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I'm trying to figure out the proper way to go about determining the ring homomorphisms between $\mathbb{Z}[i]$, the gaussian integers, and $\mathbb{Z}$.

In particular, I'm almost certain there are no nontrivial homomorphisms between these two rings, but I can't quite figure out how to prove it.

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You're right. There are no nontrivial homomorphisms $\mathbb{Z}[i] \to \mathbb{Z}$.

Suppose for contradiction that there exists a nontrivial homomorphism $\phi:\mathbb{Z}[i] \to \mathbb{Z}$. Note that $\phi(1) = \phi(1\cdot1) = \phi(1)\phi(1) = \phi(1)^2$. Thus, $\phi(1) = \phi(1)^2$ and $1 \mapsto x$ if and only if x is idempotent in $\mathbb{Z}$. Thus, $\phi(1) = 0$ or $\phi(1)=1$. If $\phi(1)=0$, $\phi$ is immediately trivial since homomorphisms are multiplicative, and we have a contradiction. Now assume that $\phi(1)=1$. Then, for any unit $u \in \mathbb{Z}[i]$, we have $\phi(u)\phi(u^{-1}) = \phi(1) = 1$, implying that the units in $\mathbb{Z}[i]$ map to the units in $\mathbb{Z}$.

Since the mapping for $1 \in \mathbb{Z}[i]$ has been taken by assumption, the mappings for the remaining 3 units in $\mathbb{Z}[i]$, namely $-1, i, -i$, remain to be determined. Since the only units in $\mathbb{Z}$ are $\pm 1$, and, moreover, these units are their own inverses, we have two binary choices for the mappings of the remaining units: $\phi(-1) = 1$ or $\phi(-1) = -1$, and $\phi(i) = \phi(-i) = 1$ or $\phi(i) = \phi(-i) = -1$.

First, suppose for contradiction that $\phi(-1) = 1$. Then, it follows that $-\phi(1) = -1 \neq 1 = \phi(-1)$. This implies that $\phi(0) = \phi(-1+1) = \phi(-1) + \phi(1) \neq 0$. However, it is true in general for ring homomorphisms that \begin{align*} \phi(0) &= \phi(0) + 0 \\ &= \phi(0) + (\phi(0) - \phi(0)) \\ &= \phi(0 + 0) - \phi(0) \\ &= \phi(0) - \phi(0) \\ &= 0 \end{align*} This is a contradiction. Hence, $\phi(-1) = -1$. However, if this holds, it follows that \begin{align*} -1 &= \phi(-1) \\ &= \phi(i^2) \\ &= \phi(i)\phi(i) \\ &\stackrel{(1)}{=} \phi(i)\phi(-i) \\ &= \phi(i\cdot(-i)) \\ &= \phi(1) \\ &= 1 \end{align*} (1): Since $\phi(i) = \phi(-i)$ regardless our second mapping choice

This is also a contradiction. It follows that neither of the two necessary choices for the mapping of -1 are feasible and we have a contradiction to the assumption that $\phi(1) = 1$. Thus, $\phi(1) = 0$ and $\phi$ must be trivial as shown previously.

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We have $$\phi(1)=\phi(1^2)=\phi(1)^2\ ,$$ so $\phi(1)=1$ or $\phi(1)=0$. If $\phi(1)=1$ then we also get $$\phi(i)^2=\phi(i^2)=\phi(-1)=-1$$ (the last step is easy to prove) and this is impossible since $\phi(i)\in{\Bbb Z}$.

Hence $\phi(1)=0$; it follows easily that $\phi(a)=0$ for all $a\in{\Bbb Z}$ and so $$\phi(a+bi)=\phi(a)+\phi(b)\phi(i)=0\ .$$

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In general we have for a commutative ring $R$ with unity:

$$\operatorname{Hom}(\mathbb Z[i],R) = \{f \in \operatorname{Hom}(\mathbb Z[X],R) ~ | ~ f(X)^2+1=0 \} = \{r \in R ~|~ r^2+1=0\}.$$

Homomorphisms from $\mathbb Z[i]$ to any ring are the same as solutions of the equation $r^2+1=0$.

The first equality ist he fundamental homomorphism theorem: A map factors over a quotient if and only the corresponding ideal is mapped to zero.

The second equality is the universal property of the polynomial ring. A map from a polynomial ring over $S$ to an $S$-algebra is the same as the choice of an element of the $S$-algebra.

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Take a ring homomorphism $\phi:\Bbb Z[i] \to \Bbb Z$. We need $\phi(1)$ to be either $0$ or $1$, since $\phi(1) = \phi(1^2) = \phi(1)^2$. If $\phi(1) = 0$, then the homomorphism is trivial, since $$\phi(a + bi) = \phi(1(a + bi)) = \phi(1)\phi(a+bi) = 0$$ So, if $\phi$ is to be non-trivial, then we need $\phi(1) = 1$.

Next up, $\phi(0) = 0$, since any ring homomorphism is a group homomorhism on the group of addition in the ring. This means that $$0 = \phi(0) = \phi(1-1) = \phi(1) + \phi(-1)$$ which implies that $\phi(-1) = -1$.

Then we have $\phi(i)$. We know that $\phi(i)^2 = \phi(i^2) = -1$, but there is no such element in $\Bbb Z$, so this is impossible.

Therefore $\phi$ must be the trivial homomorphism.

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  • $\begingroup$ I think you mean to write $\phi(-1)=-1$ after your second equation. Otherwise this is a nice way to do it! $\endgroup$ – User12345 Feb 15 '16 at 9:13
  • $\begingroup$ @User12345 Yes, that is what I meant. $\endgroup$ – Arthur Feb 15 '16 at 12:37
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$0=\varphi(0)=\varphi(1+i^2)=\varphi(1)+\varphi(i)^2$

In particular $-\varphi(i)^2=\varphi(1)$ is idempotent, so either $\varphi(i)^2=0$ or $\varphi(i)^2=-1$. The second case is impossible, thus $\varphi(i)=0$ and therefore $\varphi(1)=0$ too.

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