Is There an Isomorphism Between ℤ and ℤ[x]?

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Discussion Overview

The discussion centers on the question of whether there exists an isomorphism between the ring of integers ℤ and the ring of polynomials ℤ[x]. Participants explore the implications of such an isomorphism, including mappings and properties of ring homomorphisms.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that no isomorphism exists between ℤ and ℤ[x], questioning the possibility of a mapping that satisfies the properties of a ring isomorphism.
  • One participant suggests that while all elements of ℤ can be found in ℤ[x], the reverse is not true, indicating a fundamental difference in structure.
  • Another participant critiques an earlier attempt at establishing an isomorphism, implying that the existence of a better isomorphism has not been ruled out.
  • A participant discusses the requirement for a ring isomorphism to map the multiplicative identity correctly and explores the implications of such a mapping on the integers.
  • There is a consideration of the injectivity of a proposed mapping from ℤ[x] to ℤ, raising questions about how such a mapping could be constructed while maintaining injective properties.

Areas of Agreement / Disagreement

Participants generally disagree on the existence of an isomorphism between ℤ and ℤ[x], with multiple competing views presented regarding the implications and properties of such a mapping.

Contextual Notes

Participants express uncertainty about the injectivity of mappings and the implications of ring properties, indicating that assumptions about the structure of the rings may be critical to the discussion.

Bachelier
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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?
 
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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.
 
Bachelier said:
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?

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.
 
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?
 

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