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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!

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  • $\begingroup$ The third paragraph of your question indicates to me that you misread your professor's definition. The definition says "group homomorphism under addition such that $f(ab)=f(a)f(b)$ ...", but you seem to have read it as "group homomorphism under addition, which means that $f(ab)=f(a)f(b)$ ...". The former is correct; the latter is, as you pointed out, nonsense. $\endgroup$ – Andreas Blass Jun 8 '13 at 19:58
  • $\begingroup$ Thank you Andreas. You were absolutely right. I did misinterpret his question! $\endgroup$ – erik7970 Jun 8 '13 at 20:14
  • $\begingroup$ BTW @erik7970 , they both are isomorphic free abelian groups of rank two... $\endgroup$ – DonAntonio Jun 8 '13 at 20:16
  • $\begingroup$ Dear erik: Since you're fairly new here at math.se, you might like to know that for each question you ask, you can accept exactly one answer, and we encourage you to do so if its been helpful. To accept an answer, simply click on the $\large \checkmark$ to the left of the answer you'd like to accept. You get $2$ reputation points when you accept a question. You can also upvote as many answers as you'd like. $\endgroup$ – amWhy Jun 12 '13 at 23:03
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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 $\,-1.$

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  • $\begingroup$ Would it basically be all of the elements in the circle group? $\endgroup$ – erik7970 Jun 8 '13 at 18:27
  • $\begingroup$ The cartestian product between to pairs consisting of integers would be a complex number? $\endgroup$ – erik7970 Jun 8 '13 at 18:37
  • $\begingroup$ Ah. Ok I see. I'll see what properties between the two distinguishes one from the other. Thank you! $\endgroup$ – erik7970 Jun 8 '13 at 18:41
  • $\begingroup$ Now it's making much more sense to me. Thank you Key Ideas! $\endgroup$ – erik7970 Jun 8 '13 at 18:44
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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.

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