Answers to the exercises of Meng Daoyi's mathematical basics 2.1 "subfield"

1. What is a shard of $ \ mathbb F $?

As the formula field of $ \ mathbb F $ contains the smallest domain, and $ \ mathbb F $ itself is a domain, the formula field of $ \ mathbb F $ is itself.

 

2. It is proved that gsuss Integer Ring $ \ mathbb Z [\ SQRT {-1}] $ is the entire ring exchanged and Its shard field?

Verify that because $ \ mathbb Z [\ SQRT {-1}] =\{ a + B \ SQRT {-1} |, B \ In \ mathbb Z \} \ subset \ mathbb C $, it is easy to verify that it meets the two conditions of the child ring, therefore, $ \ mathbb Z [\ SQRT {-1}] $ is the child ring of $ \ mathbb C $, and it is easy to know. it can be used as a shard.

Set the format of $ \ mathbb Z [\ SQRT {-1}] $ to $ \ mathbb F $, certificate $ \ mathbb F =\mathbb Q [\ SQRT {-1}] $, apparently $ \ mathbb Q [\ SQRT {-1}] \ subset \ mathbb F $, ing

\ Begin {Align *} \ Phi: \ mathbb F & \ To \ mathbb C \ frac {M _ {1} + N _ {1} \ SQRT {-1 }}{ M _ {2} + n _ {2} \ SQRT {-1 }}& \ mapsto \ frac {M _ {1} M _ {2} + N _ {1} n _ {2} + (M _ {2} n _ {1}-M _ {1} n _ {2 }) \ SQRT {-1 }}{ M _ {2} ^ {2} + N _ {2} ^ {2 }}\ end {Align *}

(The elements in $ \ mathbb F $ are equivalence classes and cannot be directly computed.) that is, $ \ mathbb f \ subset \ mathbb Q [\ SQRT {-1}] $. therefore, $ F = \ mathbb Q [\ SQRT {-1}] $. (Of course, in the homogeneous sense)

 

3. It is proved that $ \ mathbb Z [\ SQRT {2}] $ is an integral ring, and its shard field is required.

It proves similar to the above question that $ \ mathbb Z [\ SQRT {2}] $ is an exchange integral ring, and its shard field is $ \ mathbb Q [\ sqrt2] $.

 

4. set $ d $ to switch the entire ring, $ m, n \ In \ mathbb N, (m, n) = 1 $. proof: Set $ a, B \ In d $ to meet $ A ^ m = B ^ m, the required and sufficient conditions for a ^ n = B ^ N $ are $ A = B $.

Proving adequacy is obvious. Re-evidence is necessary. $ U, V \ In \ mathbb Z $ makes $ UM + VN = 1 $

Set the format of $ d $ to $ \ mathbb F $, and $ d \ subset \ mathbb F $ to make any element of $ d $ reversible in $ \ mathbb F $, therefore, in $ \ mathbb F $

\ Begin {Align *} A = a ^ 1 & = a ^ {UM + vn} \ & = \ left (a ^ m \ right) ^ U \ left (a ^ n \ right) ^ V \ & = B ^ {UM + vn} \ tag {1} \ & = B \ end {Align *}

Note: (1) the formula uses reversible conditions. If you only consider the ring $ d $, then (1) is not true.

 

5. It is proved that an exchange semi-group can be embedded into an Abel group if it satisfies the elimination law. Is this proposition true for an exchange semi-group meeting the elimination law?

It is proved that $ G $ is an interchanging monoid (the operation is recorded as multiplication), and a forward product $ g \ times g =\{ (a, B) |, B \ in G \} $

And define multiplication $ (A, B) (c, d) = (AC, BD) $

Define the link $ "\ SIM" $, set $ (a, B) \ SIM (c, d) \ leftrightarrow A = C, B = d $

It is easy to verify that $ "\ SIM" $ is an equivalent relationship, and the multiplication in $ g \ times G $ is a coequal relationship. therefore, it can be used as a collection of suppliers $ (g \ times G)/\ SIM $, which is easy to verify as an Abel group. fixed $ A \ in G $. Consider its subset $ H =\{ (GA, A) | A, G \ in G \} $, define ing \ begin {Align *} \ Phi: G & \ to h \ G & \ mapsto (G, e) \ end {Align *}

Yi Zhi $ \ Phi $ is homogeneous, so that $ G $ is embedded into an Abel group.

The proof process only applies to the elimination law. Therefore, the conclusion still applies to the exchange semi-groups that satisfy the elimination law.

 

6. set $ d $ to the integral ring. If $ \ forall a, B \ In d ^ * $, $ A _ {1} exists }, B _ {1} \ In d ^ * $ to make $ AB _ {1} = Ba _ {1} $, it is said that $ d $ meets the common property, and $ M = AB _ {1} = Ba _ {1} $ is $ A, B $ is a right public factor. set $ d $ to the entire ring that satisfies the right common times. define the link $ "\ SIM" $: $ (a, B) \ SIM (c, d) in $ d \ times d ^ * $) \ leftrightarrow $ if $ D _ {1}, B _ {1} \ In d ^ * $ makes $ db _ {1} = BD _ {1} $, $ ad _ {1} = CB _ {1} $. test the following propositions:

(1) The relationship $ "\ SIM" $ is equivalent;

(2) Equivalent Class with $ \ frac {A} {B} $ as $ (a, B) $, if the operator set $ F = d \ times d ^ */\ SIM $ defines addition and multiplication as follows: \ begin {Align *} \ frac {A} {B} + \ frac {c} {d} & =\ frac {ad _ {1} + CB _ {1 }} {m }, (M = BD _ {1} = dB _ {1 }) \\\ frac {A} {B} \ cdot \ frac {c} {d} & =\ frac {ac _ {2 }}{ dB _ {2 }}, (B _ {2} \ In d ^ *, C _ {2} \ In D, CB _ {2} = BC _ {2}) \ end {Align *}

Then $ F $ is an individual;

(3) $ d $ is the same as $ d $;

(4) $ \ forall x \ In F $, $ a, B \ In d' $ makes $ x = AB ^ {-1} $.

Proof (This question is too difficult !!!! But the question is really good.) (1) first, we will prove that the definition of $ \ SIM $ is non-conflicting. That is, it is irrelevant to the selection of $ D _ {1} and B _ {1} $.

Set $ BD _ {1} = dB _ {1} \ rightarrow ad _ {1} = CB _ {1} $. if $ BD _ {2} = dB _ {2} $, Let's explain $ ad _ {2} = CD _ {2 }. $

Starting from the right public factor, $ U, V \ In d ^ * $ causes $ D _ {1} u = D _ {2} S $, thus \ begin {Align *} BD _ {2} v = BD _ {1} U & = dB _ {1} u = dB _ {2} V \ rightarrow B _ {1} U & = B _ {2} V \ end {Align *}

In addition, \ begin {Align *} ad _ {2} V & = AD _ {1} u = CD _ {1} u = CB _ {2} V \ rightarrow ad _ {2} & = CB _ {2} \ end {Align *}

Therefore, $ \ SIM $ is indeed a link. It is easy to verify that $ \ SIM $ is equivalent to the opposite body, symmetry, and transmission.

(2) First, we must prove that the rule defined in the question is indeed an operation. Taking addition as an example, we need to prove two points:

1 ). the Operation has nothing to do with the selection of $ M $ (this is actually proved by the previous question); 2) the operation has nothing to do with the selection of the representative yuan of the equivalence class $ \ frac {A} {B} $.

Article 2nd): Set $ \ frac {a'} {B '}=\ frac {A} {B} $ to pass the certificate

$ \ Frac {A} {B} + \ frac {c} {d} =\frac {a'} {B '} + \ frac {c} {d} $ $

$ \ Frac {A} {B} + \ frac {c} {d }=\ frac {ad _ {1} + CB _ {1 }}{ m }, M = BD _ {1} = dB _ {1 }$, $ D _ {2}, B _ {2}, U, V \ In d ^ * $ makes $ B 'd _ {2} = dB _ {2 }, AD _ {1} u = A 'd _ {2} V $

Based on $ \ frac {A} {B }=\ frac {a'} {B '}$, that is, $ (a, B) \ SIM (A', B ') $ know $ BD _ {1} u = B 'D _ {2} V $ SO \ begin {Align *} dB _ {1} u = BD _ {1} U & = B 'D _ {2} v = dB _ {2} V \ rightarrow B _ {1} U & = B _ {2} V \ end {align *}

Thus \ begin {Align *} (AD _ {1} + CB _ {1 }) U & = AD _ {1} U + CB _ {1} U \ & = A 'd _ {2} V + CB _ {2} V \ & = (A 'd _ {2} + CB _ {2 }) V \ end {Align *}

Add $ BD _ {1} u = B 'D _ {2} V $ to get \ begin {Align *} \ frac {ad _ {1} + CB _ {1}} {BD _ {1 }}&=\ frac {A 'd _ {2} + CB _ {2 }}{ B 'D _ {2 }}\\ rightarrow \ frac {A} {B} + \ frac {c} {d} & =\ frac {a'} {B '} + \ frac {c} {d} \ end {Align *}

Therefore, the addition definition has nothing to do with the selection of representative elements.

Similar methods can be used to authenticate multiplication. It is easy to verify that $ F = d \ times d ^ */\ SIM $ forms the body.

(3) Perform a subset of $ F $ d' =\{\ frac {A} {1} | A \ In d \} $

It is easy to verify that $ d' $ is the child ring of $ F $ and map \ begin {Align *} \ Phi: D & \ To d' \ A & \ mapsto \ frac {A} {1} \ end {Align *}

It is also easy to verify that $ \ Phi $ is homogeneous.

(4) In the domain $ F $, we have $ \ frac {A} {B} =\ frac {A} {1} \ cdot \ frac {1} {B} (B \ neq0) $

And $ \ frac {1} {B} \ cdot \ frac {B} {1 }=\ frac {1} {1 }$, that is, $ \ frac {1} {B }=\ left (\ frac {B} {1} \ right) ^ {-1} $

Therefore, $ \ forall x =\frac {q} {p} (P \ neq0) \ In F $, $ A =\frac {q} {1 }, B = \ frac {p} {1} \ In d' $ make $ x = AB ^ {-1 }. $

 

7. set $ r$ to an exchange ring, and $ S $ to a multiplication suboid of $ r$, and any element in $ S $ is not a zero factor. in $ r \ times S $, we can define the same join relationship as in Theorem 2.1.1. the business set name is $ Rs ^ {-1} $. trial certificate:

(1) $ Rs ^ {-1} $ is an exchange ring;

(2) $ r$ can be embedded in $ Rs ^ {-1} $;

(3) $ \ forall A \ in S \ subset Rs ^ {-1 }$, $ A $ is reversible.

Proof (1) it is easy to verify that $ \ {Rs ^ {-1}; + \} $ is an Abel Group and $ \ {Rs ^ {-1 }; \ cdot \} $ is an interchanging monoid and an allocation law is established. therefore, $ Rs ^ {-1} $ is an exchange ring.

(2) Take the subset of $ Rs ^ {-1} $ \ overline {r }=\{\ frac {AB} {B} | A \ In R, B \ In s ^ * \} $

E-certificate $ r \ simeq \ overline {r} $. $ r$ is embedded into $ Rs ^ {-1} $.

(3) The question should be $ A \ In s ^ * $, (2) the homogeneous $ \ Phi $ in the question maps it to $ \ frac {AB} {B} $, because $ S $ has no zero factor, therefore, $ AB \ NEQ $ contains the inverse $ \ frac {B} {AB} $ in $ Rs ^ {-1} $, because $ \ frac {AB} {B} \ cdot \ frac {B} {AB }=\ frac {AB ^ 2} {AB ^ 2 }=\ frac {B} {B} $

$ Rs ^ {-1} $.

 

8. Make $ R =\mathbb Z _ {4}, s =\{ 1, 3 \} $. find $ Rs ^ {-1} $.

Obviously, $ \ frac {0} {1}, \ frac {1} {1}, \ frac {1} {1}, \ frac {2} {1 }, \ frac {3} {1}, \ frac {1} {3}, \ frac {2} {3} $

In $ \ mathbb Z _ {4} $, \ begin {Align *} \ frac {2} {1} & =\ frac {6} {3} =\ frac {2} {3} \ frac {3} {1} & =\ frac {9} {3 }=\ frac {1} {3} \ end {Align *}

Therefore, $ Rs ^ {-1 }=\{\ frac {0} {1}, \ frac {1} {1}, \ frac {2} {1 }, \ frac {3} {1 }\}$ $

Apparently $ Rs ^ {-1} \ simeq \ mathbb Z _ {4} $.

 

9. Make $ r = \ mathbb Z, s =\{ 2 ^ n | n \ In \ mathbb n \}$. find $ Rs ^ {-1} $.

Obviously, $ Rs ^ {-1 }=\{\ frac {m} {2 ^ n} | (m, 2) = 1, m \ In \ mathbb Z, n \ In \ mathbb n \} $

 

10. make $ r = 3 \ mathbb Z, s =\{ 6 ^ n | n \ In \ mathbb n \} $, prove the sub-rings $ \ {\ frac {m} {6 ^ n}, m \ In \ mathbb Z, between $ Rs ^ {-1} $ and $ \ mathbb r$, n \ In \ mathbb Z ^ + \} $ homogeneous.

Verify that the sub-ring above is $ R' $, and the corresponding link \ begin {Align *} \ Phi: RS ^ {-1} & \ To R' \\\ frac {3 m} {6 ^ n} & \ mapsto \ frac {3 m} {6 ^ n} \ end {Align *}

$ \ Phi $ is a ing, if $ \ frac {3 m} {6 ^ n} =\ frac {3 p} {6 ^ Q} $ (here is the equivalence class ), apparently $ \ frac {m} {6 ^ n} =\ frac {p} {6 ^ Q} $

Here is the equality in $ \ mathbb r$. therefore, $ \ Phi $ is indeed a ing. easy to verify, its ticket, full. and \ begin {Align *} \ Phi \ left (\ frac {3 m} {6 ^ n} + \ frac {3 p} {6 ^ Q} \ right) & =\ Phi \ left (\ frac {3 \ left (m6 ^ q + P6 ^ n \ right)} {6 ^ {n + q }}\ right) \\& =\ frac {3m6 ^ q + 3p6 ^ n} {6 ^ {n + q }}\\& =\ Phi \ left (\ frac {3 m }{ 6 ^ n} \ right) + \ Phi \ left (\ frac {3 p} {6 ^ Q} \ right) \ Phi \ left (\ frac {3 m} {6 ^ n} \ cdot \ frac {3 p} {6 ^ Q} \ right) & =\ Phi \ left (\ frac {3 m} {6 ^ n} \ right) \ cdot \ Phi \ left (\ frac {3 p} {6 ^ Q} \ right) \ end {Align *}

Therefore, $ Rs ^ {-1} \ simeq R' $.

 

11. set $ r$ to an exchange ring, and $ R _ {1} $ to a set of non-zero factors $ r$. if another swap ring $ k \ supset r$ and $ \ forall A \ In R _ {1} $, $ A $ has a reverse element in $ K $, it proves that $ RR _ {1} ^ {-1} $ must be homogeneous with the sub-ring in $ K $.

Prove that the subset of $ K _ {1 }=\{ AB ^ {-1} | A \ In R, B \ In R _ {1 }\}$ $

It is easy to verify that $ K _ {1} $ is the child ring of $ K $, and the following can be proved to correspond to \ begin {Align *} \ Phi: rr _ {1} ^ {-1} & \ to K _ {1} \ frac {A} {B} & \ mapsto AB ^ {-1} \ end {align *}

Is homogeneous.