = 1,2,\cdots, s ... | Proof. Let \( {\mathfrak{G}}_{i} \) be the subgroup of order \( {p}_{i}{}^{{e}_{i}} \) and set \( {\mathfrak{G}}^{\prime } = \) \( {\mathfrak{G}}_{1}{\mathfrak{G}}_{2}\cdots {\mathfrak{G}}_{s} \) . This subgroup has order \( {n}^{\prime } \) divisible by \( {p}_{i}{}^{{e}_{i}} \) since \( {\mathfrak{G}}^{\prime } \sups... | Yes |
Theorem 7. If \( \mathfrak{G} \) contains \( M \) -subgroups \( {\mathfrak{G}}_{i}, i = 1,2,\cdots, n \) , such that (1) \( {a}_{i}{a}_{j} = {a}_{j}{a}_{i} \) for any \( {a}_{i}\varepsilon {\mathfrak{G}}_{i} \) and any \( {a}_{j}\varepsilon {\mathfrak{G}}_{j}, i \neq j \) , and (2) every element of \( \mathfrak{G} \) c... | Proof. We note first that each \( {\mathcal{B}}_{i} \) is invariant in \( \mathcal{B} \) ; for, if \( {g}_{i} \) e \( {\mathfrak{G}}_{i} \) and \( a = {a}_{1}{a}_{2}\cdots {a}_{n},{a}_{j} \) e \( {\mathfrak{G}}_{j} \), then\n\n\[ \n{a}^{-1}{g}_{i}a = {a}_{n}^{-1}\cdots {a}_{2}{}^{-1}{a}_{1}{}^{-1}{g}_{i}{a}_{1}{a}_{2}\... | Yes |
Theorem 10. Let \( \mathfrak{G} \) be an \( M \) -group that satisfies the descending and the ascending chain conditions for invariant \( M \) -subgroups. Then if \( \eta \) is a normal M-endomorphism, \( \eta \) is an automorphism if either (1) \( \eta \) is \( 1 - 1 \) or \( \left( 2\right) \otimes \eta = \otimes \) ... | Proof. Assume that \( \eta \) is \( 1 - 1 \) . Then if \( \mathcal{G}{{\eta }^{r}}^{-1} = \mathcal{G}{\eta }^{r} \) for some \( r = 1,2,\cdots \), any \( {y\varepsilon }\circlearrowleft {\eta }^{r - 2} \) has the property that \( {y\eta } = x{\eta }^{r} = \) \( \left( {x{\eta }^{r - 1}}\right) \eta \) for a suitable el... | Yes |
Theorem 11 (Fitting’s lemma). Let \( \\mathfrak{G} \) be an \( M \) -group that satisfies the chain conditions for invariant \( M \) -subgroups and let \( \\eta \) be a normal M-endomorphism of \( \\mathfrak{G} \) . Then \( \\mathfrak{G} = \\Re \\times \\mathfrak{H} \) where \( \\Re \) is the radical of \( \\eta \) and... | Proof. We have the descending chain of invariant \( M \) -subgroups \( \\mathfrak{G} \\supseteq \\mathfrak{G}\\eta \\supseteq {\\mathfrak{G}}_{\\eta }{}^{2} \\supseteq \\cdots \) . Hence there is an integer \( r \) such that \( \\circlearrowleft {\\eta }^{r} = \\circlearrowleft {\\eta }^{r + 1} \) . Then \( \\circlearr... | Yes |
Corollary 2. Let \( \mathfrak{G} \) be as in Corollary 1 and let \( {\eta }_{1} \) and \( {\eta }_{2} \) be normal nilpotent \( M \) -endomorphisms, then, if \( {\eta }_{1} + {\eta }_{2} \) is an endomorphism, \( {\eta }_{1} + {\eta }_{2} \) is nilpotent. | Proof. According to Corollary 1, if \( \eta = {\eta }_{1} + {\eta }_{2} \) is not nilpotent, then it is an automorphism. Let \( {\eta }^{-1} \) be its inverse. Evidently this mapping is a normal \( M \) -endomorphism and we have \( {\eta }_{1}{\eta }^{-1} + {\eta }_{2}{\eta }^{-1} = 1 \), or \( {\lambda }_{1} + {\lambd... | Yes |
Theorem 1. If \( X \) is a set of generators for a unitary module \( \mathfrak{M} \) , then every element of \( \mathfrak{M} \) can be written in the form\n\n(7)\n\n\[ \n{a}_{1}{x}_{1} + {a}_{2}{x}_{2} + \cdots + {a}_{r}{x}_{r} \n\]\n\nwhere the \( {a}_{i}\varepsilon \mathfrak{A} \) and the \( {x}_{i}{\varepsilon X} \)... | Proof. Let \( x \) be any element of \( \mathfrak{M} \) and write \( x = \sum {a}_{i}{y}_{i} \) for suitable \( {a}_{i} \) in \( \mathfrak{A},{y}_{i} \) in \( \mathfrak{M} \) . Then there exist elements \( {x}_{j} \) in \( X \) such that\n\n\[ \n{y}_{i} = \sum {m}_{ij}{x}_{j} + \sum {a}_{ij}{x}_{j},\;{m}_{ij}\text{ e }... | Yes |
Theorem 2. A module \( \mathfrak{M} \) satisfies the ascending chain condition for submodules if and only if every submodule of \( \mathfrak{M} \) is finitely generated. | Proof. We assume first that the ascending chain condition holds and we let \( \mathfrak{N} \) be any submodule of \( \mathfrak{M} \) . If \( \mathfrak{N} = 0 \), then \( \mathfrak{N} \) is generated by 0 . If \( \mathfrak{N} \neq 0 \), let \( {u}_{1} \) be any non-zero element of \( \mathfrak{N} \) and let \( \left( {u... | Yes |
Lemma 1. If \( \mathfrak{N} \subseteq \mathfrak{P} \) and \( {\mathfrak{J}}_{j}\left( \mathfrak{N}\right) = {\mathfrak{J}}_{j}\left( \mathfrak{P}\right) \) for all \( j \), then \( \mathfrak{N} = \mathfrak{P} \) . | Proof. Let \( y = {b}_{1}{x}_{1} + {b}_{2}{x}_{2} + \cdots + {b}_{r}{x}_{r} \) be any element of 9. Then \( {b}_{1} \) e \( {\Im }_{1}\left( \mathfrak{P}\right) = {\Im }_{1}\left( \mathfrak{N}\right) \) . Hence, there is an element \( {y}^{\prime } \) in \( \mathfrak{N} \) of the form \( {b}_{1}{x}_{1} + {b}_{2}{}^{\pr... | Yes |
Theorem 3. If \( \mathfrak{A} \) is a ring that satisfies the ascending (descending) chain condition for left ideals, then any finitely generated unitary \( \mathfrak{A} \) -module \( \mathfrak{M} \) satisfies the ascending (descending) chain condition for submodules. | The proof of this result for descending chains is similar to the above. | No |
Lemma 1. If \( {\mathfrak{Q}}_{1} \) and \( {\mathfrak{Q}}_{2} \) are primary ideals that have the same radical \( \mathfrak{P} \), then \( {\mathfrak{Q}}_{1} \cap {\mathfrak{Q}}_{2} \) is primary. | Proof. We know that \( \Re \left( {{\mathfrak{Q}}_{1} \cap {\mathfrak{Q}}_{2}}\right) = \Re \left( {\mathfrak{Q}}_{1}\right) \cap \Re \left( {\mathfrak{Q}}_{2}\right) \) . Hence \( \Re \left( {{\mathfrak{Q}}_{1} \cap {\mathfrak{Q}}_{2}}\right) = \mathfrak{P} \) . Now let \( a \) be a zero-divisor modulo \( {\mathfrak{Q... | Yes |
Lemma 2. Let \( \mathfrak{Q} \) be a primary ideal and let \( {\mathfrak{P}}^{\prime } \) be a prime ideal containing \( \mathfrak{Q} \) . Then \( {\mathfrak{P}}^{\prime } \supseteq \mathfrak{P} = \mathfrak{R}\left( \mathfrak{Q}\right) \) . | Proof. If \( z \equiv 0\left( {\;\operatorname{mod}\;\mathfrak{P}}\right) ,{z}^{r} \equiv 0\left( {\;\operatorname{mod}\;\mathfrak{Q}}\right) \) for some integer \( r \) . Hence \( {z}^{r} \equiv 0\left( {\;\operatorname{mod}\;{\mathfrak{P}}^{\prime }}\right) \) . Since \( {\mathfrak{P}}^{\prime } \) is prime, \( z \eq... | Yes |
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