Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Theorem 3.3. Ifa group \( \mathrm{G} \) satisfies either the ascending or descending chain condition on normal subgroups, then \( \mathrm{G} \) is the direct product of a finite number of indecomposable subgroups. | SKETCH OF PROOF. Suppose \( G \) is not a finite direct product of indecomposable subgroups. Let \( S \) be the set of all normal subgroups \( H \) of \( G \) such that \( H \) is a direct factor of \( G \) (that is, \( G = H \times {T}_{H} \) for some subgroup \( {T}_{H} \) of \( G \) ) and \( H \) is not a finite dir... | No |
Lemma 3.4. Let \( \mathrm{G} \) be a group that satisfies the ascending [resp. descending] chain condition on normal subgroups and \( \mathrm{f} \) a [normal] endomorphism of \( \mathrm{G} \). Then \( \mathrm{f} \) is an automorphism if and only if \( \mathbf{f} \) is an epimorphism [resp. monomorphism]. | PROOF. Suppose \( G \) satisfies the ACC and \( f \) is an epimorphism. The ascending chain of normal subgroups \( \langle e\rangle < \operatorname{Ker}f < \operatorname{Ker}{f}^{2} < \cdots \) (where \( {f}^{k} = {ff}\cdots f \) ) must become constant, say \( \operatorname{Ker}{f}^{n} = \operatorname{Ker}{f}^{n + 1} \... | Yes |
Lemma 3.5. (Fitting) If \( \mathrm{G} \) is a group that satisfies both the ascending and descending chain conditions on normal subgroups and \( \mathrm{f} \) is a normal endomorphism of \( \mathrm{G} \), then for some \( \mathrm{n} \geq 1,\mathrm{G} = \operatorname{Ker}{\mathrm{f}}^{\mathrm{n}} \times \operatorname{Im... | PROOF. Since \( f \) is a normal endomorphism each \( \operatorname{Im}{f}^{k}\left( {k \geq 1}\right) \) is normal in \( G \) . Hence we have two chains of normal subgroups:\n\n\[ G > \operatorname{Im}f > \operatorname{Im}{f}^{2} > \cdots \;\text{ and }\;\langle e\rangle < \operatorname{Ker}f < \operatorname{Ker}{f}^{... | Yes |
Corollary 3.6. If \( \mathrm{G} \) is an indecomposable group that satisfies both the ascending and descending chain conditions on normal subgroup’s and \( \mathbf{f} \) is a normal endomorphism of \( \mathbf{G} \) , then either \( \mathrm{f} \) is nilpotent or \( \mathrm{f} \) is an automorphism. | PROOF. For some \( n \geq 1, G = \operatorname{Ker}{f}^{n} \times \operatorname{Im}{f}^{n} \) by Fitting’s Lemma. Since \( G \) is indecomposable either \( \operatorname{Ker}{f}^{n} = \langle e\rangle \) or \( \operatorname{Im}{f}^{n} = \langle e\rangle \) . The latter implies that \( f \) is nilpotent. If \( \operator... | Yes |
Corollary 3.7. Let \( \mathrm{G}\left( { \neq \langle \mathrm{e}\rangle }\right) \) be an indecomposable group that satisfies both the ascending and descending chain conditions on normal subgroups. If \( {\mathrm{f}}_{1},\ldots ,{\mathrm{f}}_{\mathrm{n}} \) are normal nilpotent endomorphisms of \( \mathrm{G} \) such th... | SKETCH OF PROOF. Since each \( {f}_{{i}_{1}} + \cdots + {f}_{{i}_{r}} \) is an endomorphism that is normal (Exercise 8(c)), the proof will follow by induction once the case \( n = 2 \) is established. If \( {f}_{1} + {f}_{2} \) is not nilpotent, it is an automorphism by Corollary 3.6. Verify that the inverse \( g \) of... | No |
Theorem 3.8. (Krull-Schmidt) Let \( \mathrm{G} \) be a group that satisfies both the ascending and descending chain conditions on normal subgroups. If \( \mathrm{G} = {\mathrm{G}}_{1} \times {\mathrm{G}}_{2} \times \cdots \times {\mathrm{G}}_{\mathrm{s}} \) and \( \mathrm{G} = {\mathrm{H}}_{1} \times {\mathrm{H}}_{2} \... | SKETCH OF PROOF OF 3.8. Let \( P\left( 0\right) \) be the statement \( G = {H}_{1} \times \cdots \times {H}_{t} \) . For \( 1 \leq r \leq \min \left( {s, t}\right) \) let \( P\left( r\right) \) be the statement: there is a reindexing of \( {H}_{1},\ldots ,{H}_{t} \) such that \( {G}_{i} \cong {H}_{i} \) for \( i = 1,2,... | Yes |
Theorem 4.2. Let \( \mathrm{G} \) be a group that acts on a set \( \mathrm{S} \) .\n\n(i) The relation on \( \mathrm{S} \) defined by\n\n\[ \mathrm{x} \sim {\mathrm{x}}^{\prime } \Leftrightarrow \mathrm{{gx}} = {\mathrm{x}}^{\prime }\;\text{ for some }\;\mathrm{g}\varepsilon \mathrm{G} \]\n\nis an equivalence relation.... | PROOF. Exercise. | No |
Theorem 4.3. If a group \( \mathrm{G} \) acts on a set \( \mathrm{S} \), then the cardinal number of the orbit of \( \mathrm{x}\varepsilon \mathrm{S} \) is the index \( \left\lbrack {\mathrm{G} : {\mathrm{G}}_{\mathrm{x}}}\right\rbrack \) . | PROOF. Let \( g, h \in G \) . Since\n\n\[ \n{gx} = {hx} \Leftrightarrow {g}^{-1}{hx} = x \Leftrightarrow {g}^{-1}h \in {G}_{x} \Leftrightarrow h{G}_{x} = g{G}_{x},\n\]\n\nit follows that the map given by \( g{G}_{x} \mapsto {gx} \) is a well-defined bijection of the set of cosets of \( {G}_{x} \) in \( G \) onto the or... | Yes |
Corollary 4.4. Let \( \mathrm{G} \) be a finite group and \( \mathrm{K} \) a subgroup of \( \mathrm{G} \). (i) The number of elements in the conjugacy class of \( \mathrm{x}\varepsilon \mathrm{G} \) is \( \left\lbrack {\mathrm{G} : {\mathrm{C}}_{\mathrm{G}}\left( \mathrm{x}\right) }\right\rbrack \), which divides \( \l... | PROOF. (i) and (iii) follow immediately from the preceding Theorem and Lagrange’s Theorem I.4.6. Since conjugacy is an equivalence relation on \( G \) (Theorem 4.2), \( G \) is the disjoint union of the conjugacy classes \( {\bar{x}}_{1},\ldots ,{\bar{x}}_{n} \), whence (ii) follows from (i). | Yes |
Theorem 4.5. If a group \( \mathrm{G} \) acts on a set \( \mathrm{S} \), then this action induces a homomorphism \( \mathrm{G} \rightarrow \mathrm{A}\left( \mathrm{S}\right) \), where \( \mathrm{A}\left( \mathrm{S}\right) \) is the group of all permutations of \( \mathrm{S} \) . | PROOF. If \( g \in G \), define \( {\tau }_{g} : S \rightarrow S \) by \( x \mapsto {gx} \) . Since \( x = g\left( {{g}^{-1}x}\right) \) for all \( x \in S \) , \( {\tau }_{\theta } \) is surjective. Similarly \( {gx} = {gy}\left( {x,{y\varepsilon S}}\right) \) implies \( x = {g}^{-1}\left( {gx}\right) = {g}^{-1}\left(... | Yes |
Corollary 4.6. (Cayley) If \( \mathrm{G} \) is a group, then there is a monomorphism \( \mathrm{G} \rightarrow \mathrm{A}\left( \mathrm{G}\right) \) . Hence every group is isomorphic to a group of permutations. In particular every finite group is isomorphic to a subgroup of \( {\mathrm{S}}_{\mathrm{n}} \) with \( \math... | PROOF. Let \( G \) act on itself by left translation and apply Theorem 4.5 to obtain a homomorphism \( \tau : G \rightarrow A\left( G\right) \) . If \( \tau \left( g\right) = {\tau }_{g} = {1}_{G} \), then \( {gx} = {\tau }_{g}\left( x\right) = x \) for all \( x \in G \) . In particular \( {ge} = e \), whence \( g = e ... | Yes |
Proposition 4.8. Let \( \mathrm{H} \) be a subgroup of a group \( \mathrm{G} \) and let \( \mathrm{G} \) act on the set \( \mathrm{S} \) of all left cosets of \( \mathrm{H} \) in \( \mathrm{G} \) by left translation. Then the kernel of the induced homomorphism \( \mathrm{G} \rightarrow \mathrm{A}\left( \mathrm{S}\right... | PROOF. The induced homomorphism \( G \rightarrow A\left( S\right) \) is given by \( g \mapsto {\tau }_{g} \), where \( {\tau }_{g} : S \rightarrow S \) and \( {\tau }_{g}\left( {xH}\right) = {gxH} \) . If \( g \) is in the kernel, then \( {\tau }_{g} = {1}_{S} \) and \( {gxH} = {xH} \) for all \( x \in G \) ; in partic... | Yes |
Corollary 4.9. If \( \mathrm{H} \) is a subgroup of index \( \mathrm{n} \) in a group \( \mathrm{G} \) and no nontrivial normal subgroup of \( \mathrm{G} \) is contained in \( \mathrm{H} \), then \( \mathrm{G} \) is isomorphic to a subgroup of \( {\mathrm{S}}_{\mathrm{n}} \) . | PROOF. Apply Proposition 4.8 to \( H \) ; the kernel of \( G \rightarrow A\left( S\right) \) is a normal subgroup of \( G \) contained in \( H \) and must therefore be \( \langle e\rangle \) by hypothesis. Hence, \( G \rightarrow A\left( S\right) \) is a monomorphism. Therefore \( G \) is isomorphic to a subgroup of th... | Yes |
Corollary 4.10. If \( \\mathrm{H} \) is a subgroup of a finite group \( \\mathrm{G} \) of index \( \\mathrm{p} \), where \( \\mathrm{p} \) is the smallest prime dividing the order of \( \\mathbf{G} \), then \( \\mathbf{H} \) is normal in \( \\mathbf{G} \). | PROOF. Let \( S \) be the set of all left cosets of \( H \) in \( G \). Then \( A\\left( S\\right) \\cong {S}_{p} \) since \( \\left\\lbrack {G : H}\\right\\rbrack = p \). If \( K \) is the kernel of the homomorphism \( G \\rightarrow A\\left( S\\right) \) of Proposition 4.8, then \( K \) is normal in \( G \) and conta... | Yes |
Lemma 5.1. If a group \( \mathrm{H} \) of order \( {\mathrm{p}}^{\mathrm{n}} \) (p prime) acts on a finite set \( \mathrm{S} \) and if \( {\mathrm{S}}_{0} = \left\{ {\mathrm{x}\varepsilon \mathrm{S} \mid \mathrm{{hx}} = \mathrm{x}}\right. \) for all \( \mathrm{h}\varepsilon \mathrm{H} \) ), then \( \left| \mathrm{S}\ri... | PROOF OF 5.1. An orbit \( \bar{x} \) contains exactly one element if and only if \( x \in {S}_{0} \) . Hence \( S \) can be written as a disjoint union \( S = {S}_{0} \cup {\bar{x}}_{1} \cup {\bar{x}}_{2} \cup \cdots \cup {\bar{x}}_{n} \), with \( \left| {\bar{x}}_{i}\right| > 1 \) for all \( i \) . Hence \( \left| S\r... | Yes |
Theorem 5.2. (Cauchy) If \( \mathrm{G} \) is a finite group whose order is divisible by a prime \( \mathrm{p} \) , then \( \mathbf{G} \) contains an element of order \( \mathbf{p} \) . | PROOF. (J. H. McKay) Let \( S \) be the set of \( p \) -tuples of group elements \( \left\{ {\left( {{a}_{1},{a}_{2},\ldots ,{a}_{p}}\right) \mid {a}_{i} \in G\text{and}{a}_{1}{a}_{2}\cdots {a}_{p} = e}\right\} \) . Since \( {a}_{p} \) is uniquely determined as \( {\left( {a}_{1}{a}_{2}\cdots {a}_{p - 1}\right) }^{-1} ... | Yes |
Corollary 5.3. A finite group \( \mathrm{G} \) is a p-group if and only if \( \left| \mathrm{G}\right| \) is a power of \( \mathrm{p} \) . | PROOF. If \( G \) is a \( p \) -group and \( q \) a prime which divides \( \left| G\right| \), then \( G \) contains an element of order \( q \) by Cauchy’s Theorem. Since every element of \( G \) has order a power of \( p, q = p \) . Hence \( \left| G\right| \) is a power of \( p \) . The converse is an immediate cons... | Yes |
Corollary 5.4. The center \( \mathrm{C}\left( \mathrm{G}\right) \) of a nontrivial finite \( \mathrm{p} \) -group \( \mathrm{G} \) contains more than one element. | PROOF. Consider the class equation of \( G \) (see page 91):\n\n\[ \left| G\right| = \left| {C\left( G\right) }\right| + \sum \left\lbrack {G : {C}_{G}\left( {x}_{i}\right) }\right\rbrack \]\n\nSince each \( \left\lbrack {G : {C}_{G}\left( {x}_{i}\right) }\right\rbrack > 1 \) and divides \( \left| G\right| = {p}^{n}\le... | Yes |
Lemma 5.5. If \( \mathrm{H} \) is a p-subgroup of a finite group \( \mathrm{G} \), then \( \left\lbrack {{\mathrm{N}}_{\mathrm{G}}\left( \mathrm{H}\right) : \mathrm{H}}\right\rbrack \equiv \left\lbrack {\mathrm{G} : \mathrm{H}}\right\rbrack \) \( \left( {\;\operatorname{mod}\;\mathrm{p}}\right) \) . | PROOF. Let \( S \) be the set of left cosets of \( H \) in \( G \) and let \( H \) act on \( S \) by (left) translation. Then \( \left| S\right| = \left\lbrack {G : H}\right\rbrack \) . Also,\n\n\[ \n{xH\varepsilon }{S}_{0} \Leftrightarrow {hxH} = {xH}\text{ for all }{h\varepsilon H} \n\]\n\n\( \Leftrightarrow {x}^{-1}... | Yes |
Corollary 5.6. If \( \mathrm{H} \) is p-subgroup of a finite group \( \mathrm{G} \) such that \( \mathrm{p} \) divides \( \left\lbrack {\mathrm{G} : \mathrm{H}}\right\rbrack \), then \( {\mathrm{N}}_{\mathrm{G}}\left( \mathrm{H}\right) \neq \mathrm{H} \) . | PROOF. \( 0 \equiv \left\lbrack {G : H}\right\rbrack \equiv \left\lbrack {{N}_{G}\left( H\right) : H}\right\rbrack \left( {\;\operatorname{mod}\;p}\right) \) . Since \( \left\lbrack {{N}_{G}\left( H\right) : H}\right\rbrack \geq 1 \) in any case, we must have \( \left\lbrack {{N}_{G}\left( H\right) : H}\right\rbrack > ... | Yes |
Theorem 5.7. (First Sylow Theorem) Let \( \mathrm{G} \) be a group of order \( {\mathrm{p}}^{\mathrm{n}}\mathrm{m} \), with \( \mathrm{n} \geq 1 \) , \( \mathrm{p} \) prime, and \( \left( {\mathrm{p},\mathrm{m}}\right) = 1 \) . Then \( \mathrm{G} \) contains a subgroup of order \( {\mathrm{p}}^{\mathrm{i}} \) for each ... | PROOF. Since \( p\left| \right| G \mid, G \) contains an element \( a \), and therefore, a subgroup \( \langle a\rangle \) of order \( p \) by Cauchy’s Theorem. Proceeding by induction assume \( H \) is a subgroup of \( G \) of order \( {p}^{i}\left( {1 \leq i < n}\right) \) . Then \( p \mid \left\lbrack {G : H}\right\... | Yes |
Corollary 5.8. Let \( \mathrm{G} \) be a group of order \( {\mathrm{p}}^{\mathrm{n}}\mathrm{m} \) with \( \mathrm{p} \) prime, \( \mathrm{n} \geq 1 \) and \( \left( {\mathrm{m},\mathrm{p}}\right) = 1 \) . Let \( \mathrm{H} \) be a p-subgroup of \( \mathrm{G} \) .\n\n(i) \( \mathrm{H} \) is a Sylow p-subgroup of \( \mat... | SKETCH OF PROOF. (i) Corollaries I.4.6 and 5.3 and Theorem 5.7. | No |
Theorem 5.9. (Second Sylow Theorem) If \( \mathrm{H} \) is a p-subgroup of a finite group \( \mathrm{G} \), and \( \mathrm{P} \) is any Sylow \( \mathrm{p} \) -subgroup of \( \mathrm{G} \), then there exists \( \mathrm{x}\varepsilon \mathrm{G} \) such that \( \mathrm{H} < \mathrm{{xP}}{\mathrm{x}}^{-1} \) . In particul... | PROOF. Let \( S \) be the set of left cosets of \( P \) in \( G \) and let \( H \) act on \( S \) by (left) translation. \( \left| {S}_{0}\right| \equiv \left| S\right| = \left\lbrack {G : P}\right\rbrack \left( {\;\operatorname{mod}\;p}\right) \) by Lemma 5.1. But \( p \nmid \left\lbrack {G : P}\right\rbrack \) ; ther... | Yes |
Theorem 5.10. (Third Sylow Theorem) If \( \mathrm{G} \) is a finite group and \( \mathrm{p} \) a prime, then the number of Sylow p-subgroups of \( \mathrm{G} \) divides \( \left| \mathrm{G}\right| \) and is of the form \( \mathrm{{kp}} + 1 \) for some \( \mathrm{k} \geq 0 \) . | PROOF. By the second Sylow Theorem the number of Sylow \( p \) -subgroups is the number of conjugates of any one of them, say \( P \) . But this number is \( \left\lbrack {G : {N}_{G}\left( P\right) }\right\rbrack \), a divisor of \( \left| G\right| \), by Corollary 4.4. Let \( S \) be the set of all Sylow \( p \) -sub... | Yes |
Theorem 5.11. If \( \mathrm{P} \) is a Sylow p-subgroup of a finite group \( \mathrm{G} \), then \( {\mathrm{N}}_{\mathrm{G}}\left( {{\mathrm{N}}_{\mathrm{G}}\left( \mathrm{P}\right) }\right) \) \( = {\mathrm{N}}_{\mathrm{G}}\left( \mathrm{P}\right) \) . | PROOF. Every conjugate of \( P \) is a Sylow \( p \) -subgroup of \( G \) and of any subgroup of \( G \) that contains it. Since \( P \) is normal in \( N = {N}_{G}\left( P\right), P \) is the only Sylow \( p \) -subgroup of \( N \) by Theorem 5.9. Therefore,\n\n\[ x \in {N}_{G}\left( N\right) \Rightarrow {xN}{x}^{-1} ... | Yes |
Corollary 6.2. If \( \mathrm{p} \) is an odd prime, then every group of order \( 2\mathrm{p} \) is isomorphic either to the cyclic group \( {\mathbf{Z}}_{2\mathrm{p}} \) or the dihedral group \( {\mathbf{D}}_{\mathrm{p}} \) . | PROOF. Apply Proposition 6.1 with \( q = 2 \) . If \( G \) is not cyclic, the conditions on \( s \) imply \( s \equiv - 1\left( {\;\operatorname{mod}\;p}\right) \) . Hence \( G = \langle c, d\rangle ,\left| d\right| = 2,\left| c\right| = p \), and \( {dc} = {c}^{-1}d \) by Theorem I.3.4(v). Therefore, \( G \cong {D}_{p... | Yes |
Proposition 6.3. There are (up to isomorphism) exactly two distinct nonabelian groups of order 8: the quaternion group \( {\mathrm{Q}}_{8} \) and the dihedral group \( {\mathrm{D}}_{4} \). | SKETCH OF PROOF OF 6.3. Verify that \( {D}_{4} \cong {Q}_{8} \) (Exercise 10). If a group \( G \) of order 8 is nonabelian, then it cannot contain an element of order 8 or have every nonidentity element of order 2 (Exercise I.1.13). Hence \( G \) contains an element \( a \) of order 4. The group \( \langle a\rangle \) ... | No |
Proposition 6.4. There are (up to isomorphism) exactly three distinct nonabelian groups of order 12: the dihedral group \( {\mathrm{D}}_{6} \), the alternating group \( {\mathrm{A}}_{4} \), and a group \( \mathrm{T} \) generated by elements \( \mathrm{a},\mathrm{b} \) such that \( \left| \mathrm{a}\right| = 6,{\mathrm{... | SKETCH OF PROOF. Verify that there is a group \( T \) of order 12 as stated (Exercise 5) and that no two of \( {D}_{6},{A}_{4}, T \) are isomorphic (Exercise 6). If \( G \) is a non-abelian group of order 12, let \( P \) be a Sylow 3-subgroup of \( G \) . Then \( \left| P\right| = 3 \) and \( \left\lbrack {G : P}\right... | No |
Theorem 7.3. The direct product of a finite number of nilpotent groups is nilpotent. | PROOF. Suppose for convenience that \( G = H \times K \), the proof for more than two factors being similar. Assume inductively that \( {C}_{i}\left( G\right) = {C}_{i}\left( H\right) \times {C}_{i}\left( K\right) \) (the case \( i = 1 \) is obvious). Let \( {\pi }_{H} \) be the canonical epimorphism \( H \rightarrow H... | Yes |
Lemma 7.4. If \( \mathrm{H} \) is a proper subgroup of a nilpotent group \( \mathrm{G} \), then \( \mathrm{H} \) is a proper subgroup of its normalizer \( {\mathrm{N}}_{\mathrm{G}}\left( \mathrm{H}\right) \) . | PROOF. Let \( {C}_{0}\left( G\right) = \langle e\rangle \) and let \( n \) be the largest index such that \( {C}_{n}\left( G\right) < H \) ; (there is such an \( n \) since \( G \) is nilpotent and \( H \) a proper subgroup). Choose \( a \in {C}_{n + 1}\left( G\right) \) with \( a \notin H \) . Then for every \( h \in ... | Yes |
Proposition 7.5. A finite group is nilpotent if and only if it is the direct product of its Sylow subgroups. | PROOF. If \( G \) is the direct product of its Sylow \( p \) -subgroups, then \( G \) is nilpotent by Theorems 7.2 and 7.3. If \( G \) is nilpotent and \( P \) is a Sylow \( p \) -subgroup of \( G \) for some prime \( p \), then either \( P = G \) (and we are done) or \( P \) is a proper subgroup of \( G \) . In the la... | Yes |
Corollary 7.6. If \( \mathrm{G} \) is a finite nilpotent group and \( \mathrm{m} \) divides \( \left| \mathrm{G}\right| \), then \( \mathrm{G} \) has a subgroup of order \( \mathrm{m} \) . | PROOF. Exercise. | No |
If \( \mathrm{G} \) is a group, then \( {\mathrm{G}}^{\prime } \) is a normal subgroup of \( \mathrm{G} \) and \( \mathrm{G}/{\mathrm{G}}^{\prime } \) is abelian. If \( \mathbf{N} \) is a normal subgroup of \( \mathbf{G} \), then \( \mathbf{G}/\mathbf{N} \) is abelian if and only if \( \mathbf{N} \) contains \( {\mathb... | Let \( f : G \rightarrow G \) be any automorphism. Then\n\n\[ f\left( {{ab}{a}^{-1}{b}^{-1}}\right) = f\left( a\right) f\left( b\right) f{\left( a\right) }^{-1}f{\left( b\right) }^{-1}\varepsilon {G}^{\prime }.\]\n\nIt follows that \( f\left( {G}^{\prime }\right) < {G}^{\prime } \) . In particular, if \( f \) is the au... | Yes |
Proposition 7.10. Every nilpotent group is solvable. | PROOF. Since by the definition of \( {C}_{i}\left( G\right) {C}_{i}\left( G\right) /{C}_{i - 1}\left( G\right) = C\left( {G/{C}_{i - 1}\left( G\right) }\right) \) is abelian, \( {C}_{i}{\left( G\right) }^{\prime } < {C}_{i - 1}\left( G\right) \) for all \( i > 1 \) and \( {C}_{1}{\left( G\right) }^{\prime } = C{\left( ... | Yes |
Theorem 7.11. (i) Every subgroup and every homomorphic image of a solvable group is solvable. | SKETCH OF PROOF. (i) If \( f : G \rightarrow H \) is a homomorphism [epimorphism], verify that \( f\left( {G}^{\left( i\right) }\right) < {H}^{\left( i\right) }\left\lbrack {f\left( {G}^{\left( i\right) }\right) = {H}^{\left( i\right) }}\right\rbrack \) for all \( i \) . Suppose \( f \) is an epimorphism, and \( G \) i... | No |
Corollary 7.12. If \( \mathrm{n} \geq 5 \), then the symmetric group \( {\mathrm{S}}_{\mathrm{n}} \) is not solvable. | PROOF. If \( {S}_{n} \) were solvable, then \( {A}_{n} \) would be solvable. Since \( {A}_{n} \) is nonabelian, \( {A}_{n}{}^{\prime } \neq \left( 1\right) \) . Since \( {A}_{n}{}^{\prime } \) is normal in \( {A}_{n} \) (Theorem 7.8) and \( {A}_{n} \) is simple (Theorem I.6.10), we must have \( {A}_{n}{}^{\prime } = {A... | Yes |
Theorem 8.4. (i) Every finite group \( \mathbf{G} \) has a composition series. | PROOF. (i) Let \( {G}_{1} \) be a maximal normal subgroup of \( G \) ; then \( G/{G}_{1} \) is simple by Corollary I.5.12. Let \( {G}_{2} \) be a maximal normal subgroup of \( {G}_{1} \), and so on. Since \( G \) is finite, this process must end with \( {G}_{n} = \langle e\rangle \) . Thus \( G > {G}_{1} > \cdots > {G}... | Yes |
Proposition 8.6. A finite group \( \mathrm{G} \) is solvable if and only if \( \mathrm{G} \) has a composition series whose factors are cyclic of prime order. | PROOF. A (composition) series with cyclic factors is a solvable series. Conversely, assume \( G = {G}_{0} > {G}_{1} > \cdots > {G}_{n} = \langle e\rangle \) is a solvable series for \( G \) . If \( {G}_{0} \neq {G}_{1} \) , let \( {H}_{1} \) be a maximal normal subgroup of \( G = {G}_{0} \) which contains \( {G}_{1} \)... | Yes |
Lemma 8.8. If \( \mathrm{S} \) is a composition series of a group \( \mathrm{G} \), then any refinement of \( \mathrm{S} \) is equivalent to \( \mathrm{S} \) . | PROOF. Let \( S \) be denoted \( G = {G}_{0} > {G}_{1} > \cdots > {G}_{n} = \langle e\rangle \) . By Theorem 8.4 (iii) \( S \) has no proper refinements. This implies that the only possible refinements of \( S \) are obtained by inserting additional copies of each \( {G}_{i} \) . Consequently any refinement of \( S \) ... | Yes |
Lemma 8.9. (Zassenhaus) Let \( {\mathrm{A}}^{ * },\mathrm{\;A},{\mathrm{\;B}}^{ * },\mathrm{\;B} \) be subgroups of a group \( \mathrm{G} \) such that \( {\mathrm{A}}^{ * } \) is normal in \( \mathrm{A} \) and \( {\mathrm{B}}^{ * } \) is normal in \( \mathrm{B} \). (i) \( {\mathrm{A}}^{ * }\left( {\mathrm{\;A} \cap {\m... | PROOF. Since \( {B}^{ * } \) is normal in \( B, A \cap {B}^{ * } = \left( {A \cap B}\right) \cap {B}^{ * } \) is a normal subgroup of \( A \cap B \) (Theorem I.5.3 (i)); similarly \( {A}^{ * } \cap B \) is normal in \( A \cap B \) . Consequently \( D = \left( {{A}^{ * } \cap B}\right) \left( {A \cap {B}^{ * }}\right) \... | Yes |
Theorem 8.11. (Jordan-Hölder) Any two composition series of a group \( \mathrm{G} \) are equivalent. Therefore every group having a composition series determines a unique list of simple groups. | PROOF OF 8.11. Since composition series are subnormal series, any two composition series have equivalent refinements by the Theorem 8.10. But every refinement of a composition series \( S \) is equivalent to \( S \) by Lemma 8.8. It follows that any two composition series are equivalent. | Yes |
Theorem 1.2. Let \( \mathrm{R} \) be a ring. Then\n\n(i) \( 0\mathrm{a} = \mathrm{a}0 = 0 \) for all \( \mathrm{a}\varepsilon \mathrm{R} \) ;\n\n(ii) \( \left( {-\mathrm{a}}\right) \mathrm{b} = \mathrm{a}\left( {-\mathrm{b}}\right) = - \left( \mathrm{{ab}}\right) \) for all \( \mathrm{a},\mathrm{b}\varepsilon \mathrm{R... | SKETCH OF PROOF. (i) \( {0a} = \left( {0 + 0}\right) a = {0a} + {0a} \), whence \( {0a} = 0 \) . (ii) \( {ab} + \left( {-a}\right) b = \left( {a + \left( {-a}\right) }\right) b = {0b} = 0 \), whence \( \left( {-a}\right) b = - \left( {ab}\right) \) by Theorem I.1.2(iii). (ii) implies (iii). (v) is proved by induction a... | No |
Theorem 1.6. (Binomial Theorem). Let \( \mathbf{R} \) be a ring with identity, \( \mathbf{n} \) a positive integer, and \( \mathrm{a},\mathrm{b},{\mathrm{a}}_{1},{\mathrm{a}}_{2},\ldots ,{\mathrm{a}}_{\mathrm{s}} \in \mathrm{R} \) . (i) If \( \mathrm{{ab}} = \mathrm{{ba}} \), then \( {\left( \mathrm{a} + \mathrm{b}\rig... | (i) Use induction on \( n \) and the fact that \( \left( \begin{array}{l} n \\ k \end{array}\right) + \left( \begin{matrix} n \\ k + 1 \end{matrix}\right) \) \( = \left( \begin{array}{l} n + 1 \\ k + 1 \end{array}\right) \) for \( k < n \) (Exercise 10(c)); the distributive law and the commutativity of \( a \) and \( b... | No |
Theorem 1.9. Let \( \mathrm{R} \) be a ring with identity \( {1}_{\mathrm{R}} \) and characteristic \( \mathrm{n} > 0 \) .\n\n(i) If \( \varphi : \mathbf{Z} \rightarrow \mathbf{R} \) is the map given by \( \mathrm{m} \mapsto {\mathrm{{m1}}}_{\mathbf{R}} \), then \( \varphi \) is a homomorphism of rings with kernel \( \... | SKETCH OF PROOF. (ii) If \( k \) is the least positive integer such that \( k{1}_{R} = 0 \) , then for all \( {a\varepsilon R} : {ka} = k\left( {{1}_{R}a}\right) = \left( {k{1}_{R}}\right) a = 0 \cdot a = 0 \) by Theorem 1.2. (iii) If \( n = {kr} \) with \( 1 < k < n,1 < r < n \), then \( 0 = n{1}_{R} = \left( {kr}\rig... | No |
Theorem 1.10. Every ring \( \mathrm{R} \) may be embedded in a ring \( \mathrm{S} \) with identity. The ring \( \mathrm{S} \) (which is not unique) may be chosen to be either of characteristic zero or of the same characteristic as \( \mathrm{R} \) . | SKETCH OF PROOF. Let \( S \) be the additive abelian group \( R \oplus \mathbf{Z} \) and define multiplication in \( S \) by\n\n\[ \left( {{r}_{1},{k}_{1}}\right) \left( {{r}_{2},{k}_{2}}\right) = \left( {{r}_{1}{r}_{2} + {k}_{2}{r}_{1} + {k}_{1}{r}_{2},{k}_{1}{k}_{2}}\right) ,\left( {{r}_{i} \in R;{k}_{i} \in \mathbf{... | No |
Theorem 2.2. A nonempty subset I of a ring \( \mathrm{R} \) is a left [resp. right] ideal if and only if for all \( \mathrm{a},\mathrm{b}\varepsilon \mathrm{I} \) and \( \mathrm{r}\varepsilon \mathrm{R} \) :\n\n(i) \( \mathrm{a},\mathrm{b}\varepsilon \mathrm{I} \Rightarrow \mathrm{a} - \mathrm{b}\varepsilon \mathrm{I} ... | PROOF. Exercise; see Theorem I.2.5. | No |
Corollary 2.3. Let \( \left\{ {{\mathrm{A}}_{\mathrm{i}} \mid \mathrm{i} \in \mathbf{l}}\right\} \) be a family of \( \left\lbrack {left}\right\rbrack \) ideals in a ring \( \mathbf{R} \) . Then \( \mathop{\bigcap }\limits_{{i \in I}}{\mathrm{\;A}}_{\mathrm{i}} \) is also a [left] ideal. | PROOF. Exercise. | No |
Theorem 2.5. Let \( \mathrm{R} \) be a ring \( \mathrm{a}\varepsilon \mathrm{R} \) and \( \mathrm{X} \subset \mathrm{R} \) . (i) The principal ideal (a) consists of all elements of the form \( \mathrm{{ra}} + \mathrm{{as}} + \mathrm{{na}} + \) \( \mathop{\sum }\limits_{{\mathrm{i} = 1}}^{\mathrm{m}}{\mathrm{r}}_{\mathr... | SKETCH OF PROOF OF 2.5. (i) Show that the set\n\n\[ I = \left\{ {{ra} + {as} + {na} + \mathop{\sum }\limits_{{i = 1}}^{m}{r}_{i}a{s}_{i} \mid r, s,{r}_{i},{s}_{i} \in R;n \in \mathbf{Z};m \in {\mathbf{N}}^{ * }}\right\} \]\n\n is an ideal containing \( a \) and contained in every ideal containing \( a \) . Then \( I = ... | No |
Theorem 2.6. Let \( \mathrm{A},{\mathrm{A}}_{1},{\mathrm{\;A}}_{2},\ldots ,{\mathrm{A}}_{\mathrm{n}},\mathrm{B} \) and \( \mathrm{C} \) be [left] ideals in a ring \( \mathrm{R} \). (i) \( {\mathrm{A}}_{1} + {\mathrm{A}}_{2} + \cdots + {\mathrm{A}}_{\mathrm{n}} \) and \( {\mathrm{A}}_{1}{\mathrm{\;A}}_{2}\cdots {\mathrm... | SKETCH OF PROOF. Use Theorem 2.2 for (i). | No |
Theorem 2.7. Let \( \mathrm{R} \) be a ring and \( \mathrm{I} \) an ideal of \( \mathrm{R} \) . Then the additive quotient group \( \mathrm{R}/\mathrm{I} \) is a ring with multiplication given by\n\n\[ \left( {a + I}\right) \left( {b + I}\right) = {ab} + I \]\n\nIf \( \mathbf{R} \) is commutative or has an identity, th... | SKETCH OF PROOF OF 2.7. Once we have shown that multiplication in \( R/I \) is well defined, the proof that \( R/I \) is a ring is routine. (For example, if \( R \) has identity \( {1}_{R} \), then \( {1}_{R} + I \) is the identity in \( R/I \) .) Suppose \( a + I = {a}^{\prime } + I \) and \( b + I = {b}^{\prime } + I... | No |
Theorem 2.8. Iff \( : \mathrm{R} \rightarrow \mathrm{S} \) is a homomorphism of rings, then the kernel of \( \mathrm{f} \) is an ideal in R. Conversely if \( \mathrm{I} \) is an ideal in \( \mathrm{R} \), then the map \( \pi : \mathrm{R} \rightarrow \mathrm{R}/\mathrm{I} \) given by \( \mathrm{r} \mapsto \mathrm{r} + \... | PROOF OF 2.8. Ker \( f \) is an additive subgroup of \( R \) . If \( x \in \operatorname{Ker}f \) and \( r \in R \), then \( f\left( {rx}\right) = f\left( r\right) f\left( x\right) = f\left( r\right) 0 = 0 \), whence \( {rx\varepsilon }\operatorname{Ker}f \) . Similarly, \( {xr\varepsilon }\operatorname{Ker}f \) . Ther... | Yes |
Theorem 2.9. If \( \mathrm{f} : \mathrm{R} \rightarrow \mathrm{S} \) is a homomorphism of rings and \( \mathrm{I} \) is an ideal of \( \mathrm{R} \) which is contained in the kernel of \( \mathrm{f} \), then there is a unique homomorphism of rings \( \widehat{\mathrm{f}} : \mathrm{R}/\mathrm{I} \rightarrow \mathrm{S} \... | # PROOF. Exercise; see Theorem I.5.6. | No |
Corollary 2.10. (First Isomorphism Theorem) If \( \mathrm{f} : \mathrm{R} \rightarrow \mathrm{S} \) is a homomorphism of rings, then \( \mathrm{f} \) induces an isomorphism of rings \( \mathrm{R}/\operatorname{Ker}\mathrm{f} \cong \operatorname{Im}\mathrm{f} \) . | PROOF. Exercise; see Corollary I.5.7. | No |
Corollary 2.11. If \( \mathrm{f} : \mathrm{R} \rightarrow \mathrm{S} \) is a homomorphism of rings, I is an ideal in \( \mathrm{R} \) and \( \mathrm{J} \) is an ideal in \( \mathrm{S} \) such that \( \mathrm{f}\left( \mathrm{I}\right) \subset \mathrm{J} \), then \( \mathrm{f} \) induces a homomorphism of rings \( \over... | PROOF. Exercise; see Corollary I.5.8. | No |
Theorem 2.12. Let \( \mathrm{I} \) and \( \mathrm{J} \) be ideals in a ring \( \mathrm{R} \). (i) (Second Isomorphism Theorem) There is an isomorphisms of rings \( \mathrm{I}/\left( {\mathrm{I} \cap \mathrm{J}}\right) \cong \left( {\mathrm{I} + \mathrm{J}}\right) /\mathrm{J} \) | PROOF. Exercise; see Corollaries I.5.9 and I.5.10. | No |
Theorem 2.13. If \( \mathrm{I} \) is an ideal in a ring \( \mathrm{R} \), then there is a one-to-one correspondence between the set of all ideals of \( \mathrm{R} \) which contain \( \mathrm{I} \) and the set of all ideals of \( \mathrm{R}/\mathrm{I} \), given by \( \mathrm{J} \mapsto \mathrm{J}/\mathrm{I} \) . Hence e... | PROOF. Exercise; see Theorem I.5.11, Corollary I.5.12 and Exercise 13. | No |
Theorem 2.15. If \( \mathrm{P} \) is an ideal in a ring \( \mathrm{R} \) such that \( \mathrm{P} \neq \mathrm{R} \) and for all \( \mathrm{a},\mathrm{b}\varepsilon \mathrm{R} \)\n\n\[ \mathrm{{ab}}\varepsilon \mathrm{P} \Rightarrow \mathrm{a}\varepsilon \mathrm{P}\text{ or }\mathrm{b}\varepsilon \mathrm{P}, \]\n\n(1)\n... | PROOF OF 2.15. If \( A \) and \( B \) are ideals such that \( {AB} \subset P \) and \( A ⊄ P \), then there exists an element \( a \in A - P \) . For every \( b \in B,{ab} \in {AB} \subset P \), whence \( a \in P \) or \( b \in P \) . Since \( a \nmid P \), we must have \( b \in P \) for all \( b \in B \) ; that is, \(... | Yes |
Theorem 2.16. In a commutative ring \( \mathrm{R} \) with identity \( {1}_{\mathrm{R}} \neq 0 \) an ideal \( \mathrm{P} \) is prime if and only if the quotient ring \( \mathrm{R}/\mathrm{P} \) is an integral domain. | PROOF. \( R/P \) is a commutative ring with identity \( {1}_{R} + P \) and zero element \( 0 + P = P \) by Theorem 2.7. If \( P \) is prime, then \( {1}_{R} + P \neq P \) since \( P \neq R \) . Furthermore, \( R/P \) has no zero divisors since\n\n\[ \left( {a + P}\right) \left( {b + P}\right) = P \Rightarrow {ab} + P =... | Yes |
Theorem 2.18. In a nonzero ring \( \mathrm{R} \) with identity maximal [left] ideals always exist. In fact every [left] ideal in \( \mathrm{R} \) (except \( \mathrm{R} \) itself) is contained in a maximal [left] ideal. | PROOF. Since 0 is an ideal and \( 0 \neq R \), it suffices to prove the second statement. The proof is a straightforward application of Zorn’s Lemma. If \( A \) is a [left] ideal in \( R \) such that \( A \neq R \), let \( \mathcal{S} \) be the set of all [left] ideals \( B \) in \( R \) such that \( A \subset B \neq R... | Yes |
Theorem 2.19. If \( \mathrm{R} \) is a commutative ring such that \( {\mathrm{R}}^{2} = \mathrm{R} \) (in particular if \( \mathrm{R} \) has an identity), then every maximal ideal \( \mathrm{M} \) in \( \mathrm{R} \) is prime. | PROOF OF 2.19. Suppose \( {ab} \in M \) but \( a \notin M \) and \( b \notin M \) . Then each of the ideals \( M + \left( a\right) \) and \( M + \left( b\right) \) properly contains \( M \) . By maximality \( M + \left( a\right) = R = M + \left( b\right) \) . Since \( R \) is commutative and \( {ab} \in M \), Theorem 2... | Yes |
Theorem 2.20. Let \( \mathrm{M} \) be an ideal in a ring \( \mathrm{R} \) with identity \( {1}_{\mathrm{R}} \neq 0 \). (i) If \( \mathrm{M} \) is maximal and \( \mathrm{R} \) is commutative, then the quotient ring \( \mathrm{R}/\mathrm{M} \) is a field. | PROOF OF 2.20. (i) If \( M \) is maximal, then \( M \) is prime (Theorem 2.19), whence \( R/M \) is an integral domain by Theorem 2.16. Thus we need only show that if \( a + M \neq M \), then \( a + M \) has a multiplicative inverse in \( R/M \). Now \( a + M \neq M \) implies that \( a \notin M \), whence \( M \) is p... | Yes |
Corollary 2.21. The following conditions on a commutative ring \( \mathrm{R} \) with identity \( {1}_{\mathrm{R}} \neq 0 \) are equivalent.\n\n(i) \( \mathrm{R} \) is a field;\n\n(ii) \( \mathrm{R} \) has no proper ideals;\n\n(iii) 0 is a maximal ideal in \( \mathrm{R} \) ;\n\n(iv) every nonzero homomorphism of rings \... | PROOF OF 2.21. This result may be proved directly (Exercise 7) or as follows. \( R \cong R/0 \) is a field if and only if 0 is maximal by Theorem 2.20. But clearly 0 is maximal if and only if \( R \) has no proper ideals. Finally, for every ideal \( I\left( { \neq R}\right) \) the canonical map \( \pi : R \rightarrow R... | No |
Theorem 2.22. Let \( \\left\\{ {{\\mathrm{R}}_{\\mathrm{i}} \\mid \\mathrm{i}\\varepsilon \\mathrm{I}}\\right\\} \) be a nonempty family of rings and \( \\mathop{\\prod }\\limits_{{i \\in I}}{\\mathrm{R}}_{\\mathrm{i}} \) the direct product of the additive abelian groups \( {\\mathbf{R}}_{\\mathbf{i}} \) ;\n\n(i) \( \\... | PROOF. Exercise. | No |
Theorem 2.23. Let \( \\left\\{ {{\\mathrm{R}}_{\\mathrm{i}} \\mid \\mathrm{i}\\varepsilon \\mathrm{I}}\\right\\} \) be a nonempty family of rings, \( \\mathrm{S} \) a ring and \( \\left\\{ {{\\varphi }_{\\mathrm{i}} : \\mathrm{S} \\rightarrow {\\mathrm{R}}_{\\mathrm{i}} \\mid \\mathrm{i}\\varepsilon \\mathrm{I}}\\right... | SKETCH OF PROOF. By Theorem I.8.2 there is a unique homomorphism of groups \( \\varphi : S \\rightarrow \\mathop{\\prod }\\limits_{{i \\in I}}{R}_{i} \) such that \( {\\pi }_{i}\\varphi = {\\varphi }_{i} \) for all \( {i\\varepsilon I} \) . Verify that \( \\varphi \) is also a ring homomorphism. Thus \( \\mathop{\\prod... | No |
Theorem 2.24. Let \( {\mathrm{A}}_{1},{\mathrm{\;A}}_{2},\ldots ,{\mathrm{A}}_{\mathrm{n}} \) be ideals in a ring \( \mathrm{R} \) such that (i) \( {\mathrm{A}}_{1} + {\mathrm{A}}_{2} + \cdots + \) \( {\mathrm{A}}_{\mathrm{n}} = \mathrm{R} \) and \( \left( \mathrm{{ii}}\right) \) for each \( \mathrm{k}\left( {1 \leq \m... | PROOF. By the proof of Theorem I.8.6 the map \( \varphi : {A}_{1} \times {A}_{2} \times \cdots \times {A}_{n} \rightarrow R \) given by \( \left( {{a}_{1},\ldots ,{a}_{n}}\right) \mapsto {a}_{1} + {a}_{2} + \cdots + {a}_{n} \) is an isomorphism of additive abelian groups. We need only verify that \( \varphi \) is a rin... | Yes |
Theorem 2.25. (Chinese Remainder Theorem) Let \( {\mathrm{A}}_{1},\ldots ,{\mathrm{A}}_{\mathrm{n}} \) be ideals in a ring \( \mathrm{R} \) such that \( {\mathrm{R}}^{2} + {\mathrm{A}}_{\mathrm{i}} = \mathrm{R} \) for all \( \mathrm{i} \) and \( {\mathrm{A}}_{\mathrm{i}} + {\mathrm{A}}_{\mathrm{j}} = \mathrm{R} \) for ... | SKETCH OF PROOF OF 2.25. Since \( {A}_{1} + {A}_{2} = R \) and \( {A}_{1} + {A}_{3} = R \) ,\n\n\[ {R}^{2} = \left( {{A}_{1} + {A}_{2}}\right) \left( {{A}_{1} + {A}_{3}}\right) = {A}_{1}{}^{2} + {A}_{1}{A}_{3} + {A}_{2}{A}_{1} + {A}_{2}{A}_{3} \]\n\n\[ \subset {A}_{1} + {A}_{2}{A}_{3} \subset {A}_{1} + \left( {{A}_{2} ... | Yes |
Corollary 2.26. Let \( {\mathrm{m}}_{1},{\mathrm{\;m}}_{2},\ldots ,{\mathrm{m}}_{\mathrm{n}} \) be positive integers such that \( \left( {{\mathrm{m}}_{\mathrm{i}},{\mathrm{m}}_{\mathrm{j}}}\right) = 1 \) for \( \mathrm{i} \neq \mathrm{j} \) . If \( {\mathrm{b}}_{1},{\mathrm{\;b}}_{2},\ldots ,{\mathrm{b}}_{\mathrm{n}} ... | SKETCH OF PROOF. Let \( {A}_{i} = \left( {m}_{i}\right) \) ; then \( \mathop{\bigcap }\limits_{{i = 1}}^{n}{A}_{i} = \left( m\right) \) . Show that \( \left( {{m}_{i},{m}_{j}}\right) = 1 \) implies \( {A}_{i} + {A}_{j} = \mathbf{Z} \) and apply Theorem 2.25. ∎ | No |
Corollary 2.27. If \( {\mathrm{A}}_{1},\ldots ,{\mathrm{A}}_{\mathrm{n}} \) are ideals in a ring \( \mathrm{R} \), then there is a monomorphism of rings\n\n\[ \theta : \mathrm{R}/\left( {{\mathrm{A}}_{1} \cap \cdots \cap {\mathrm{A}}_{\mathrm{n}}}\right) \rightarrow \mathrm{R}/{\mathrm{A}}_{1} \times \mathrm{R}/{\mathr... | SKETCH OF PROOF. By Theorem 2.23 the canonical epimorphisms \( {\pi }_{k} : R \rightarrow \) \( R/{A}_{k}\left( {k = 1,\ldots, n}\right) \) induce a homomorphism of rings \( {\theta }_{1} : R \rightarrow R/{A}_{1} \times \cdots \times R/{A}_{n} \) with \( {\theta }_{1}\left( r\right) = \left( {r + {A}_{1},\ldots, r + {... | No |
Lemma 3.6. If \( \mathrm{R} \) is a principal ideal ring and \( \left( {\mathrm{a}}_{1}\right) \subset \left( {\mathrm{a}}_{2}\right) \subset \cdots \) is a chain of ideals in \( \mathrm{R} \), then for some positive integer \( \mathrm{n},\left( {\mathrm{a}}_{\mathrm{j}}\right) = \left( {\mathrm{a}}_{\mathrm{n}}\right)... | PROOF. Let \( A = \mathop{\bigcup }\limits_{{i \geq 1}}\left( {a}_{i}\right) \) . We claim that \( A \) is an ideal. If \( b, c \in A \), then \( b \in \left( {a}_{i}\right) \) and \( c \in \left( {a}_{j}\right) \) . Either \( i \leq j \) or \( i \geq j \) ; say \( i \geq j \) . Consequently \( \left( {a}_{j}\right) \s... | Yes |
Theorem 3.9. Every Euclidean ring \( \mathrm{R} \) is a principal ideal ring with identity. Consequently every Euclidean domain is a unique factorization domain. | PROOF OF 3.9. If \( I \) is a nonzero ideal in \( R \), choose \( a \in I \) such that \( \varphi \left( a\right) \) is the least integer in the set of nonnegative integers \( \{ \varphi \left( x\right) \mid x \neq 0;{x\varepsilon I}\} \) . If \( {b\varepsilon I} \), then \( b = {qa} + r \) with \( r = 0 \) or \( r \ne... | Yes |
Theorem 3.11. Let \( {\mathrm{a}}_{1},\ldots ,{\mathrm{a}}_{\mathrm{n}} \) be elements of a commutative ring \( \mathrm{R} \) with identity.\n\n(i) \( \mathrm{d}\varepsilon \mathrm{R} \) is a greatest common divisor of \( \left\{ {{\mathrm{a}}_{1},\ldots ,{\mathrm{a}}_{\mathrm{n}}}\right\} \) such that \( \mathrm{d} = ... | SKETCH OF PROOF OF 3.11. (i) Use Definition 3.10 and Theorem 2.5. | No |
Theorem 4.2. Let \( \mathrm{S} \) be a multiplicative subset of a commutative ring \( \mathrm{R} \) . The relation defined on the set \( \mathrm{R} \times \mathrm{S} \) by\n\n\[ \left( {\mathrm{r},\mathrm{s}}\right) \sim \left( {{\mathrm{r}}^{\prime },{\mathrm{s}}^{\prime }}\right) \; \Leftrightarrow \;{\mathrm{s}}_{1}... | PROOF. Exercise. | No |
Theorem 4.3. Let \( \mathrm{S} \) be a multiplicative subset of a commutative ring \( \mathrm{R} \) and let \( {\mathrm{S}}^{-1}\mathrm{R} \) be the set of equivalence classes of \( \mathrm{R} \times \mathrm{S} \) under the equivalence relation of Theorem 4.2.\n\n(i) \( {\mathrm{S}}^{-1}\mathrm{R} \) is a commutative r... | SKETCH OF PROOF. (i) Once we know that addition and multiplication in \( {S}^{-1}R \) are well-defined binary operations (independent of the choice of \( r, s,{r}^{\prime },{s}^{\prime } \) ), the rest of the proof of (i) is routine. In particular, for all \( s,{s}^{\prime }{\varepsilon S},0/s = 0/{s}^{\prime } \) and ... | Yes |
Theorem 4.5. Let \( \mathrm{S} \) be a multiplicative subset of a commutative ring \( \mathrm{R} \) and let \( \mathrm{T} \) be any commutative ring with identity. If \( \mathrm{f} : \mathrm{R} \rightarrow \mathrm{T} \) is a homomorphism of rings such that \( \mathrm{f}\left( \mathrm{s}\right) \) is a unit in \( \mathr... | SKETCH OF PROOF. Verify that the map \( \bar{f} : {S}^{-1}R \rightarrow T \) given by \( \bar{f}\left( {r/s}\right) \) \( = f\left( r\right) f{\left( s\right) }^{-1} \) is a well-defined homomorphism of rings such that \( \bar{f}{\varphi }_{S} = f \) . If\n\n\( {}^{3} \) For the noncommutative analogue, see Definition ... | No |
Corollary 4.6. Let \( \mathrm{R} \) be an integral domain considered as a subring of its quotient field \( \mathrm{F} \) . If \( \mathrm{E} \) is a field and \( \mathrm{f} : \mathrm{R} \rightarrow \mathrm{E} \) a monomorphism of rings, then there is a unique monomorphism of fields \( \bar{\mathrm{f}} : \mathrm{F} \righ... | SKETCH OF PROOF. Let \( S \) be the set of all nonzero elements of \( R \) and apply Theorem 4.5 to \( f : R \rightarrow E \) . Then there is a homomorphism \( \bar{f} : {S}^{-1}R = F \rightarrow E \) such that \( \bar{f}{\varphi }_{S} = f \) . Verify that \( \bar{f} \) is a monomorphism. Since \( R \) is identified wi... | No |
Theorem 4.7. Let \( \mathrm{S} \) be a multiplicative subset of a commutative ring \( \mathbf{R} \). (i) If \( \mathrm{I} \) is an ideal in \( \mathrm{R} \), then \( {\mathrm{S}}^{-1}\mathrm{I} = \{ \mathrm{a}/\mathrm{s} \mid \mathrm{a}\varepsilon \mathrm{I};\mathrm{s}\varepsilon \mathrm{S}\} \) is an ideal in \( {\mat... | SKETCH OF PROOF OF 4.7. Use the facts that in \( {S}^{-1}R,\mathop{\sum }\limits_{{i = 1}}^{n}\left( {{c}_{i}/s}\right) = \left( {\mathop{\sum }\limits_{{i = 1}}^{n}{c}_{i}}\right) /s;\mathop{\sum }\limits_{{j = 1}}^{m}\left( {{a}_{j}{b}_{j}/s}\right) = \mathop{\sum }\limits_{{j = 1}}^{m}\left( {{a}_{j}/s}\right) \left... | No |
Theorem 4.8. Let \( \mathrm{S} \) be a multiplicative subset of a commutative ring \( \mathrm{R} \) with identity and let \( \mathrm{I} \) be an ideal of \( \mathrm{R} \). Then \( {\mathrm{S}}^{-1}\mathrm{I} = {\mathrm{S}}^{-1}\mathrm{R} \) if and only if \( \mathrm{S} \cap \mathrm{I} \neq \varnothing \). | PROOF. If \( s \in S \cap I \), then \( {1}_{{S}^{-1}R} = s/s \in {S}^{-1}I \) and hence \( {S}^{-1}I = {S}^{-1}R \). Conversely, if \( {S}^{-1}I = {S}^{-1}R \), then \( {\varphi }_{S}{}^{-1}\left( {{S}^{-1}I}\right) = R \) whence \( {\varphi }_{S}\left( {1}_{R}\right) = a/s \) for some \( {a\varepsilon I},{s\varepsilo... | Yes |
Lemma 4.9. Let \( \mathrm{S} \) be a multiplicative subset of a commutative ring \( \mathrm{R} \) with identity and let \( \mathrm{I} \) be an ideal in \( \mathrm{R} \). (i) \( \mathrm{I} \subset \varphi {\mathrm{s}}^{-1}\left( {{\mathrm{\;S}}^{-1}\mathrm{I}}\right) \). | PROOF. (i) If \( a \in I \), then \( {as} \in I \) for every \( s \in S \). Consequently, \( {\varphi }_{S}\left( a\right) = {as}/s \in {S}^{-1}I \), whence \( a \) e \( {\varphi }_{S}{}^{-1}\left( {{S}^{-1}I}\right) \). Therefore, \( I \subset {\varphi }_{S}{}^{-1}\left( {{S}^{-1}I}\right) \). | Yes |
Theorem 4.10. Let \( \mathrm{S} \) be a multiplicative subset of a commutative ring \( \mathrm{R} \) with identity. Then there is a one-to-one correspondence between the set \( \mathcal{U} \) of prime ideals of \( \mathbf{R} \) which are disjoint from \( \mathrm{S} \) and the set \( \mathcal{U} \) of prime ideals of \(... | PROOF. By Lemma 4.9(iii) the assignment \( P \mapsto {S}^{-1}P \) defines an injective map \( \mathcal{U} \rightarrow \mathcal{U} \) . We need only show that it is surjective as well. Let \( J \) be a prime ideal of \( {S}^{-1}R \) and let \( P = {\varphi }_{S}{}^{-1}\left( J\right) \) . Since \( {S}^{-1}P = J \) by Le... | Yes |
Theorem 4.11. Let \( \mathrm{P} \) be a prime ideal in a commutative ring \( \mathrm{R} \) with identity.\n\n(i) There is a one-to-one correspondence between the set of prime ideals of \( \mathbf{R} \) which are contained in \( \mathrm{P} \) and the set of prime ideals of \( {\mathrm{R}}_{\mathrm{P}} \), given by \( \m... | PROOF. Since the prime ideals of \( R \) contained in \( P \) are precisely those which are disjoint from \( S = R - P \) ,(i) is an immediate consequence of Theorem 4.10. If \( M \) is a maximal ideal of \( {R}_{P} \), then \( M \) is prime by Theorem 2.19, whence \( M = {Q}_{P} \) for some prime ideal \( Q \) of \( R... | Yes |
Theorem 4.13. If \( \mathrm{R} \) is a commutative ring with identity then the following conditions are equivalent.\n\n(i) \( \mathrm{R} \) is a local ring;\n\n(ii) all nonunits of \( \mathrm{R} \) are contained in some ideal \( \mathrm{M} \neq \mathrm{R} \) ;\n\n(iii) the nonunits of \( \mathrm{R} \) form an ideal. | SKETCH OF PROOF. If \( I \) is an ideal of \( R \) and \( {a\varepsilon I} \), then \( \left( a\right) \subset I \) by Theorem 2.5. Consequently, \( I \neq R \) if and only if \( I \) consists only of nonunits (Theorem 3.2(iv)). (ii) \( \Rightarrow \) (iii) and (iii) \( \Rightarrow \) (i) follow from this fact. (i) \( ... | No |
Theorem 5.1. Let \( \mathrm{R} \) be a ring and let \( R\left\lbrack \mathrm{x}\right\rbrack \) denote the set of all sequences of elements of \( \mathrm{R}\left( {{\mathrm{a}}_{0},{\mathrm{a}}_{1},\ldots }\right) \) such that \( {\mathrm{a}}_{\mathrm{i}} = 0 \) for all but a finite number of indices \( \mathrm{i} \). ... | PROOF. Exercise. If \( R \) has an identity \( {1}_{R} \), then \( \left( {{1}_{R},0,0,\ldots }\right) \) is an identity in \( R\left\lbrack x\right\rbrack \) . Observe that if \( \left( {{a}_{0},{a}_{1},\ldots }\right) ,\left( {{b}_{0},{b}_{1},\ldots }\right) \in R\left\lbrack x\right\rbrack \) and \( k \) [resp. \( j... | No |
Theorem 5.2. Let \( \mathrm{R} \) be a ring with identity and denote by \( \mathrm{x} \) the element \( \left( {0,{1}_{\mathrm{R}},0,0,\ldots }\right) \) of \( \mathrm{R}\left\lbrack \mathrm{x}\right\rbrack \) .\n\n(i) \( {\mathrm{x}}^{\mathrm{n}} = \left( {0,0,\ldots ,0,{1}_{\mathrm{R}},0,\ldots }\right) \), where \( ... | SKETCH OF PROOF. Use induction for (i) and straightforward computation for (ii). (iii) If \( f = \left( {{a}_{0},{a}_{1},\ldots }\right) {\varepsilon R}\left\lbrack x\right\rbrack \), there must be a largest index \( n \) such that \( {a}_{n} \neq 0 \) . Then \( {a}_{0},{a}_{1},\ldots ,{a}_{n} \in R \) are the desired ... | No |
Theorem 5.3. Let \( \mathrm{R} \) be a ring and denote by \( \mathrm{R}\left\lbrack {{\mathrm{x}}_{1},\ldots ,{\mathrm{x}}_{\mathrm{n}}}\right\rbrack \) the set of all functions \( \mathrm{f} : {\mathrm{N}}^{\mathrm{n}} \rightarrow \mathrm{R} \) such that \( \mathrm{f}\left( \mathrm{u}\right) \neq 0 \) for at most a fi... | PROOF. Exercise. | No |
Theorem 5.4. Let \( \mathrm{R} \) be a ring with identity and \( \mathrm{n} \) a positive integer. For each \( \mathrm{i} = 1,2,\ldots ,\mathrm{n} \) let \( {\mathrm{x}}_{\mathrm{i}} \in \mathrm{R}\left\lbrack {{\mathrm{x}}_{1},\ldots ,{\mathrm{x}}_{\mathrm{n}}}\right\rbrack \) be defined by \( {\mathrm{x}}_{\mathrm{i}... | SKETCH OF PROOF. (v) Let \( {a}_{{k}_{1}},\ldots ,{}_{{k}_{n}} = f\left( {{k}_{1},\ldots ,{k}_{n}}\right) \) . | No |
Theorem 5.5. Let \( \mathrm{R} \) and \( \mathrm{S} \) be commutative rings with identity and \( \varphi : \mathrm{R} \rightarrow \mathrm{S} \) a homomorphism of rings such that \( \varphi \left( {1}_{\mathrm{R}}\right) = {1}_{\mathrm{S}} \) . If \( {\mathrm{s}}_{1},{\mathrm{\;s}}_{2},\ldots ,{\mathrm{s}}_{\mathrm{n}} ... | SKETCH OF PROOF. If \( {f\varepsilon R}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \), then\n\n\[ f = \mathop{\sum }\limits_{{i = 0}}^{m}{a}_{i}{x}_{1}^{{k}_{i1}}\cdots {x}_{n}^{{k}_{in}}\left( {{a}_{i}{\varepsilon R};{k}_{ij}\varepsilon \mathbf{N}}\right) \]\n\nby Theorem 5.4. The map \( \bar{\varphi } \) give... | No |
Corollary 5.6. If \( \varphi : \mathrm{R} \rightarrow \mathrm{S} \) is a homomorphism of commutative rings and \( {\mathrm{s}}_{1},{\mathrm{\;s}}_{2},\ldots ,{\mathrm{s}}_{\mathrm{n}} \in \mathrm{S} \), then the map \( \mathrm{R}\left\lbrack {{\mathrm{x}}_{1},\ldots ,{\mathrm{x}}_{\mathrm{n}}}\right\rbrack \rightarrow ... | SKETCH OF PROOF OF 5.6. The proof of Theorem 5.5 showing that the assignment \( f \mapsto {\varphi f}\left( {{s}_{1},\ldots ,{s}_{n}}\right) \) defines a homomorphism is valid even when \( R \) and \( S \) do not have identities. | No |
Corollary 5.7. Let \( \mathrm{R} \) be a commutative ring with identity and \( \mathrm{n} \) a positive integer. For each \( \mathrm{k}\left( {1 \leq \mathrm{k} < \mathrm{n}}\right) \) there are isomorphisms of rings \( \mathbf{R}\left\lbrack {{\mathrm{x}}_{1},\ldots ,{\mathrm{x}}_{\mathrm{k}}}\right\rbrack \left\lbrac... | PROOF. The corollary may be proved by directly constructing the isomorphisms or by using the universal mapping property of Theorem 5.5 as follows. Given a homomorphism \( \varphi : R \rightarrow S \) of commutative rings with identity and elements \( {s}_{1},\ldots ,{s}_{n} \in S \), there exists a homomorphism \( \bar... | Yes |
Proposition 5.8. Let \( \mathrm{R} \) be a ring and denote by \( \mathrm{R}\left\lbrack \left\lbrack \mathrm{x}\right\rbrack \right\rbrack \) the set of all sequences of elements of \( \mathrm{R}\left( {{\mathrm{a}}_{0},{\mathrm{a}}_{1},\ldots }\right) \). (i) \( \mathrm{R}\left\lbrack \left\lbrack \mathrm{x}\right\rbr... | ## PROOF. Exercise; see Theorem 5.1. | No |
(i) \( f \) is a unit in \( R\left\lbrack \left\lbrack x\right\rbrack \right\rbrack \) if and only if its constant term \( {a}_{0} \) is a unit in \( R \) . | If there exists \( g = \sum {b}_{i}{x}^{i}{\varepsilon R}\left\lbrack \left\lbrack x\right\rbrack \right\rbrack \) such that\n\n\[ \n{fg} = {gf} = {1}_{R} \in R\left\lbrack \left\lbrack x\right\rbrack \right\rbrack ,\n\]\n\nit follows immediately that \( {a}_{0}{b}_{0} = {b}_{0}{a}_{0} = {1}_{R} \), whence \( {a}_{0} \... | Yes |
If \( \mathrm{R} \) is a division ring, then the units in \( \mathrm{R}\left\lbrack \left\lbrack \mathrm{x}\right\rbrack \right\rbrack \) are precisely those power series with nonzero constant term. The principal ideal (x) consists precisely of the nonunits in \( \mathrm{R}\left\lbrack \left\lbrack \mathrm{x}\right\rbr... | The first statement follows from Proposition 5.9 (i) and the fact that every nonzero element of \( R \) is a unit. Since \( x \) is in the center of \( R\left\lbrack \left\lbrack x\right\rbrack \right\rbrack \) ,\n\n\[ \left( x\right) = \{ {xf} \mid {f\varepsilon R}\left\lbrack \left\lbrack x\right\rbrack \right\rbrack... | Yes |
Theorem 6.1. Let \( \mathrm{R} \) be a ring and \( \mathrm{f},\mathrm{g} \in \mathrm{R}\left\lbrack {{\mathrm{x}}_{1},\ldots ,{\mathrm{x}}_{\mathrm{n}}}\right\rbrack \) . (i) \( \deg \left( {\mathrm{f} + \mathrm{g}}\right) \leq \max \left( {\deg \mathrm{f},\deg \mathrm{g}}\right) \) . (ii) \( \deg \left( \mathrm{{fg}}\... | SKETCH OF PROOF OF 6.1. Since we shall apply this theorem primarily when \( n = 1 \) we shall prove only that case. (i) is easy (ii) is trivial if \( f = 0 \) or \( g = 0 \) . If \( 0 \neq f = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{x}^{i} \) has degree \( n \) and \( 0 \neq g = \mathop{\sum }\limits_{{i = 0}}^{m}{b... | No |
Theorem 6.2. (The Division Algorithm) Let \( \mathrm{R} \) be a ring with identity and \( \mathrm{f},\mathrm{g} \in \mathrm{R}\left\lbrack \mathrm{x}\right\rbrack \) nonzero polynomials such that the leading coefficient of \( \mathrm{g} \) is a unit in \( \mathrm{R} \) . Then there exist unique polynomials \( \mathrm{q... | PROOF. If \( \deg g > \deg f \), let \( q = 0 \) and \( r = f \) . If \( \deg g \leq \deg f \), then \( f = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{x}^{i} \) , \( g = \mathop{\sum }\limits_{{i = 0}}^{m}{b}_{i}{x}^{i} \), with \( {a}_{n} \neq 0,{b}_{m} \neq 0, m \leq n \), and \( {b}_{m} \) a unit in \( R \) . Procee... | Yes |
Corollary 6.3. (Remainder Theorem) Let \( \mathrm{R} \) be a ring with identity and \n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{a}_{i}{x}^{i}{\varepsilon R}\left\lbrack x\right\rbrack \]\n\nFor any \( \mathrm{c}\varepsilon \mathrm{R} \) there exists a unique \( \mathrm{q}\left( \mathrm{x}\right) \vare... | PROOF. If \( f = 0 \) let \( q = 0 \) . Suppose then that \( f \neq 0 \) . Theorem 6.2 implies that there exist unique polynomials \( q\left( x\right), r\left( x\right) \) in \( R\left\lbrack x\right\rbrack \) such that \( f\left( x\right) = q\left( x\right) \left( {x - c}\right) + r\left( x\right) \) and \( \deg r\lef... | No |
If \( \mathrm{F} \) is a field, then the polynomial ring \( \mathrm{F}\left\lbrack \mathrm{x}\right\rbrack \) is a Euclidean domain, whence \( \mathrm{F}\left\lbrack \mathrm{x}\right\rbrack \) is a principal ideal domain and a unique factorization domain. The units in \( \mathrm{F}\left\lbrack \mathrm{x}\right\rbrack \... | SKETCH OF PROOF. \( F\left\lbrack x\right\rbrack \) is an integral domain by Theorem 5.1. Define \( \varphi : F\left\lbrack x\right\rbrack - \{ 0\} \rightarrow \mathbf{N} \) by \( \varphi \left( f\right) = \deg f \) . Since every nonzero element of \( F \) is a unit, Theorems 6.1(iv) and 6.2 imply that \( F\left\lbrack... | No |
Theorem 6.6. Let \( \mathbf{R} \) be a commutative ring with identity and \( \mathbf{f} \in \mathbf{R}\left\lbrack \mathbf{x}\right\rbrack \) . Then \( \mathbf{c} \in \mathbf{R} \) is a root of \( \mathrm{f} \) if and only if \( \mathrm{x} - \mathrm{c} \) divides \( \mathrm{f} \) . | SKETCH OF PROOF. We have \( f\left( x\right) = q\left( x\right) \left( {x - c}\right) + f\left( c\right) \) by Corollary 6.3. If \( x - c \mid f\left( x\right) \), then \( h\left( x\right) \left( {x - c}\right) = f\left( x\right) = q\left( x\right) \left( {x - c}\right) + f\left( c\right) \) with \( h \in R\left\lbrack... | No |
Theorem 6.7. If \( \mathrm{D} \) is an integral domain contained in an integral domain \( \mathrm{E} \) and \( \mathrm{f} \in \mathrm{D}\left\lbrack \mathrm{x}\right\rbrack \) has degree \( \mathrm{n} \), then \( \mathrm{f} \) has at most \( \mathrm{n} \) distinct roots in \( \mathrm{E} \) . | SKETCH OF PROOF. Let \( {c}_{1},{c}_{2},\ldots \) be the distinct roots of \( f \) in \( E \) . By Theorem 6.6 \( f\left( x\right) = {q}_{1}\left( x\right) \left( {x - {c}_{1}}\right) \), whence \( 0 = f\left( {c}_{2}\right) = {q}_{1}\left( {c}_{2}\right) \left( {{c}_{2} - {c}_{1}}\right) \) by Corollary 5.6. Since \( ... | Yes |
Proposition 6.8. Let \( \\mathrm{D} \) be a unique factorization domain with quotient field \( \\mathrm{F} \) and let \( \\mathrm{f} = \\mathop{\\sum }\\limits_{{i = 0}}^{n}{\\mathrm{a}}_{\\mathrm{i}}{\\mathrm{x}}^{\\mathrm{i}}\\varepsilon \\mathrm{D}\\left\\lbrack \\mathrm{x}\\right\\rbrack \) . If \( \\mathrm{u} = \\... | SKETCH OF PROOF. \( f\\left( u\\right) = 0 \) implies that \( {a}_{0}{d}^{n} = c\\left( {\\mathop{\\sum }\\limits_{{i = 1}}^{n}\\left( {-{a}_{i}}\\right) {c}^{i - 1}{d}^{n - i}}\\right) \) and \( - {a}_{n}{c}^{n} = \\left( {\\mathop{\\sum }\\limits_{{i = 0}}^{{n - 1}}{c}^{i}{d}^{n - i - 1}}\\right) d \) . Consequently,... | No |
Lemma 6.9. Let \( \mathrm{D} \) be an integral domain and \( \mathrm{f} = \mathop{\sum }\limits_{{i = 0}}^{n}{\mathrm{a}}_{\mathrm{i}}{\mathrm{x}}^{\mathrm{i}}\varepsilon \mathrm{D}\left\lbrack \mathrm{x}\right\rbrack \) . Let \( {\mathrm{f}}^{\prime }\varepsilon \mathrm{D}\left\lbrack \mathrm{x}\right\rbrack \) be the... | PROOF. Exercise. | No |
Let \( \mathrm{D} \) be an integral domain which is a subring of an integral domain E. Let \( \mathrm{f}\varepsilon \mathrm{D}\left\lbrack \mathrm{x}\right\rbrack \) and \( \mathrm{c}\varepsilon \mathrm{E} \). (i) \( \mathrm{c} \) is a multiple root of \( \mathrm{f} \) if and only if \( \mathrm{f}\left( \mathrm{c}\righ... | PROOF. (i) \( f\left( x\right) = {\left( x - c\right) }^{m}g\left( x\right) \) where \( m \) is the multiplicity of \( f\left( {m \geq 0}\right) \) and \( g\left( c\right) \neq 0 \) . By Lemma 6.9 \( {f}^{\prime }\left( x\right) = m{\left( x - c\right) }^{m - 1}g\left( x\right) + {\left( x - c\right) }^{m}{g}^{\prime }... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.