Question

Let \( \mathbb{Z}[x]\) be the set of all polynomials p(x) in the variable x with with integer coefficients. Show that the function \( g: \mathbb{Z}[x] \to \mathbb{Z} \) defined by \(g(p(x))=p(0)\) is a ring homomorphism

Answer 1

hmmmm so would i show this as follows:
\(g(p(x)+q(x))=(a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2+...+(a_n+b_n)x^n\) \(=(a_0+a_1x+a_2x^2+...+a_nx^n)+(b_0+b_1x+b_2x^2+...+b_nx^n)=g(p(x))+g(q(x))\)

\(g(p(x)*q(x))=(a_0*b_0)+(a_1*b_1)x+(a_2*b_2)x^2+...+(a*n*b_n)x^n\) \(=(a_0+a_1x+a_2x^2+...+a_nx^n)*(b_0+b_1x+b_2x^2+...+b_nx^n)=g(p(x))*g(q(x))\)
I am not sure how g(p(x))=p(0) comes into play though

Answer 2

@experimentX can u check this?

Answer 3

Just ...