Determine if the Gaussian Integers $\mathbb{Z}[i]$ and $\mathbb{Z}\times\mathbb{Z}$ are isomorphic rings

Question

Yet again I am having difficulty with another exercise from my abstract algebra class. It is a homework question that my professor came up with himself and is as follows, word by word:

Let $R$ and $T$ be rings. A function $f:R\to T$ is a ring homomorphism if it is a group homomorphism under addition such that $f(ab) = f(a)f(b)$. Furthermore, if $f$ is also a bijection, then we say that $f$ is a ring homomorphism and that $R$ and $T$ are isomorphic rings, denoted $R\cong T$. Determine if the Gaussian integers $\mathbb{Z}[i]$ and $\mathbb{Z}\times\mathbb{Z}$ are isomorphic rings.

What initially throws me off is he wrote "a group homomorphism under addition such that $f(ab) = f(a)f(b)$" where the latter part appears, at least to me, to be multiplication, not addition. I checked online to see what conditions needed to be satisfied for something to be considered a ring homomorphism and they were: $f(a+b) = f(a) + f(b)$ and $f(ab) = f(a)f(b)$ for all $a,b\in R$. I'm wondering if what he wrote as part of the question was some sort of typo on his part or if I'm perhaps simply missing something?

And, finally, to show that $\mathbb{Z}[i]$ and $\mathbb{Z}\times\mathbb{Z}$ are isomorphic rings, I understand I would need to first show that there exists a function from $\mathbb{Z}[i]$ to $\mathbb{Z}\times\mathbb{Z}$ that is a group homomorphism under addition and multiplication, correct? And it would be up to me to define the function however I choose as long as its a group homomorphism?

I apologize for writing so much as I'm very confused with this exercise. Thank you guys so much for the help!

Answer 1

Hint: $\ (0,1)\cdot (1,0) = (0,0)\,$ in $\, \Bbb Z\times \Bbb Z$

If an isomorphism exists, what does it imply when used to map the above equation into $\,\Bbb Z[i]?\ $

Alternatively, $\ \smash[t] {0 = 1+i^2\ \stackrel{f}\Rightarrow\ (0,0) = (1,1)+(a,b)^2 = (1+a^2,1+b^2)},\, $ hence we've deduced that $\ 1+x^2\ $ has a root in $\,\Bbb Z,\,$ contradiction

The key idea is to show no isomorphism exists by finding some ring-theoretic property one ring has but the other doesn't, where ring-theoretic property means a property preserved by isomorphisms. The first property is: being a domain, i.e. no nontrivial zero-divisors, i.e. $\ x,y \ne 0\,\Rightarrow\, xy\ne 0,\ $ and the second property is: $\, x^2 + 1\,$ has a root, $ $ i.e. the squareness of $\,-1.$

Answer 2

From Key Ideas' hint, $\mathbb{Z} \times \mathbb{Z}$ has zero divisors. Recall that a zero divisor of a ring $R$ is a nonzero element $a$ such that $ab = 0$ for some nonzero $b \in R$. However, since $\mathbb{Z}[i] \subseteq \mathbb{C}$ and $\mathbb{C}$ is a field, $\mathbb{Z}[i]$ has no zero divisors.

Also, a ring isomorphism is a bijective ring homomorphism. Use what I said to prove that no such isomorphism exists.