Answer to Question #215209 in Linear Algebra for Majhar

Question

 Let P_k = {p(x)|p(x) is a polynomial of 

degree ≤ k with real coefficients}, for k €N.

Apply the Fundamental Theorem of 

Homomorphism to prove that p₅/p₃ is isomorphic to p₁.

Answer 1

Here P_k={p(x):f(x) is polynomial of degree $\le$ k with real coefficient}

Let

f=p₅$\to$ P₁ with the function defined by f(ax⁵+bx⁴+cx³+dx²+ex+1)=ax+b

for all ax⁵+bx⁴+cx³+dx²+ex+f $\in$ P

where a,b,c,d,e,f are arbitary constants.

then it is clear that, f is homomorphism.

Because,

let p(x) (a₁x⁵+b₁x⁴+c₁x³+d₁x²+e₁x+f₁)$\in$ P₅

q(x)=(a₂x⁵+b₂x⁴+c₂x³+d₂x²+e₂x+f₂)$\in$ P₅

now,F(f(x)+q(x))=

f{(a₁+a₂)x⁵+(b₁+b₂)x⁴+(c₁+c₂)x³+(d₁+d₂)x²+(e₁+e₂)x+f₁+f₂}

=(a1+a2)x+(b1+b2)

=(a₁x+b₁)+(a₂x+b₂)

=f(p(x))+f(q(x)) $\forall$ p(x),q(x)$\in$ P₅

f is homomorphism.

Also,

kerf={P(x)+P₅:f(p(x))=0}

f(p(x))=0

a₁x+b=0 $\forall$ x$\in$ R

a₁=0,b₁=0

p(x)=c₁x³+d₁x²+e₁x+f $\in$ P₃

ker f=p₃

also,

since f is a onto mapping

by homomorphism fundamental therom

P₅\kerf=P₁

P₅/P₃=P₁