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Proposition 7.5.5\n\n\\[\\frac{\\partial C}{\\partial t} = \\frac{\\sigma }{2\\sqrt{t}}s{\\Phi }^{\\prime }\\left( \\omega \\right) + {Kr}{e}^{-{rt}}\\Phi \\left( {\\omega - \\sigma \\sqrt{t}}\\right) .\\] | Proof.\n\n\\[\\frac{\\partial }{\\partial t}\\left\\lbrack {{e}^{-{rt}}\\left( {S\\left( t\\right) - K}\\right) }\\right\\rbrack = {e}^{-{rt}}\\frac{\\partial S\\left( t\\right) }{\\partial t} - r{e}^{-{rt}}S\\left( t\\right) + {Kr}{e}^{-{rt}}\\]\n\n\\[ = {e}^{-{rt}}S\\left( t\\right) \\left( {r - \\frac{{\\sigma }^{2}... | Yes |
Corollary 7.5.1 \( C\left( {s, t, K,\sigma, r}\right) \) is\n\n(a) decreasing and convex in \( K \) ;\n\n(b) increasing and convex in \( s \) ;\n\n(c) increasing, but neither convex nor concave, in \( r,\sigma \), and \( t \) . | Proof. (a) From Proposition 7.5.1, we have \( \frac{\partial C}{\partial K} < 0 \), and\n\n\[ \frac{{\partial }^{2}C}{\partial {K}^{2}} = - {e}^{-{rt}}{\Phi }^{\prime }\left( {\omega - \sigma \sqrt{t}}\right) \frac{\partial \omega }{\partial K} \]\n\n\[ = {e}^{-{rt}}{\Phi }^{\prime }\left( {\omega - \sigma \sqrt{t}}\ri... | Yes |
Theorem 8.4.1 If the jumps have a lognormal distribution with mean parameter \( {\mu }_{0} \) and variance parameter \( {\sigma }_{0}^{2} \), then the no-arbitrage cost of a European call option having strike price \( K \) and expiration time \( t \) is as follows: | \[ \text{ no-arbitrage cost } = \mathop{\sum }\limits_{{n = 0}}^{\infty }{e}^{-{\lambda tE}\left\lbrack J\right\rbrack }\frac{{\left( \lambda tE\left\lbrack J\right\rbrack \right) }^{n}}{n!}C\left( {s, t, K,\sigma \left( n\right), r\left( n\right) }\right) ,\] where \[ {\sigma }^{2}\left( n\right) = {\sigma }^{2} + n{\... | Yes |
Example 1.3. Take \( S = \mathbb{Z} \), and let \( \sim \) be the relation defined by\n\n\[ a \sim b \Leftrightarrow a - b\\text{is even.} \]\n\nThen \( \mathbb{Z}/ \sim \) consists of two equivalence classes: | Indeed, every integer \( b \) is either even (and hence \( b - 0 \) is even, so \( b \sim 0 \), and \( b \in {\\left\\lbrack 0\\right\\rbrack }_{ \\sim } \) ) or odd (and hence \( b - 1 \) is even, so \( b \sim 1 \), and \( b \in {\\left\\lbrack 1\\right\\rbrack }_{ \\sim } \) ). | Yes |
Proposition 2.1. Assume \( A \neq \varnothing \), and let \( f : A \rightarrow B \) be a function. Then\n\n(1) \( f \) has a left-inverse if and only if it is injective. | Proof. Let's prove (1).\n\n\( \left( \Rightarrow \right) \) If \( f : A \rightarrow B \) has a left-inverse, then there exists a \( g : B \rightarrow A \) such that \( g \circ f = {\operatorname{id}}_{A} \) . Now assume that \( {a}^{\prime } \neq {a}^{\prime \prime } \) are arbitrary different elements in \( A \) ; the... | Yes |
Proposition 2.3. A function is injective if and only if it is a monomorphism. | Proof. \( \left( \Rightarrow \right) \) By Proposition 2.1, if a function \( f : A \rightarrow B \) is injective, then it has a left-inverse \( g : B \rightarrow A \) . Now assume that \( {\alpha }^{\prime },{\alpha }^{\prime \prime } \) are arbitrary functions from another set \( Z \) to \( A \) and that\n\n\[ f \circ... | Yes |
Let \( A, B \) be sets. Then there are natural projections \( {\pi }_{A},{\pi }_{B} \) | defined by \[ {\pi }_{A}\left( \left( {a, b}\right) \right) \mathrel{\text{:=}} a,\;{\pi }_{B}\left( \left( {a, b}\right) \right) \mathrel{\text{:=}} b \] for all \( \left( {a, b}\right) \in A \times B \) . Both of these maps are (clearly) surjective. | Yes |
Similarly, there are natural injections from \( A \) and \( B \) to the disjoint union: | obtained by sending \( a \in A \) (resp., \( b \in B \) ) to the corresponding element in the isomorphic copy \( {A}^{\prime } \) of \( A \) (resp., \( {B}^{\prime } \) of \( B \) ) in \( A \coprod B \) . | No |
If \( \sim \) is an equivalence relation on a set \( A \), there is a (clearly surjective) canonical projection | \[ A \rightarrow A/ \sim \] obtained by sending every \( a \in A \) to its equivalence class \( {\left\lbrack a\right\rbrack }_{ \sim } \) | Yes |
Theorem 2.7. Let \( f : A \rightarrow B \) be any function, and define \( \sim \) as above. Then \( f \) decomposes as follows:\n\n\n\nwhere the first function is the canonical projection \( A \rightarrow A/ \sim \) (a... | Proof. Spelling out the first item discussed above, we have to verify that, for all \( {a}^{\prime },{a}^{\prime \prime } \) in \( A \), \n\n\[ {\left\lbrack {a}^{\prime }\right\rbrack }_{ \sim } = {\left\lbrack {a}^{\prime \prime }\right\rbrack }_{ \sim } \Rightarrow f\left( {a}^{\prime }\right) = f\left( {a}^{\prime ... | Yes |
It is hopefully crystal clear by now that sets (as objects), together with set-functions (as morphisms), form a category; if not, the reader must stop here and go no further until this assertion sheds any residual mystery 17. | - \( \operatorname{Obj}\left( \operatorname{Set}\right) = \) the class of all sets;\n\n- for \( A, B \) in \( \operatorname{Obj}\left( \operatorname{Set}\right) \) (that is, for \( A, B \) sets) \( {\operatorname{Hom}}_{\operatorname{Set}}\left( {A, B}\right) = {B}^{A} \) . | No |
Suppose \( S \) is a set and \( \sim \) is a relation on \( S \) satisfying the reflexive and transitive properties. Then we can encode this data into a category: | - objects: the elements of \( S \) ;\n\n- morphisms: if \( a, b \) are objects (that is, if \( a, b \in S \) ), then let \( \operatorname{Hom}\left( {a, b}\right) \) be the set consisting of the element \( \left( {a, b}\right) \in S \times S \) if \( a \sim b \), and let \( \operatorname{Hom}\left( {a, b}\right) = \var... | No |
Let \( S \) again be a set. Define a category \( \widehat{\mathrm{S}} \) by setting\n\n- \( \operatorname{Obj}\left( \widehat{\mathrm{S}}\right) = \mathcal{P}\left( S\right) \), the power set \( S \) (cf. \( \$ \underline{1.2} \) and Exercise 2.11);\n\n- for \( A, B \) objects of \( \widehat{\mathrm{S}} \) (that is, \(... | Checking the axioms specified in [3.1] should be routine (make sure this is the case!). | No |
For the sake of concreteness, let's apply the construction given in Example 3.5 to the category constructed in Example 3.3, say for \( S = \mathbb{Z} \) and \( \sim \) the relation \( \leq \) . Call \( \mathrm{C} \) this category, and choose an object \( A \) of \( \mathrm{C} \) -that is, an integer, for example, \( A ... | \[ \left( {m,3}\right) \rightarrow \left( {n,3}\right) \] if and only if \( m \leq n \) . In this case \( {\mathrm{C}}_{A} \) may be harmlessly identified with the ’subcategory’ of integers \( \leq 3 \), with ’the same’ morphisms as in \( \mathrm{C} \). | Yes |
It is useful to contemplate a few more 'abstract' examples in the style of Examples 3.5 and 3.7 These will be essential ingredients in the promised revisitation of some of the operations mentioned in [1.3] Their definition will appear disappointedly simple-minded to the reader who has mastered Examples 3.5 and 3.7.\n\n... | I will leave to the reader the task of formalizing this rough description. This example is really nothing more than a mixture of \( {\mathrm{C}}_{A} \) and \( {\mathrm{C}}_{B} \), where the two structures interact because of the stringent requirement that the same \( \sigma \) must make both sides of the diagram commut... | No |
Proposition 4.2. The inverse of an isomorphism is unique. | Proof. We have to verify that if both \( {g}_{1} \) and \( {g}_{2} : B \rightarrow A \) act as inverses of a given isomorphism \( f : A \rightarrow B \), then \( {g}_{1} = {g}_{2} \) . The standard trick for this kind of verification is to compose \( f \) on the left by one of the morphisms, and on the right by the oth... | Yes |
Proposition 4.3. With notation as above:\n\n- Each identity \( {1}_{A} \) is an isomorphism and is its own inverse.\n\n- If \( f \) is an isomorphism, then \( {f}^{-1} \) is an isomorphism and further \( {\left( {f}^{-1}\right) }^{-1} = f \) .\n\n- If \( f \in {\operatorname{Hom}}_{\mathrm{C}}\left( {A, B}\right), g \i... | Proof. These all 'prove themselves'. For example, it is immediate to verify that \( {f}^{-1}{g}^{-1} \) is a left-inverse of \( {gf} \) : indeed 20,\n\n\[ \left( {{f}^{-1}{g}^{-1}}\right) \left( {gf}\right) = {f}^{-1}\left( {\left( {{g}^{-1}g}\right) f}\right) = {f}^{-1}\left( {{1}_{B}f}\right) = {f}^{-1}f = {1}_{A}. \... | No |
An automorphism of an object \( A \) of a category \( \mathrm{C} \) is an isomorphism from \( A \) to itself. The set of automorphisms of \( A \) is denoted \( {\operatorname{Aut}}_{\mathrm{C}}\left( A\right) \) ; it is a subset of \( {\operatorname{End}}_{\mathrm{C}}\left( A\right) \) . | By Proposition 4.3, composition confers on \( {\operatorname{Aut}}_{\mathrm{C}}\left( A\right) \) a remarkable structure:\n\n- the composition of two elements \( f, g \in {\operatorname{Aut}}_{\mathrm{C}}\left( A\right) \) is an element \( {gf} \in {\operatorname{Aut}}_{\mathrm{C}}\left( A\right) \) ;\n\n- composition ... | Yes |
Example 4.9. As proven in Proposition 2.3, in the category Set the monomorphisms are precisely the injective functions. | The reader should have by now checked that, likewise, in Set the epimorphisms are precisely the surjective functions (cf. Exercise 2.5). Thus, while the definitions given in 82.6 may have looked counterintuitive at first, they work as natural 'categorical counterparts' of the ordinary notions of injective/surjective fu... | No |
In the categories of Example 3.3, every morphism is both a monomorphism and an epimorphism. | Indeed, recall that there is at most one morphism between any two objects in these categories; hence the conditions defining monomorphisms and epimorphisms are vacuous. | Yes |
The category obtained by endowing \( \mathbb{Z} \) with the relation \( \leq \) (see Example 3.3) has no initial or final object. | Indeed, an initial object in this category would be an integer \( i \) such that \( i \leq a \) for all integers \( a \) ; there is no such integer. Similarly, a final object would be an integer \( f \) larger than every integer, and there is no such thing. | Yes |
If \( {I}_{1},{I}_{2} \) are both initial objects in \( \mathrm{C} \), then \( {I}_{1} \cong {I}_{2} \) . | Proof. Recall that (by definition of category!) for every object \( A \) of \( \mathrm{C} \) there is at least one element in \( {\operatorname{Hom}}_{\mathrm{C}}\left( {A, A}\right) \), namely the identity \( {1}_{A} \) . If \( I \) is initial, then there is a unique morphism \( I \rightarrow I \), which therefore mus... | Yes |
Claim 5.5. Denoting by \( \pi \) the ’canonical projection’ defined in Example 2.6, the pair \( \left( {\pi, A/ \sim }\right) \) is an initial object of this category. | Proof. Consider any \( \left( {\varphi, Z}\right) \) as above. We have to prove that there exists a unique morphism \( \left( {\pi, A/ \sim }\right) \rightarrow \left( {\varphi, Z}\right). | No |
Proposition 5.6. The disjoint union is a coproduct in Set. | Proof. Recall (§1.4) that the disjoint union \( A \coprod B \) is defined as the union of two disjoint isomorphic copies \( {A}^{\prime },{B}^{\prime } \) of \( A, B \), respectively; for example, we may let \( {A}^{\prime } = \{ 0\} \times A,{B}^{\prime } = \{ 1\} \times B \) . The functions \( {i}_{A},{i}_{B} \) are ... | Yes |
Since we explicitly require \( G \) to be nonempty, the most economical way to concoct a group is by letting \( G = \{ e\} \) be a singleton. There is only one function \( G \times G \rightarrow G \) in this case, so there is only one possible binary operation on \( G \), defined by\n\n\[ e \bullet e \mathrel{\text{:=}... | The three axioms trivially hold for this example, so \( \{ e\} \) is equipped with a unique group structure.\n\nThis is usually called the trivial group; purists should call any such group \( a \) trivial group, since every singleton gives rise to one. | Yes |
The reader should now check that \( 2 \times 2 \) matrices\n\n\[ \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \]\n\nwith real entries, and such that \( {ad} - {bc} \neq 0 \), form a group under the ordinary matrix multiplication:\n\n\[ \left( \begin{array}{ll} {a}_{1} & {b}_{1} \\ {c}_{1} & {d}_{1} \end{a... | Since, for example,\n\n\[ \left( \begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right) \left( \begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right) = \left( \begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right) \neq \left( \begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right) = \left( \begin{array}{ll} 1 & 0 \\ 1 & 1 \end{arra... | No |
Proposition 1.6. If \( h \in G \) is an identity of \( G \), then \( h = {e}_{G} \) . | Proof. Using first that \( {e}_{G} \) is an identity and then that \( h \) is an identity, one get.\n\n\[ h = {e}_{G}h = {e}_{G} \]\n\n(Amusingly, this argument only uses that \( {e}_{G} \) is a ’left’ identity and \( h \) is a ’right’ identity.) | Yes |
Proposition 1.7. The inverse is also unique: if \( {h}_{1},{h}_{2} \) are both inverses of \( g \) in \( G \) , then \( {h}_{1} = {h}_{2} \) . | Proof. This actually follows from Proposition 1.4.2 (by viewing \( G \) as the set of isomorphisms of a groupoid with a single object). The reader should construct a stand-alone proof, using the same trick, but carefully hiding any reference to morphisms. | No |
Proposition 1.8. Let \( G \) be a group. Then \( \forall a, g, h \in G \)\n\n\[ \n{ga} = {ha} \Rightarrow g = h,\;{ag} = {ah} \Rightarrow g = h.\n\] | Proof. Both statements are proven by multiplying (on the appropriate side) by \( {a}^{-1} \) and applying associativity. For example,\n\n\[ \n{ga} = {ha} \Rightarrow \left( {ga}\right) {a}^{-1} = \left( {ha}\right) {a}^{-1} \Rightarrow g\left( {a{a}^{-1}}\right) = h\left( {a{a}^{-1}}\right) \Rightarrow g{e}_{G} = h{e}_... | Yes |
Lemma 1.10. If \( {g}^{n} = e \) for some positive integer \( n \), then \( \left| g\right| \) is a divisor of \( n \) . | Proof. As observed, \( n \geq \left| g\right| \) by definition of order, that is, \( n - \left| g\right| \geq 0 \) . There must then exist a positive integer \( m \) such that\n\n\[ r = n - \left| g\right| \cdot m \geq 0\;\text{ and }\;n - \left| g\right| \cdot \left( {m + 1}\right) < 0, \]\n\nthat is, \( r < \left| g\... | Yes |
Proposition 1.13. Let \( q \in G \) be an element of finite order. Then \( {g}^{m} \) has finite order \( \forall m \geq 0 \), and in fact \[ \left| {g}^{m}\right| = \frac{\operatorname{lcm}\left( {m,\left| g\right| }\right) }{m} = \frac{\left| g\right| }{\gcd \left( {m,\left| g\right| }\right) }.\] | Proof. The equality of the two numbers \( \frac{\operatorname{lcm}\left( {m,\left| g\right| }\right) }{m} \) and \( \frac{\left| g\right| }{\gcd \left( {m,\left| g\right| }\right) } \) follows from elementary properties of \( \gcd \) and \( \operatorname{lcm} : \operatorname{lcm}\left( {a, b}\right) = {ab}/\gcd \left( ... | Yes |
Proposition 1.14. If \( {gh} = {hg} \), then \( \left| {gh}\right| \) divides \( \operatorname{lcm}\left( {\left| g\right| ,\left| h\right| }\right) \) . | Proof. Let \( \left| g\right| = m,\left| h\right| = n \) . If \( N \) is any common multiple of \( m \) and \( n \), then \( {g}^{N} = {h}^{N} = e \) by Corollary 1.11 Since \( g \) and \( h \) commute,\n\n\[{\left( gh\right) }^{N} = \underset{N\text{ times }}{\underbrace{\left( {gh}\right) \left( {gh}\right) \cdots \c... | Yes |
Lemma 2.2. If \( a \equiv {a}^{\prime }{\;\operatorname{mod}\;n} \) and \( b \equiv {b}^{\prime }{\;\operatorname{mod}\;n} \), then\n\n\[ \left( {a + b}\right) \equiv \left( {{a}^{\prime } + {b}^{\prime }}\right) {\;\operatorname{mod}\;n}. \]\n | Proof. By hypothesis \( n \mid \left( {{a}^{\prime } - a}\right) \) and \( n \mid \left( {{b}^{\prime } - b}\right) \) ; therefore \( \exists k,\ell \in \mathbb{Z} \) such that\n\n\[ \left( {{a}^{\prime } - a}\right) = {kn},\;\left( {{b}^{\prime } - b}\right) = \ell n. \]\n\nThen\n\n\[ \left( {{a}^{\prime } + {b}^{\pri... | Yes |
Proposition 2.3. The order of \( {\left\lbrack m\right\rbrack }_{n} \) in \( \mathbb{Z}/n\mathbb{Z} \) is 1 if \( n \mid m \), and more generally\n\n\[ \left| {\left\lbrack m\right\rbrack }_{n}\right| = \frac{n}{\gcd \left( {m, n}\right) }.\] | Proof. If \( n \mid m \), then \( {\left\lbrack m\right\rbrack }_{n} = {\left\lbrack 0\right\rbrack }_{n} \) . If \( n \) does not divide \( m \), observe again that \( {\left\lbrack m\right\rbrack }_{n} = m{\left\lbrack 1\right\rbrack }_{n} \) and apply Proposition 1.13. | No |
Proposition 2.6. Multiplication makes \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) into a group. | Proof. Simple properties of gcd’s show that if \( \gcd \left( {{m}_{1}, n}\right) = \gcd \left( {{m}_{2}, n}\right) = 1 \), then \( \gcd \left( {{m}_{1}{m}_{2}, n}\right) = 1 \) . (For example, if a prime integer divided both \( n \) and \( {m}_{1}{m}_{2} \), then it would necessarily divide \( {m}_{1} \) or \( {m}_{2}... | Yes |
Proposition 3.2. Let \( \varphi : G \rightarrow H \) be a group homomorphism. Then\n\n- \( \varphi \left( {e}_{G}\right) = {e}_{H} \) ;\n\n- \( \forall g \in G,\varphi \left( {g}^{-1}\right) = \varphi {\left( g\right) }^{-1} \) . | Proof. The first item follows from the definition of homomorphism and cancellation: since \( {e}_{H} = {e}_{H} \cdot {e}_{H} \), \n\n\[ \n{e}_{H} \cdot \varphi \left( {e}_{G}\right) = \varphi \left( {e}_{G}\right) = \varphi \left( {{e}_{G} \cdot {e}_{G}}\right) = \varphi \left( {e}_{G}\right) \cdot \varphi \left( {e}_{... | Yes |
Proposition 3.3. Trivial groups are both initial and final in Grp. | Proof. It should be clear that trivial groups are final: there is only one function from a set to a singleton, that is, the constant function; this is vacuously a group homomorphism.\n\nTo see that trivial groups are initial, let \( T = \{ e\} \) be a trivial group; for any group \( G \), define \( \varphi : T \rightar... | Yes |
Proposition 3.4. With operation defined componentwise, \( G \times H \) is a product in Grp. | Proof. Recall (§ 15.4) that this means that \( G \times H \) satisfies the following universal property: for any group \( A \) and any choice of group homomorphisms \( {\varphi }_{G} : A \rightarrow G \), \( {\varphi }_{H} : A \rightarrow H \), there exists a unique group homomorphism \( {\varphi }_{G} \times {\varphi ... | Yes |
Proposition 4.1. Let \( \varphi : G \rightarrow H \) be a group homomorphism, and let \( g \in G \) be an element of finite order. Then \( \left| {\varphi \left( g\right) }\right| \) divides \( \left| g\right| \) . | Proof. As observed, \( \varphi {\left( g\right) }^{\left| g\right| } = {e}_{H} \) ; applying Lemma 1.10 gives the statement. | No |
There are no nontrivial homomorphisms \( \mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z} \) : indeed, the image of every element of \( \mathbb{Z}/n\mathbb{Z} \) must have finite order, and the only element with finite order in \( \left( {\mathbb{Z}, + }\right) \) is 0 . | Indeed, the image of every element of \( \mathbb{Z}/n\mathbb{Z} \) must have finite order, and the only element with finite order in \( \left( {\mathbb{Z}, + }\right) \) is 0. | Yes |
Proposition 4.3. Let \( \varphi : G \rightarrow H \) be a group homomorphism. Then \( \varphi \) is an isomorphism of groups if and only if it is a bijection. | Proof. One implication is immediate, as pointed out above. For the other implication, assume \( \varphi : G \rightarrow H \) is a bijective group homomorphism. As a bijection, \( \varphi \) has an inverse in Set:\n\n\[{\varphi }^{-1} : H \rightarrow G\]\n\nwe simply need to check that this is a group homomorphism. Let ... | Yes |
Proposition 4.8. Let \( \varphi : G \rightarrow H \) be an isomorphism.\n\n\[ \text{-}\left( {\forall g \in G}\right) : \left| {\varphi \left( g\right) }\right| = \left| g\right| \text{;} \] | Proof. The first assertion follows from Proposition 4.1 the order of \( \varphi \left( g\right) \) divides the order of \( g \), and on the other hand the order of \( g = {\varphi }^{-1}\left( {\varphi \left( g\right) }\right) \) must divide the order of \( \varphi \left( g\right) \) ; thus the two orders must be equal... | Yes |
Lemma 5.1. If \( w \in W\left( A\right) \) has length \( n \), then \( 2{2}^{n\left\lfloor \frac{n}{2}\right\rfloor }\left( w\right) \) is a reduced word. | Proof. Indeed, either \( r\left( w\right) = w \) or the length of \( r\left( w\right) \) is less than the length of \( w \) ; but one cannot decrease the length of \( w \) more than \( n/2 \) times, since each nonidentity application of \( r \) decreases the length by two. | No |
Proposition 5.2. The pair \( \left( {j, F\left( A\right) }\right) \) satisfies the universal property for free groups on \( A \) . | Proof. This is also essentially evident, once one has absorbed all the notation. Any function \( f : A \rightarrow G \) to a group extends uniquely to a map \( \varphi : F\left( A\right) \rightarrow \) \( G \), determined by the homomorphism condition and by the requirement that the diagram commutes, which fixes its va... | Yes |
Example 5.3. It is easy to ’visualize’ \( F\left( {\{ a\} }\right) \cong \mathbb{Z} \) ; but it is already somewhat challenging for the free group on two generators, \( F\left( {\{ x, y\} }\right) \) . The best I can do is the following: behold the infinite graph \( {}^{23} \) | This is an example of the Cayley graph of a group (cf. Exercise 8.6): a graph whose vertices correspond to the elements of the group and whose edges connect vertices according to the action of generators.\n\nobtained by starting at a point (the center of the picture), then branching out in four directions by a length o... | No |
Proposition 5.6. For every set \( A,{F}^{ab}\left( A\right) \cong {\mathbb{Z}}^{\oplus A} \) . | Proof. The key point is again that every element of \( {\mathbb{Z}}^{\oplus A} \) may be written uniquely as a finite sum\n\n\[ \mathop{\sum }\limits_{{a \in A}}{m}_{a}j\left( a\right) ,\;{m}_{a} \neq 0\text{ for only finitely many }a; \]\n\nonce this is understood, the argument is precisely the same as for Claim 5.4\n... | No |
Proposition 6.2. A nonempty subset \( H \) of a group \( G \) is a subgroup if and only if\n\n\[ \left( {\forall a, b \in H}\right) : \;a{b}^{-1} \in H. \] | Proof. It is clear that if \( H \) is a subgroup, then the stated condition holds: indeed, if \( b \in H \), then the inverse of \( b \) must also be in \( H \) and \( H \) is closed under the operation of \( G \) .\n\nConversely, assume the stated condition holds; we have to check that \( H \) is closed under the oper... | Yes |
Lemma 6.3. If \( {\left\{ {H}_{\alpha }\right\} }_{\alpha \in A} \) is any family of subgroups of a group \( G \), then\n\n\[ H = \mathop{\bigcap }\limits_{{\alpha \in A}}{H}_{\alpha } \]\n\nis a subgroup of \( G \) . | Proof. This follows right away from Proposition 6.2. \( H \) is nonempty, because \( e \in {H}_{\alpha } \) for all \( \alpha \), so \( e \in H \) ; and\n\n\[ a, b \in H \Rightarrow \left( {\forall \alpha \in A}\right) : a, b \in {H}_{\alpha } \Rightarrow \left( {\forall \alpha \in A}\right) : a{b}^{-1} \in {H}_{\alpha... | Yes |
Lemma 6.4. Let \( \varphi : G \rightarrow {G}^{\prime } \) be a group homomorphism, and let \( {H}^{\prime } \) be a subgroup of \( {G}^{\prime } \) . Then \( {\varphi }^{-1}\left( {H}^{\prime }\right) \) is a subgroup of \( G \) . | Proof. Recall (end of [12.5] that \( {\varphi }^{-1}\left( {H}^{\prime }\right) \) consists of all \( g \in G \) such that \( \varphi \left( g\right) \in \) \( {H}^{\prime } \) . Since \( \varphi \left( {e}_{G}\right) = {e}_{{G}^{\prime }} \in {H}^{\prime } \), this set is nonempty. If \( a, b \in {\varphi }^{-1}\left(... | Yes |
Proposition 6.6. Let \( \varphi : G \rightarrow {G}^{\prime } \) be a homomorphism. Then the inclusion \( i \) : \( \ker \varphi \hookrightarrow G \) is final in the category 26 of group homomorphisms \( \alpha : K \rightarrow G \) such that \( \varphi \circ \alpha \) is the trivial map. | Proof. If \( \alpha : K \rightarrow G \) is such that \( \varphi \circ \alpha \) is the trivial map, then \( \forall k \in K \)\n\n\[ \varphi \circ \alpha \left( k\right) = \varphi \left( {\alpha \left( k\right) }\right) = {e}_{{G}^{\prime }}, \]\n\nthat is, \( \alpha \left( k\right) \in \ker \varphi \) . We can (and m... | Yes |
Proposition 6.9. Let \( G \subseteq \mathbb{Z} \) be a subgroup. Then \( G = d\mathbb{Z} \) for some \( d \geq 0 \) . | Proof of Proposition 6.9. If \( G = \{ 0\} \), then \( G = 0\mathbb{Z} \) . If not, note that \( G \) must contain positive integers: indeed, if \( a \in G \) and \( a \leq 0 \), then \( - a \in G \) and \( - a > 0 \) . We can then let \( d \) be the smallest positive integer 29 in \( G \), and I claim \( G = d\mathbb{... | Yes |
Proposition 6.11. Let \( n > 0 \) be an integer and let \( G \subseteq \mathbb{Z}/n\mathbb{Z} \) be a subgroup. Then \( G \) is the cyclic subgroup of \( \mathbb{Z}/n\mathbb{Z} \) generated by \( {\left\lbrack d\right\rbrack }_{n} \), for some divisor \( d \) of \( n \) . | Proof. Let \( {\pi }_{n} : \mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z} \) be the quotient map, and consider \( {G}^{\prime } \mathrel{\text{:=}} {\pi }_{n}^{-1}\left( G\right) \) . By Lemma 6.4. \( {G}^{\prime } \) is a subgroup of \( \mathbb{Z} \) ; by Proposition 6.9, \( {G}^{\prime } \) is a cyclic subgroup of \( ... | Yes |
Proposition 6.12. The following are equivalent:\n\n(a) \( \\varphi \) is a monomorphism;\n\n(b) \( \\ker \\varphi = \\left\\{ {e}_{G}\\right\\} \) ;\n\n(c) \( \\varphi : G \\rightarrow {G}^{\\prime } \) is injective (as a set-function). | Proof. (a) \( \\Rightarrow \) (b): Assume (a) holds, and consider the two parallel compositions\n\n\[ \n\\ker \\varphi \\xrightarrow[e]{i}G\\overset{\\varphi }{ \\rightarrow }{G}^{\\prime }\n\]\nwhere \( i \) is the inclusion and \( e \) is the trivial map. Both \( \\varphi \\circ i \) and \( \\varphi \\circ e \) are t... | Yes |
Lemma 7.2. If \( \varphi : G \rightarrow {G}^{\prime } \) is any group homomorphism, then \( \ker \varphi \) is a normal subgroup of \( G \) . | Proof. We already know that \( \ker \varphi \) is a subgroup of \( G \) ; to verify it is normal note that \( \forall g \in G,\forall n \in \ker \varphi \)\n\n\[ \varphi \left( {{gn}{g}^{-1}}\right) = \varphi \left( g\right) \varphi \left( n\right) \varphi \left( {g}^{-1}\right) = \varphi \left( g\right) {e}_{{G}^{\pri... | Yes |
Proposition 7.4. Let \( \sim \) be an equivalence relation on a group \( G \), satisfying \( \left( \dagger \right) \) . Then\n\n- the equivalence class of \( {e}_{G} \) is a subgroup \( H \) of \( G \) ; and\n\n- \( a \sim b \Leftrightarrow {a}^{-1}b \in H \Leftrightarrow {aH} = {bH} \) . | Proof. Let \( H \subseteq G \) be the equivalence class of the identity; \( H \neq \varnothing \) as \( {e}_{G} \in H \) . For \( a, b \in H \), we have \( {e}_{G} \sim b \) and hence \( {b}^{-1} \sim {e}_{G} \) (applying \( \left( \dagger \right) \), multiplying on the left by \( {b}^{-1} \) ); hence \( a{b}^{-1} \sim... | Yes |
Proposition 7.6. If \( H \) is any subgroup of a group \( G \), the relation \( { \sim }_{L} \) defined by\n\n\[ \n\left( {\forall a, b \in G}\right) : \;a{ \sim }_{L}b \Leftrightarrow {a}^{-1}b \in H \n\]\n\nis an equivalence relation satisfying \( \left( \dagger \right) \) . | Proof. This is straightforward and is mostly left to the reader (Exercise 7.8). To see that the relation satisfies \( \left( \dagger \right) \), note that\n\n\[ \na{ \sim }_{L}b \Rightarrow {a}^{-1}b \in H \Rightarrow {a}^{-1}\left( {{g}^{-1}g}\right) b \in H \Rightarrow {\left( ga\right) }^{-1}\left( {gb}\right) \in H... | No |
Proposition 7.8. There is a one-to-one correspondence between subgroups of \( G \) and equivalence relations on \( G \) satisfying \( \left( {\dagger \dagger }\right) \) ; for the relation \( { \sim }_{R} \) corresponding to a subgroup \( H, G/{ \sim }_{R} \) may be described as the set of right-cosets \( {Ha} \) of \(... | The relation corresponding to \( H \) in this second way is defined by\n\n\[ \n a{ \sim }_{R}b \Leftrightarrow a{b}^{-1} \in H \Leftrightarrow {Ha} = {Hb}. \n\] | No |
Let \( G = {S}_{3} \), and let \( H \) be the subgroup consisting of the identity and the \( 1 \leftrightarrow 2 \) switch:\n\n\[ H = \left\{ {\left( \begin{array}{lll} 1 & 2 & 3 \\ 1 & 2 & 3 \end{array}\right) ,\left( \begin{array}{lll} 1 & 2 & 3 \\ 2 & 1 & 3 \end{array}\right) }\right\} \]\n\nThen\n\n\[ \left( \begin... | This state of affairs simply reflects the fact that the two conditions \( \left( \dagger \right) \) and \( \left( {\dagger \dagger }\right) \) are different: there is no reason to expect that if one holds, the other one should also hold (unless \( G \) is commutative, of course). Once more, keep in mind that both have ... | No |
Proposition 7.10. The relations \( { \sim }_{L},{ \sim }_{R} \) corresponding to a subgroup \( H \) coincide if and only if \( H \) is normal. | Proof. Two relations coincide if the corresponding partitions agree. Therefore\n\n\( { \sim }_{L} = { \sim }_{R} \Leftrightarrow \) left- and right-cosets of \( H \) coincide \( \Leftrightarrow \left( {\forall g \in G}\right) : {gH} = {Hg}. \)\n\nBut this is one of the equivalent conditions defining the notion of norma... | Yes |
Theorem 7.12. Let \( H \) be a normal subgroup of a group \( G \) . Then for every group homomorphism \( \varphi : G \rightarrow {G}^{\prime } \) such that \( H \subseteq \ker \varphi \) there exists a unique group homomorphism \( \widetilde{\varphi } : G/H \rightarrow {G}^{\prime } \) so that the diagram\n\n![23387543... | Proof. We only need to match the stated universal property with the one we proved in Proposition 7.3, and indeed,\n\n\[ H \subseteq \ker \varphi \Leftrightarrow \left( {\forall h \in H}\right) : \varphi \left( h\right) = {e}_{{G}^{\prime }} \]\n\nis equivalent to\n\n\[ \left( {\forall a, b \in G}\right) : a{b}^{-1} \in... | Yes |
Theorem 8.1. Every group homomorphism \( \varphi : G \rightarrow {G}^{\prime } \) may be decomposed as follows: \n\nwhere the isomorphism \( \widetilde{\varphi } \) in the middle is the homomorphism induced by \( \va... | It is important that the reader agree that we have already proved anything that deserves to be proven here. We know that the projection on the left and the inclusion on the right are homomorphisms and \( \widetilde{\varphi } \) comes from Theorem 7.12. The decomposition is the same one obtained at the level of set-func... | No |
Corollary 8.2. Suppose \( \varphi : G \rightarrow {G}^{\prime } \) is a surjective group homomorphism. Then\n\n\[ \n{G}^{\prime } \cong \frac{G}{\ker \varphi }.\n\] | Proof. \( {im\varphi } = {G}^{\prime } \) in Theorem 8.1 | No |
Claim 8.4. If \( {H}_{1} \subseteq {G}_{1} \) and \( {H}_{2} \subseteq {G}_{2} \) are normal subgroups, then \( {H}_{1} \times {H}_{2} \) is a normal subgroup of the group \( {G}_{1} \times {G}_{2} \) and\n\n\[ \frac{{G}_{1} \times {G}_{2}}{{H}_{1} \times {H}_{2}} \cong \frac{{G}_{1}}{{H}_{1}} \times \frac{{G}_{2}}{{H}... | Indeed, composing the projections\n\n\[ {\pi }_{1} : {G}_{1} \times {G}_{2} \rightarrow {G}_{1},\;{\pi }_{2} : {G}_{1} \times {G}_{2} \rightarrow {G}_{2} \]\n\nwith the morphisms to the quotients gives surjective homomorphisms\n\n\[ {\pi }_{1} : {G}_{1} \times {G}_{2} \rightarrow \frac{{G}_{1}}{{H}_{1}},\;{\pi }_{2} : ... | Yes |
Example 8.5. As a particular case of Claim 8.4, take \( {H}_{1} = \left\{ {e}_{{G}_{1}}\right\} \subseteq {G}_{1} \) and \( {H}_{2} = {G}_{2} \subseteq {G}_{2} \)\n\n\[ \n\frac{{G}_{1} \times {G}_{2}}{{G}_{2}} \cong \frac{{G}_{1}}{\left\{ {e}_{{G}_{1}}\right\} } \times \frac{{G}_{2}}{{G}_{2}} \cong {G}_{1} \n\]\n\nwher... | For instance 34 (cf. \( §\overline{4.1} \) )\n\n\[ \n\frac{{C}_{6}}{{C}_{3}} \cong \frac{{C}_{2} \times {C}_{3}}{{C}_{3}} \cong {C}_{2} \n\] | No |
The cyclic group \( {C}_{3} \) may be viewed as a subgroup of the dihedral group \( {D}_{6} \) : the rotations of a triangle give a copy of \( {C}_{3} \) inside \( {D}_{6} \) . Then \( {C}_{3} \) is normal in \( {D}_{6} \), and \[ \frac{{D}_{6}}{{C}_{3}} \cong {C}_{2} \] | This can of course be checked 'by hand'. But note that there is an evident surjective homomorphism \( {D}_{6} \rightarrow {C}_{2} \), whose kernel is \( {C}_{3} \) : map an element \( \sigma \) of \( {D}_{6} \) to the identity in \( {C}_{2} \) if it does not flip the triangle (that is, precisely when \( \sigma \in {C}_... | Yes |
One can give a circle (denoted \( {S}^{1} \) ) a group structure by identifying its points with rotations of a plane about a point and adding them accordingly. The function\n\n\[ \rho : {\mathbb{R}}^{1} \rightarrow {S}^{1} \]\n\nmapping a number \( r \) to the result of a rotation by \( {2\pi r} \) radians is then a su... | By Corollary 8.2, therefore,\n\n\[ \frac{\mathbb{R}}{\mathbb{Z}} \cong {S}^{1} \]\n\n(Cf. Exercise 1.1.6) Geometrically, this amounts to 'wrapping' \( \mathbb{R} \) infinitely many times around the circle, realizing \( \mathbb{R} \) as the ’universal cover’ of \( {S}^{1} \) ; here, \( \mathbb{Z} \) plays the role of ’f... | No |
Here is the effect of this operation on the lattice of subgroups of \( {C}_{12} \cong \mathbb{Z}/{12}\mathbb{Z} \) (labeled by generators; cf. 6.4), after quotienting by \( H = \langle \left\lbrack 6\right\rbrack \rangle \cong {C}_{2} \). | Here is why this works. First note that if \( H \subseteq K \) are subgroups of a group \( G \) and \( H \) is normal in \( G \), then \( H \) is normal in \( K \) . | No |
Proposition 8.9. Let \( H \) be a normal subgroup of a group \( G \). Then for every subgroup \( K \) of \( G \) containing \( H \), \( K/H \) may be identified with a subgroup of \( G/H \). The function \[ u : \{ \text{subgroups}K\text{of}G\text{containing}H\} \rightarrow \{ \text{subgroups of}G/H\} \] defined by \( u... | Proof. The group \( K/H \) consists of the cosets \( {aH} \in G/H \) with \( a \in K \), and in this sense it is a subset (and clearly a subgroup) of \( G/H \). It is also clear that if \( H \subseteq K \subseteq L \), then \( u\left( K\right) = K/H \subseteq L/H = u\left( L\right) \); that is, \( u \) preserves inclus... | No |
Proposition 8.10. Let \( H \) be a normal subgroup of a group \( G \), and let \( N \) be a subgroup of \( G \) containing \( H \) . Then \( N/H \) is normal in \( G/H \) if and only if \( N \) is normal in \( G \), and in this case \[ \frac{G/H}{N/H} \cong \frac{G}{N} \] | Proof. If \( N \) is normal, then consider the projection \[ G \rightarrow \frac{G}{N} : \] the subgroup \( H \) is contained in \( N \), which is the kernel of this homomorphism, so we get (by the universal property of quotients, Theorem 7.12) an induced homomorphism \[ \frac{G}{H} \rightarrow \frac{G}{N} \] The subgr... | Yes |
Proposition 8.11. Let \( H, K \) be subgroups of a group \( G \), and assume that \( H \) is normal in \( G \) . Then\n\n- \( {HK} \) is a subgroup of \( G \), and \( H \) is normal in \( {HK} \) ;\n\n- \( H \cap K \) is normal in \( K \), and\n\n\[ \frac{HK}{H} \cong \frac{K}{H \cap K} \] | Proof. To verify that \( {HK} \) is a subgroup of \( G \) when \( H \) is normal, note that \( {HK} \) is the union of all cosets \( {Hk} \), with \( k \in K \) ; that is,\n\n\[ {HK} = {\pi }^{-1}\left( {\pi \left( K\right) }\right) \]\n\nwhere \( \pi : G \rightarrow G/H \) is the canonical projection. Since \( \pi \le... | Yes |
Lemma 8.13. Let \( H \) be a subgroup of a group \( G \) . Then \( \forall g \in G \) the functions\n\n\[ H \rightarrow {gH},\;h \mapsto {gh},\]\n\n\[ H \rightarrow {Hg},\;h \mapsto {hg} \]\n\nare bijections. | Proof. Both functions are surjective by definition of coset. Cancellation implies that they are injective. | No |
Corollary 8.14 (Lagrange’s theorem). If \( G \) is a finite group and \( H \subseteq G \) is a subgroup, then \( \left| G\right| = \left\lbrack {G : H}\right\rbrack \cdot \left| H\right| \) . In particular, \( \left| H\right| \) is a divisor of \( \left| G\right| \) . | Proof. Indeed, \( G \) is the disjoint union of \( \left| {G/H}\right| \) distinct cosets \( {gH} \), and \( \left| {gH}\right| = \left| H\right| \) by Lemma 8.13 | Yes |
The order \( \left| g\right| \) of any element \( g \) of a finite group \( G \) is a divisor of \( \left| G\right| \) | indeed, \( \left| g\right| \) equals the order of the subgroup \( \langle g\rangle \) generated by \( g \) | No |
If \( \left| G\right| \) is a prime integer \( p \), then necessarily \( G \cong \mathbb{Z}/p\mathbb{Z} \) . | Indeed, let \( g \in G \) be any element other than the identity; then \( \langle g\rangle \) is a subgroup of \( G \), of order \( > 1 \) . By Lagrange’s theorem, \( \left| {\langle g\rangle }\right| = p = \left| G\right| \) ; that is, \( G \cong \langle g\rangle \) is cyclic of order \( p \), as claimed. | Yes |
Example 8.17 (Fermat’s little theorem). Let \( p \) be a prime integer, and let \( a \) be any integer. Then \( {a}^{p} \equiv a{\;\operatorname{mod}\;p} \) . | Indeed, this is immediate if \( a \) is a multiple of \( p \) ; if \( a \) is not a multiple of \( p \), then the class \( {\left\lbrack a\right\rbrack }_{p} \) modulo \( p \) is nonzero, so it is an element of the group \( {\left( \mathbb{Z}/p\mathbb{Z}\right) }^{ * } \), which has order \( p - 1 \) . Thus\n\n\[{\left... | Yes |
Proposition 8.18. Let \( \varphi : G \rightarrow {G}^{\prime } \) be a homomorphism of abelian groups. The following are equivalent:\n\n(a) \( \varphi \) is an epimorphism;\n\n(b) \( \operatorname{coker}\varphi \) is trivial;\n\n(c) \( \varphi : G \rightarrow {G}^{\prime } \) is surjective (as a set-function). | Proof. (a) \( \Rightarrow \) (b): Assume (a) holds, and consider the two parallel compositions\n\n\[ G\overset{\varphi }{ \rightarrow }{G}^{\prime }\xrightarrow[e]{\pi }\operatorname{coker}\varphi \]\n\nwhere \( \pi \) is the canonical projection and \( e \) is the trivial map. Both \( \pi \circ \varphi \) and \( e \ci... | Yes |
Every group \( G \) acts in a natural way on the underlying set \( G \) . The function \( \rho : G \times G \rightarrow G \) is simply the operation in the group: | \[ \left( {\forall g, a \in G}\right) : \;\rho \left( {g, a}\right) = {ga}. \] | Yes |
Theorem 9.5 (Cayley's theorem). Every group acts faithfully on some set. That is, every group may be realized as a subgroup of a permutation group. | Proof. Indeed, simply observe that the left-multiplication action of \( G \) on itself is manifestly faithful. | Yes |
Proposition 9.9. Every transitive left-action of \( G \) on a nonempty set \( A \) is isomorphic to the left-multiplication of \( G \) on \( G/H \), for \( H = \) the stabilizer of any \( a \in A \) . | Proof. Let \( G \) act transitively on a set \( A \), let \( a \in A \) be any element, and let \( H = {\operatorname{Stab}}_{G}\left( a\right) \) . I claim that there is an equivariant bijection\n\n\[ \varphi : G/H \rightarrow A \]\n\ndefined by\n\n\[ \varphi \left( {gH}\right) \mathrel{\text{:=}} {ga} \]\n\nfor all \... | Yes |
Corollary 9.10. If \( O \) is an orbit of the action of a finite group \( G \) on a set \( A \), then \( O \) is a finite set and\n\n\[ \left| O\right| \text{divides}\left| G\right| \text{.} \] | Proof. By Proposition 9.9 there is a bijection between \( O \) and \( G/{\operatorname{Stab}}_{G}\left( a\right) \) for any element \( a \in O \) ; thus\n\n\[ \left| O\right| \cdot \left| {{\operatorname{Stab}}_{G}\left( a\right) }\right| = \left| G\right| \]\n\nby Corollary 8.14 | Yes |
Example 9.11. There are no transitive actions of \( {S}_{3} \) on a set with 5 elements. | Indeed, 5 does not divide 6. | Yes |
Proposition 9.12. Suppose a group \( G \) acts on a set \( A \), and let \( a \in A, g \in G \) , \( b = {ga} \) . Then\n\n\[{\operatorname{Stab}}_{G}\left( b\right) = g{\operatorname{Stab}}_{G}\left( a\right) {g}^{-1}.\] | Proof. Indeed, assume \( h \in {\operatorname{Stab}}_{G}\left( a\right) \) ; then\n\n\[ \left( {{gh}{g}^{-1}}\right) \left( b\right) = {gh}\left( {{g}^{-1}g}\right) a = {gha} = {ga} = b :\]\n\nthus \( {gh}{g}^{-1} \in {\operatorname{Stab}}_{G}\left( b\right) \) . This proves the \( \supseteq \) inclusion; \( \subseteq ... | Yes |
Lemma 1.2. In a ring \( R \) ,\n\n\[ 0 \cdot r = 0 = r \cdot 0 \]\n\nfor all \( r \in R \) . | Proof. Indeed, \( 0 = 0 + 0 \) ; hence, applying distributivity,\n\n\[ r \cdot 0 = r \cdot \left( {0 + 0}\right) = r \cdot 0 + r \cdot 0, \]\n\nfrom which \( r \cdot 0 = 0 \) by cancellation (in the group \( \left( {R, + }\right) \) ). The equality \( 0 \cdot r = 0 \) is proven similarly. | Yes |
Example 1.4. More interesting examples are the number-based groups such as \( \mathbb{Z} \) or \( \mathbb{R} \), with the usual operations. These are very well known to our reader, who will realize immediately that they satisfy the requirements given in Definition 1.1 but they are very special. Why? | To begin with, note that multiplication is commutative in these examples; this is not among the requirements we have posed on rings in the official definition given above. | No |
Proposition 1.9. In a ring \( R, a \in R \) is not a left- (resp., right-) zero-divisor if and only if left (resp., right) multiplication by a is an injective function \( R \rightarrow R \) . | Proof. Let's verify the 'left' statement (the 'right' statement is of course entirely analogous). Assume \( a \) is not a left-zero-divisor and \( {ab} = {ac} \) for \( b, c \in R \) . Then, by distributivity,\n\n\[ a\left( {b - c}\right) = {ab} - {ac} = 0, \]\n\nand this implies \( b - c = 0 \) since \( a \) is not a ... | Yes |
Proposition 1.12. In a ring \( R \):\n\n- \( u \) is a left- (resp., right-) unit if and only if left- (resp., right-) multiplication by \( u \) is a surjective functions \( R \rightarrow R \) ;\n\n- if \( u \) is a left- (resp., right-) unit, then right- (resp., left-) multiplication by \( u \) is injective; that is, ... | Proof. These assertions are all straightforward. For example, denote by \( {\rho }_{u} : R \rightarrow \) \( R \) right-multiplication by \( u \), so that \( {\rho }_{u}\left( r\right) = {ru} \) . If \( u \) is a right-unit, let \( v \in R \) be such that \( {vu} = 1 \) ; then \( \forall r \in R \)\n\n\[{\rho }_{u} \ci... | No |
Proposition 1.15. Assume \( R \) is a finite commutative ring; then \( R \) is an integral domain if and only if it is a field. | Proof. One implication holds for all rings, as pointed out above; thus we only have to verify that if \( R \) is a finite integral domain, then it is a field. This amounts to verifying that if \( a \) is a non-zero-divisor in a finite (commutative) ring \( R \), then it is a unit in \( R \) .\n\nNow, if \( a \) is a no... | Yes |
The group of units in the ring \( \mathbb{Z}/n\mathbb{Z} \) is precisely the group \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) | indeed, a class \( {\left\lbrack m\right\rbrack }_{n} \) is a unit if and only if (right-) multiplication by \( {\left\lbrack m\right\rbrack }_{n} \) is surjective (by Proposition 1.12), if and only if the map \( a \mapsto a{\left\lbrack m\right\rbrack }_{n} \) is surjective, if and only if \( {\left\lbrack m\right\rbr... | Yes |
Proposition 2.1. \( \left( {i,\mathbb{Z}\left\lbrack {{x}_{1},\cdots ,{x}_{n}}\right\rbrack }\right) \) is initial in \( {\mathcal{R}}_{A} \) . | Proof. Let \( \left( {j, R}\right) \) be an arbitrary object of \( {\mathcal{R}}_{A} \) ; we have to show that there is a unique morphism \( \left( {i,\mathbb{Z}\left\lbrack {{x}_{1},\cdots ,{x}_{n}}\right\rbrack }\right) \rightarrow \left( {j, R}\right) \), that is, there exists exactly one ring homomorphism \( \varph... | Yes |
Proposition 2.4. For a ring homomorphism \( \varphi : R \rightarrow S \), the following are equivalent:\n\n(a) \( \varphi \) is a monomorphism;\n\n(b) \( \ker \varphi = \{ 0\} \) ;\n\n(c) \( \varphi \) is injective (as a set-function). | Proof. We prove (a) \( \Rightarrow \) (b), leaving the rest to the reader. Assume \( \varphi : R \rightarrow S \) is a monomorphism and \( r \in \ker \varphi \) . Applying the extension property of Example 2.2, we obtain unique ring homomorphisms \( {\operatorname{ev}}_{r} : \mathbb{Z}\left\lbrack x\right\rbrack \right... | No |
Proposition 2.7. Let \( R \) be a ring. Then the function \( r \mapsto {\lambda }_{r} \) is an injective ring homomorphism | Proof. For any \( r \in R \) and for all \( a, b \in R \), distributivity gives\n\n\[{\lambda }_{r}\left( {a + b}\right) = r\left( {a + b}\right) = {ra} + {rb} = {\lambda }_{r}\left( a\right) + {\lambda }_{r}\left( b\right) :\]\n\nthis shows that \( {\lambda }_{r} \) is indeed an endomorphism of the group \( \left( {R,... | Yes |
Let \( \varphi : R \rightarrow S \) be any ring homomorphism. Then \( \ker \varphi \) is an ideal of \( R \) . | Indeed, we know already that \( \ker \varphi \) is a subgroup; we have to verify the absorption properties. These are an immediate consequence of Lemma 1.2 for all \( r \in R \) , all \( a \in \ker \varphi \), we have\n\n\[ \varphi \left( {ra}\right) = \varphi \left( r\right) \varphi \left( a\right) = \varphi \left( r\... | Yes |
Example 3.4. It need not be, if \( I \) is an arbitrary subgroup of \( R \) . For example, take \( \mathbb{Z} \) as a subgroup of \( \mathbb{Q} \) ; then\n\n\[ 0 + \mathbb{Z} = 1 + \mathbb{Z} \]\n\n\( \left( { = \mathbb{Z}}\right) \) as elements of the group \( \mathbb{Q}/\mathbb{Z} \), and \( \frac{1}{2} + \mathbb{Z} ... | Answer: \( I \) is the kernel of \( R \rightarrow R/I \), so necessarily \( I \) must be an ideal, as seen in Example 3.3\n\nConversely, let us assume \( I \) is an ideal of \( R \), and verify that the proposed prescription for the operation in \( R/I \) is well-defined. For this, suppose\n\n\[ {a}^{\prime } + I = {a}... | Yes |
We know that all subgroups of \( \left( {\mathbb{Z}, + }\right) \) are of the form \( n\mathbb{Z} \) for a nonnegative integer \( n \) (Proposition 116.9). | It is immediately verified that all subgroups of \( \mathbb{Z} \) are in fact ideals of the ring \( \left( {\mathbb{Z},+, \cdot }\right) \) . The quotients \( \mathbb{Z}/n\mathbb{Z} \) are of course nothing but the rings so-denoted in [1.2] (and earlier). | No |
Theorem 3.8. Let \( I \) be a two-sided ideal of a ring \( R \) . Then for every ring homomorphism \( \varphi : R \rightarrow S \) such that \( I \subseteq \ker \varphi \) there exists a unique ring homomorphism \( \widetilde{\varphi } : R/I \rightarrow S \) so that the diagram commutes. | As a reminder to the lazy reader, \( \widetilde{\varphi } \) is defined by \[ \widetilde{\varphi }\left( {r + I}\right) \mathrel{\text{:=}} \varphi \left( r\right) \] (part of) the content of the theorem is that this function is well-defined (if \( I \subseteq \) \( \ker \varphi \) ), and it is a ring homomorphism. | No |
Theorem 3.9. Every ring homomorphism \( \varphi : R \rightarrow S \) may be decomposed as follows:  where the isomorphism \( \widetilde{\varphi } \) in the middle is the homomorphism induced by \( \varphi \) (as in T... | The reader will realize that this statement requires no proof at this point: the decomposition holds at the level of groups (by Theorem II18.1) and the maps are all ring homomorphisms as observed earlier in this section. | No |
Proposition 3.11. Let \( I \) be an ideal of a ring \( R \), and let \( J \) be an ideal of \( R \) containing \( I \). Then \( J/I \) is an ideal of \( R/I \), and\n\n\[ \frac{R/I}{J/I} \cong \frac{R}{J} \] | Proof. Since \( I \subseteq J = \ker \left( {R \rightarrow R/J}\right) \), we have an induced ring homomorphism\n\n\[ \varphi : R/I \rightarrow R/J \]\n\nby Theorem 3.8 Explicitly, \( \varphi \left( {r + I}\right) = r + J;\varphi \) is manifestly surjective. Since\n\n\( \ker \varphi = \{ r + I \mid \varphi \left( {r + ... | Yes |
For example, let \( R \) be a commutative ring, and let \( a, b \in R \) ; denote by \( \bar{b} \) the class of \( b \) in \( R/\left( a\right) \) . Then\n\n\[ \left( {R/\left( a\right) }\right) /\left( \bar{b}\right) \cong R/\left( {a, b}\right) . \] | Indeed, this is a particular case of Proposition 3.11 since\n\n\[ \left( \bar{b}\right) = \frac{\left( a, b\right) }{\left( a\right) } \]\n\nas ideals of \( R/\left( a\right) \) . | Yes |
Proposition 4.4. \( \mathbb{Z} \) is a PID. | Proof. Let \( I \subseteq \mathbb{Z} \) be an ideal. Since \( I \) is a subgroup, \( I = n\mathbb{Z} \) for some \( n \in \mathbb{Z} \), by Proposition 116.9 Since \( n\mathbb{Z} = \left( n\right) \), this shows that \( I \) is principal. | Yes |
Lemma 4.5. Let \( f\left( x\right) \) be a monic polynomial, and assume\n\n\[ f\left( x\right) {q}_{1}\left( x\right) + {r}_{1}\left( x\right) = f\left( x\right) {q}_{2}\left( x\right) + {r}_{2}\left( x\right) \]\n\nwith both \( {r}_{1}\left( x\right) \) and \( {r}_{2}\left( x\right) \) polynomials of degree \( < \deg ... | Proof. Indeed, we have\n\n\[ f\left( x\right) \left( {{q}_{1}\left( x\right) - {q}_{2}\left( x\right) }\right) = {r}_{2}\left( x\right) - {r}_{1}\left( x\right) \]\n\nif \( {r}_{2}\left( x\right) \neq {r}_{1}\left( x\right) \), then \( {r}_{2}\left( x\right) - {r}_{1}\left( x\right) \) has degree \( < \deg f\left( x\ri... | Yes |
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