Prove that there is a unique product on R for which (R, +,-) is a ring for which the above three rules hold. Prove that is not commutative. Prove that Q, R ER such that A = QB+ R with R=0 or deg(R) < deg (B). has a right-division algorithm, i.e. for all A, B R with B 0 there exist

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One may construct the rational numbers Q from the integers Z as equivalence classes of pairs of integers
(a, b) with b 0 for the equivalence relation (a, b)~ (a₁, b₁) ab₁ = a₁b. Equivalence classes
are written as a/b and called 'fractions.
Write C for the field of complex numbers and C[x] for the polynomial ring in the variable z over C.
As above, we consider
C(x) = {f/g: f.ge C[r], g = 0}
where 'f/g' is a new symbol for the equivalence class of a pair of polynomials (f, g) with g 0 for the
equivalence relation
f/g=fi/91 (f.g)~ (f1,91) ⇒ fg₁ = fig.
Define product and sum by
f/g-h/k= (fh)/(gk) and f/g+h/k := (fk+gh)/(gk).
We consider R = C(z) [] given as expressions of the form
A = Σa₁(x)² = a (x) + ª₁(x)ə + …..ªn(x)ən
with a (x) = C(x), n € Zzo. If A 0 and an # 0, write deg (A) := n; so this is the highest power of
> that occurs in A. Define the following partial 'rules' for addition and multiplication in R:
[definition of sum] Σª¡(x)ő¹ + Σb;(x)ởª := [(a¡(x) + b¡(x)) via the usual sum in C[z];
[definition of product of 'constants', i.e. n = 0] ao(x)bo(r) is the usual product in C(z);
[rule to multiply with from the left] We let
Əa(x) = a(x) + a'(x),
where a'(x) is the derivative of the function a(z) = C(z) with respect to z.
Prove that there is a unique product on R for which (R, +,-) is a ring for which the above
three rules hold. Prove that is not commutative.
Prove that has a right-division algorithm, i.e. for all A, B R with B 0 there exist
Q, R ER such that A = QB+ R with R = 0 or deg(R) < deg(B).
Transcribed Image Text:One may construct the rational numbers Q from the integers Z as equivalence classes of pairs of integers (a, b) with b 0 for the equivalence relation (a, b)~ (a₁, b₁) ab₁ = a₁b. Equivalence classes are written as a/b and called 'fractions. Write C for the field of complex numbers and C[x] for the polynomial ring in the variable z over C. As above, we consider C(x) = {f/g: f.ge C[r], g = 0} where 'f/g' is a new symbol for the equivalence class of a pair of polynomials (f, g) with g 0 for the equivalence relation f/g=fi/91 (f.g)~ (f1,91) ⇒ fg₁ = fig. Define product and sum by f/g-h/k= (fh)/(gk) and f/g+h/k := (fk+gh)/(gk). We consider R = C(z) [] given as expressions of the form A = Σa₁(x)² = a (x) + ª₁(x)ə + …..ªn(x)ən with a (x) = C(x), n € Zzo. If A 0 and an # 0, write deg (A) := n; so this is the highest power of > that occurs in A. Define the following partial 'rules' for addition and multiplication in R: [definition of sum] Σª¡(x)ő¹ + Σb;(x)ởª := [(a¡(x) + b¡(x)) via the usual sum in C[z]; [definition of product of 'constants', i.e. n = 0] ao(x)bo(r) is the usual product in C(z); [rule to multiply with from the left] We let Əa(x) = a(x) + a'(x), where a'(x) is the derivative of the function a(z) = C(z) with respect to z. Prove that there is a unique product on R for which (R, +,-) is a ring for which the above three rules hold. Prove that is not commutative. Prove that has a right-division algorithm, i.e. for all A, B R with B 0 there exist Q, R ER such that A = QB+ R with R = 0 or deg(R) < deg(B).
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