Answers to the 2.2 "polynomial ring" exercise of Meng daoyu's mathematical Basics

1. Set $ r$ to an integral ring of the exchange, $ R [x] $ to a polynomial ring on $ r$, $ f, g \ in R [x] $. proof:

$ {\ RM deg} f \ cdot g ={\ RM deg} f + {\ RM deg} G $

Is the above formula true for the general switch ring?

Proof: Set $ F = A _ {0} + A _ {1} X + \ cdots + A _ {n} x ^ N, g (x) = B _ {0} + B _ {1} X + \ cdots + B _ {m} x ^ m $, where $ A _ {n }, B _ {m} \ neq0 $. because $ r$ has no zero factor, $ A _ {n} B _ {m} \ neq0 $, obviously $ {\ RM deg} f \ cdot G = m + n = {\ RM deg} f + {\ RM deg} G. $

The general round-ring form is not necessarily true. For example, in $ \ mathbb Z _ {6} [x] $, consider $ f (x) = 2x ^ 4, g (x) = 3x $

Apparently $ {\ RM deg} f (x) g (x) = 0 \ NEQ 4 + 1 $.

 

2. set $ r$ to the entire ring of the exchange, $ F $ to the formula field of $ r$, and $ f [x] $ to the one-dimensional polynomial ring on $ F $. it is proved that $ R [x] $ is a $ r$ polynomial ring and $ R [x] $ has the same shard field as $ f [x] $.

It proves that because $ r \ subset F $, and $ x $ is the superobject on $ F $, it is also the superobject on $ r$, therefore, $ R [x] $ forms a polynomial ring on $ r$. set $ R [x] $ to $ A $, and $ f [x] $ to $ B $. Obviously, $ A \ subset B $. verify the other half, and choose $ \ frac {f} {g} \ in B $, $ F = A _ {0} + A _ {1} X + \ cdots + A _ {n} x ^ n \ In F [x] $, and each $ A _ {I }=\ frac {P _ {I }}{ Q _ {I }}, (P _ {I} \ In R, Q _ {I} \ In R ^ *) $

Make $ r = \ prod _ {I = 0} ^ {n} Q _ {I} \ In R ^ * $, then $ f =\frac {1} {r} \ left (A _ {0} '+ A _ {1} 'x + \ cdots + A _ {n }' x ^ n \ right ), (A _ {I} '\ In R) $

Similarly, $ G = \ frac {1} {s} \ left (B _ {0} '+ B _ {1} 'x + \ cdots + B _ {n}' x ^ n \ right ), (s \ In R ^ *, B _ {I} '\ In R) $

Therefore, $ \ frac {f} {g} =\frac {s \ left (A _ {0} '+ A _ {1} 'x + \ cdots + _{ n} 'x ^ n \ right )} {r \ left (B _ {0} '+ B _ {1} 'x + \ cdots + B _ {n}' x ^ n \ right )} \ in a $

Therefore, $ B \ subset A $. Thus $ A = B $.

 

3. set $ \ mathbb Q $ to a rational number field. proof that $ \ Omega =-\ frac {1} {2} + \ frac {\ SQRT {-3} {2} $ is the algebraic element on $ \ mathbb Q $ and $ $ \ mathbb Q [\ Omega] \ simeq \ mathbb Q [x]/<x ^ 2 + x + 1>. $

It proves that $ \ Omega ^ 3-1 = 0 $, so $ \ Omega $ is the algebraic element on $ \ mathbb Q $. ing \ begin {Align *} \ Phi: \ mathbb Q [x] & \ To \ mathbb Q [\ Omega] \ f (x) & \ mapsto F (\ Omega) \ end {Align *}

It is not difficult to verify that $ \ Phi $ is full homomorphic. let's look at its homomorphic core $ {\ RM Ker} \ Phi $, set $ F (\ Omega) = 0, f \ In \ mathbb Q [x] $, note $ \ Omega ^ 2 + \ Omega + 1 = 0 $

Obviously, $ G (\ Omega) \ neq0 $ has a rational polynomial of no more than $1 $. according to the knowledge of Higher Algebra, it is clear that $ x ^ 2 + x + 1 \ big | f (x) $, therefore, the Yizhi homomorphic kernel is the primary ideal $ <x ^ 2 + x + 1> $, therefore, according to the basic theorem of ring homomorphic, we can know $ \ mathbb Q [\ Omega] \ simeq \ mathbb Q [x]/<x ^ 2 + x + 1>. $

 

4. prove that $ u = \ sqrt2 + \ sqrt3 $ is an algebraic element on $ \ mathbb Q $, and find the ideal $ \ mathbb Q [x] $ I $ to make $ \ mathbb Q [x]/I \ simeq \ mathbb Q [u]. $

Note that $ U ^ 4-10u ^ 2 + 1 = 0 $, and $ U $ is the algebraic element on $ \ mathbb Q $. ing \ begin {Align *} \ Phi: \ mathbb Q [x] & \ To \ mathbb Q [u] \ f (x) & \ mapsto f (u) \ end {Align *}

Obviously, $ \ Phi $ is a full homomorphic shot. considering its homomorphic kernel, similar to the above question, we can see that the homomorphic kernel $ {\ RM Ker} \ Phi $ is the primary ideal $ <x ^ 4-10x ^ 2 + 1 >$, therefore, $ I = <x ^ 4-10x ^ 2 + 1> $, according to the basic theorem of ring homomorphic, $ \ mathbb Q [x]/I \ simeq \ mathbb Q [u]. $

 

5. set $ I $ to the ideal of switching the ring $ r$, so that $ I [X _ {1}, \ cdots, X _ {n}] $ is a set of polynomials in $ I $ for $ R [X _ {1}, \ cdots, X _ {n}] $. proof:

(1) $ I [X _ {1}, \ cdots, X _ {n}] $ is $ R [X _ {1}, \ cdots, X _ {n}] $ is ideal;

(2) $ R [X _ {1}, \ cdots, X _ {n}]/I [X _ {1}, \ cdots, X _ {n}] \ simeq (R/I) [Y _ {1}, \ cdots, Y _ {n}] $, where $ Y _ {1 }, \ cdots, Y _ {n} $ has nothing to do with algebra on $ R/I $.

Prove (1) any $ f, g \ In I [X _ {1}, \ cdots, X _ {n}] $, because $ I $ is ideal, obviously $ f-g \ In I $, and for any $ H \ in R [X _ {1}, \ cdots, X _ {n}] $, if \ begin {Align *} f & =\ sum _ {K _ {1}, \ cdots, K _ {n} A _ {K _ {1} \ cdots K _ {n} X _ {1} ^ {K _ {1} \ cdots X _{ n} ^ {K _ {n }}, A _ {K _ {1} \ cdots K _ {n} \ In I} \ H & =\ sum _ {L _ {1}, \ cdots, L _ {n} B _ {L _ {1} \ cdots L _ {n} X _ {1} ^ {L _ {1} \ cdots X _{ n} ^ {L _ {n }}, A _ {L _ {1} \ cdots L _ {n} \ In R} \ end {Align *}

The coefficients of $ HF $ are in the form of $ \ sum a _ {I _ {1} \ cdots I _ {n} B _ {J _ {1} \ cdots J _ {n }}\ in I $

Therefore, $ I [X _ {1}, \ cdots, X _ {n}] $ is $ R [X _ {1}, \ cdots, X _ {n}] $.

(2) ing \ begin {Align *} \ Phi: R [X _ {1}, \ cdots, X _ {n}] & \ To (R/I) [Y _ {1}, \ cdots, Y _ {n}] \ sum _ {K _ {1}, \ cdots, K _ {n} A _ {K _ {1} \ cdots K _ {n} X _ {1} ^ {K _ {1} \ cdots X _{ n} ^ {K _ {n }}& \ mapsto \ sum _ {K _ {1 }, \ cdots, K _ {n }}\ overline {A _ {K _ {1} \ cdots K _ {n} y _ {1} ^ {K _ {1 }}\ cdots Y _ {n} ^ {K _ {n} \ end {Align *}

It is easy to verify that $ \ Phi $ is a full homomorphic kernel and $ {\ RM Ker} \ Phi = I [X _ {1}, \ cdots, X _ {n}] $

According to the Basic homomorphic theorem, we know $ R [X _ {1}, \ cdots, X _ {n}]/I [X _ {1}, \ cdots, X _ {n}] \ simeq (R/I) [Y _ {1}, \ cdots, Y _ {n}]. $

 

6. set $ r$ to a ring, so $ R [[x] =\{ (A _ {0}, A _ {1}, \ cdots) = \ sum _ {n = 0} ^ {\ infty} A _ {n} x ^ n | A _ {n} \ In r \} $

And define addition and multiplication in $ R [[x] $: \ begin {Align *} \ sum _ {n = 0} ^ {\ infty} A _ {n} x ^ N + \ sum _ {n = 0} ^ {\ infty} B _ {n} x ^ N & = \ sum _ {n = 0} ^ {\ infty} (A _ {n} + B _ {n }) x ^ n \ sum _ {n = 0} ^ {\ infty} A _ {n} x ^ n \ cdot \ sum _ {n = 0} ^ {\ infty} B _ {n} x ^ N & = \ sum _ {n = 0} ^ {\ infty} \ left (\ sum _ {I + J = n} _{ i} B _ {J} \ right) x ^ n \ end {Align *}

Prove that $ R [[x] $ is a ring (called the formal idempotent ring on $ r$ ).

It turns out that $ \ {R [[x]; + \} $ constitutes an Abel group, and multiplication also has an obvious distribution law for addition, let's verify the combination law of multiplication \ begin {Align *} \ left (\ sum _ {n = 0} ^ {\ infty} A _ {n} x ^ n \ cdot \ sum _ {n = 0} ^ {\ infty} B _ {n} x ^ n \ right) \ sum _ {n = 0} ^ {\ infty} C _ {n} x ^ N & = \ sum _ {n = 0} ^ {\ infty} \ left (\ sum _ {I + J = n} A _ {I} B _ {J} \ right) x ^ n \ cdot \ sum _ {n = 0} ^ {\ infty} C _ {n} x ^ n \ & = \ sum _ {n = 0} ^ {\ infty} \ left (\ sum _ {I + J = n} \ sum _ {p + q = I} A _ {p} B _ {q} \ right) C _ {J} x ^ n \ & = \ sum _ {n = 0} ^ {\ infty} \ left (\ sum _ {p + q = n} _ {p} \ sum _ {I + J = Q} B _ {I} C _ {J} \ right) x ^ n \ & = \ sum _ {n = 0} ^ {\ infty} A _ {n} x ^ n \ left (\ sum _ {n = 0} ^ {\ infty} B _ {n} x ^ n \ cdot \ sum _ {n = 0} ^ {\ infty} C _ {n} x ^ n \ right) \ end {Align *}

Therefore, $ \ {R [[x]; \ cdot \} $ forms a semi-group. Thus, $ R [[x] $ is a ring.

 

7. set $ M $ to a semi-group, and $ r$ to an exchange ring. order $ R [m] =\{ f | F: M \ to R, | M \ setminus f ^ {-1} (0) | <\ infty \} $

In $ R [m] $, define addition and multiplication: \ begin {Align *} (F + G) (m) & = f (m) + g (m) \ (f \ cdot g) (m) & =\ sum _ {QP = m} f (p) g (q) \ end {Align *}

It is proved that $ R [m] $ is a ring (called the semi-oId ring or semi-oId algebra on $ M $ r$ ).

The proof first describes the closeness of addition and multiplication. The meaning of this question is equivalent to the value of $ f \ in R [m] $, and the value of $ F $ is not zero, obviously $ F + G \ in R [m], f \ cdot g \ in R [m] $

Easy verification $ \ {R [m]; + \} $ constitutes an Abel group. The value of 0 is $0 $. next, consider $ \ {R [m]; \ cdot \} $. The combination of laws is easy to verify. so $ R [m] $ is a ring.

 

8. set $ r$ as the swarm ring, and $ M $ as a semi-group composed of non-negative integer pairs. it is proved that the semi-Group Ring $ R [m] $ on $ r$ is homogeneous with the one-dimensional polynomial ring $ R [x] $ on $ r$.

Enable $ f \ in R [m] $ to make $ g _ {f} (X) = \ sum _ {I = 0} ^ {\ infty} f (I) x ^ I $. map \ begin {Align *} \ Phi: R [m] & \ To R [x] \ f & \ mapsto g _ {f} \ end {Align *}

Yi Zheng $ \ Phi $ is the homogeneous ring. Therefore, $ R [m] \ simeq R [X]. $