从ℤ到任意环的同态是否存在？无单位元环的反例问询

Hey there, let's break down these two ring theory questions clearly, starting with definitions to avoid confusion:

First, a quick note on definitions: when talking about ring homomorphisms, we have two common frameworks:

• Non-unital homomorphisms: Only require preserving addition (( \phi(a+b) = \phi(a) + \phi(b) )) and multiplication (( \phi(ab) = \phi(a)\phi(b) )) for all elements in the domain.
• Unital homomorphisms: Add the requirement that if the domain has a multiplicative identity ( 1 ), the homomorphism maps ( 1 ) to the multiplicative identity of the codomain.

问题1：是否存在从ℤ到任意环的同态？ #

It depends on which homomorphism definition you're using:

• If we allow non-unital homomorphisms: Yes, every ring has at least one homomorphism from ( \mathbb{Z} )—the zero homomorphism, which maps every integer to the additive identity of the target ring. This always satisfies the addition and multiplication preservation rules.
• If we restrict to unital homomorphisms: No. Any ring without a multiplicative identity can't have a unital homomorphism from ( \mathbb{Z} ), since there's no element in the target ring to map ( 1 \in \mathbb{Z} ) to (that would need to be a multiplicative identity, which the ring doesn't have).

问题2：是否存在无单位元的环R，使得不存在从ℤ到R的同态？ #

Again, this hinges on the homomorphism definition:

• For non-unital homomorphisms: No, the zero homomorphism always exists as a valid map between ( \mathbb{Z} ) and any ring ( R ), regardless of whether ( R ) has a unit.
• For unital homomorphisms: Absolutely, and every non-unital ring fits this bill. Let's take a concrete example: ( R = 2\mathbb{Z} ) (the even integers under standard addition and multiplication). This ring has no multiplicative identity (there's no even integer ( e ) such that ( e \cdot k = k ) for all even ( k )). A unital homomorphism from ( \mathbb{Z} ) to ( R ) would need to map ( 1 \in \mathbb{Z} ) to an identity element of ( R )—but since ( R ) has no identity, no such homomorphism can exist.

If you were asking about non-zero non-unital homomorphisms, we can also construct examples. For instance, take the zero ring on the additive group ( \mathbb{Q}/\mathbb{Z} ) (all elements have finite additive order). Suppose there's a non-zero homomorphism ( \phi: \mathbb{Z} \to R ). Then ( \phi(1) = x \neq 0 ), so ( \phi(n) = nx ) for all ( n \in \mathbb{Z} ). But since ( x ) has finite order ( k ), ( \phi(k) = kx = 0 = \phi(0) )—this is allowed (homomorphisms can have non-trivial kernels), but if we wanted an injective homomorphism, that's impossible here. But the question asks about any homomorphism, so the zero one still counts.

内容的提问来源于stack exchange，提问作者green frog