# ℤ and ℤ[x] isomorphism

1. Oct 4, 2011

### Bachelier

There exists none. What's the easiest way to prove this?
Can we state that all elements of ℤ are in ℤ[x] but not the other way around?

2. Oct 4, 2011

### Office_Shredder

Staff Emeritus
Just because you came up with a bad attempt at an isomorphism (the obvious inclusion of the integers) doesn't mean a better one doesn't exist.

3. Oct 4, 2011

### lavinia

I think a ring isomorphism must map 1 to 1 in Z[x] so Z must be mapped to Z in Z[x].

h(m) = h(1.m) = h(1)h(m). So h(1) = 1. I think this is right.

4. Oct 6, 2011

### Deveno

if there WAS such an isomorphism, we would also have one from Z[x] to Z, say φ.

now 1 ( = 1 + 0x + 0x^2 +....) in Z[x] is a multiplicative identity,

so say φ(f(x)) = k.

so k = φ(f(x)) = φ(1f(x)) = φ(1)φ(f(x)) = φ(1)k.

so φ(1) = 1.

by induction, for n ≥ 1, φ(n) = φ(1+1+...+1) (n times)

= φ(1) + φ(1) + ...+ φ(1) (n times)

= 1 + 1 +...+ 1 (n times)

= n.

similarly, φ(-n) = -φ(n) = -n.

given this, and the fact that we must assign an integer value to φ(x),

how can φ be injective?