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Math Case Study

Decent Essays

Problem 4. Prove that (y^3+z^3 ) x^2+yz^4 is irreducible over C[x,y,z]. Also prove that (y^3+z^3 ) x^2+y^2 z^3 is irreducible. Assume that (y^3+z^3 ) x^2+yz^4=a*b. Then one of a or b is linear in x^2 and the other doesn’t have x^2 at all because the degree of the product is the sum of the two degrees. Now we write 〖a=cx〗^2+d, so c and d have only y’s and z’s. Then (y^3+z^3 ) x^2+yz^4=(〖cx〗^2+d)*b But now b*d=yz^4, and since C[y,z] is a unique factorization domain, b and d must be monomials. But this means b is a monomial, in y and z. Since b*c=y^3+z^3 and we see that y^3+z^3 factors to be (y+z)(y^2-yz+z^2). Since b is a monomial and b*c=(y+z)(y^2-yz+z^2) Then b must be either (y+z) or (y^2-yz+z^2). However, they must not have …show more content…

Guess: Res(x-a,f(x))= b_n a^n+⋯+b_1 a+b_0. Lemma: det[■(0&-a&0@⋮&⋮&⋮@■(0@b_n )&■(0@b_(n-2) )&■(0@b_(n-3) )) ■(…&0&0@…&⋮&⋮@■(…@…)&■(1@b_1 )&■(-a@b_0 ))]=b_n a^(n-1) Base Case: n = 2 det[■(0&-a@b_2&b_0 )]=0(-b_o )-b_2 (-a)= b_2 a n = 3 det[■(0&-a&0@0&1&-a@b_3&b_1&b_0 )]=0-(-a)(0(b_0 )-b_3 (-a))=b_3 a^2 Induction: Assume the determinant is b_n a^(n-1) is true for n≥3. Now look at n+1. det[■(0&-a&0@⋮&⋮&⋮@■(0@b_(n+1) )&■(0@b_(n-2) )&■(0@b_(n-3) )) ■(…&0&0@…&⋮&⋮@■(…@…)&■(1@b_1 )&■(-a@b_0 ))]=0-(-a)det[■(0&-a&0@⋮&⋮&⋮@■(0@b_n )&■(0@b_(n-2) )&■(0@b_(n-3) )) ■(…&0&0@…&⋮&⋮@■(…@…)&■(1@b_1 )&■(-a@b_0 ))] =a(b_(n+1) 〖(a〗^(n-1)))=b_(n+1) a^n Proof: Base Case is n = 2: det[■(1&-a&0@0&1&-a@b_2&b_1&b_0 )]=1*(1*(b_0 )-b_1*(-a))-(-a)*(0*(b_0 )-b_2 (-a)) =b_0+b_1 a+b_2 a^2 Induction: Assume for n≥2, det[■(1&-a&0@■(0@⋮)&■(1@⋮)&■(-a@⋮)@■(0@b_n )&■(0@b_(n-1) )&■(0@b_(n-2) )) ■(…&0&0@■(…@…)&■(0@⋮)&■(0@⋮)@■(…@…)&■(1@b_1 )&■(-a@b_0 ))]=b_n a^n+⋯+b_1 a+b_0 Now look at n+1. det[■(1&-a&0@■(0@⋮)&■(1@⋮)&■(-a@⋮)@■(0@b_(n+1) )&■(0@b_n )&■(0@b_(n-1) )) ■(…&0&0@■(…@…)&■(0@⋮)&■(0@⋮)@■(…@…)&■(1@b_1 )&■(-a@b_0 ))] =1*det[■(1&-a&0@■(0@⋮)&■(1@⋮)&■(-a@⋮)@■(0@b_n )&■(0@b_(n-1) )&■(0@b_(n-2) )) ■(…&0&0@■(…@…)&■(0@⋮)&■(0@⋮)@■(…@…)&■(1@b_1 )&■(-a@b_0 ))]-(-a)det[■(0&-a&0@■(0@⋮)&■(1@⋮)&■(-a@⋮)@■(0@b_(n+1) )&■(0@b_(n-1) )&■(0@b_(n-2) )) ■(…&0&0@■(…@…)&■(0@⋮)&■(0@⋮)@■(…@…)&■(1@b_1 )&■(-a@b_0 ))] 〖=1*(b〗_n a^n+〖=b〗_(n-1) a^(n-1)…+b_1 a+b_0+a(b_(n+1) a^n ) By

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