# Answer to Question #5108 in Abstract Algebra for Annemarie

Question #5108

Let Z be the ring of integers. Consider the function

f : Z[x] → Z

defined by f(g(x)) = g(0). for example, f(x2 + 1) = 1.

(a) Show that f is a ring homomorpism.

(b) What is the ker(f)?

f : Z[x] → Z

defined by f(g(x)) = g(0). for example, f(x2 + 1) = 1.

(a) Show that f is a ring homomorpism.

(b) What is the ker(f)?

Expert's answer

1) we must show that f is save multiplication and addition

let h(x)=

am*x^m+...+a0,

g(x)=bn*x^n+...+b0

f(h(x)+g(x))=[h(x)+g(x)](0)=h(0)+g(0)=f(h(x))+f(g(x))

f(h(x)*g(x))=f(am*bn*x^(n+m)+...+a0*b0)=am*bn*0^(n+m)+...+a0*b0=a(0)*b(0)=f(h(x))*f(g(x))

so

f is ring homomorphism

2) Ker(f) all that polynoms such that g(0)=0 so it

have g(x)=x*h(x), where h - polynom from Z[x]

let h(x)=

am*x^m+...+a0,

g(x)=bn*x^n+...+b0

f(h(x)+g(x))=[h(x)+g(x)](0)=h(0)+g(0)=f(h(x))+f(g(x))

f(h(x)*g(x))=f(am*bn*x^(n+m)+...+a0*b0)=am*bn*0^(n+m)+...+a0*b0=a(0)*b(0)=f(h(x))*f(g(x))

so

f is ring homomorphism

2) Ker(f) all that polynoms such that g(0)=0 so it

have g(x)=x*h(x), where h - polynom from Z[x]

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