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The Gaussian integers are the ones of the form $m + ni$ where $m,n$ are both integers. I need to show that given any two Gaussian integers $a$ and $b$, $ab = 0$ must imply that $a = 0$ or $b = 0$.

I tried to show it by contradiction. Assume we have $a \neq 0 $ and $b \neq 0$ such that $ab= 0$. Then $(m_1 + n_1i)(m_2+n_2i) = 0$, hence $(m_1m_2 - n_1n_2) + (m_1n_2 + m_2n_1)i = 0$, hence $ m_1m_2 = n_1n_2$ and $m_1n_2 = -m_2n_1$. We then multiply these two equations together to get $m_1^{2}(m_2n_2) = -n_1^{2}(m_2n_2)$ but then I can't really do the cancellation law because maybe $m_2 = 0$ or $n_2 = 0$. I am stuck. Any suggestions?

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    $\begingroup$ Take complex conjugates and multiply the lot together and you get $(m_1^2+n_1^2)(m_2^2+n_2^2)=0$. $\endgroup$ – Derek Holt Nov 17 '14 at 2:57
  • $\begingroup$ but how is this relevant ? to show that there are no zero divisors $\endgroup$ – alkabary Nov 17 '14 at 3:24
  • $\begingroup$ @Alkabary it's relevant because it transfers the question into the integers, which you probably know have no zero divisors. Think about it for a bit. $\endgroup$ – rschwieb Nov 17 '14 at 3:36
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    $\begingroup$ I cannot make any sense of your last question! I am just using your own notation and showing you how to get a contradiction. Since $m_1^2+n_1^2$ and $m_2^2+n_2^2$ are both integers, one of them has to be $0$. If the first of them is $0$, then we get $m_1=n_1=0$. $\endgroup$ – Derek Holt Nov 17 '14 at 3:51
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    $\begingroup$ Only one suggestion: please learn how to spell Gaussian. It was Guassian (well, guassian) everywhere in your question initially. $\endgroup$ – KCd Nov 17 '14 at 4:21
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Here is one approach. Define the norm $N: \mathbb{Z}[i] \rightarrow \mathbb{Z}$ by $N(a+bi) = (a+bi)(a-bi) = a^{2}+b^{2}$. Here are some steps, which I leave you to prove, which will together prove your claim.

$1)$ Prove that the norm $N$ is multiplicative, i.e. for $\alpha, \beta \in \mathbb{Z}[i]$, $N(\alpha\beta) = N(\alpha)N(\beta)$.

$2)$ Prove that $N(\alpha) = 0$ if and only if $\alpha = 0$.

$3)$ Finally, using the fact that $\mathbb{Z}$ is an integral domain, use $1$ and $2$ to conclude $\mathbb{Z}[i]$ is an integral domain.

The thing I like about this approach in particular is that defining the norm $N$ can also be used to show the stronger fact that $\mathbb{Z}[i]$ is a Euclidean domain.

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Yet another way: The Gaussian integers are clearly a subring of $\mathbb C$: certainly a subset, and the rules for addition and multiplication in the two rings are the same. Since $\mathbb C$ has no zero divisors, neither does the ring of Gaussian integers.

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You can note that the ring is isomorphic to $\Bbb Z[x]/(x^2+1)$, and that $x^2+1$ is irreducible and prime, so that it generates a prime ideal, and the quotient is a domain.

Or you can do as Derek has suggested and note that for any nonzero $x$, $x\overline{x}\neq 0$ is a nonzero element of $\Bbb Z$, and use that with the equation $xy=0$ to prove one of x and y is zero.

Further hint: $xy=0$ implies $(x\overline{x})(y\overline{y})=0$. But these are two integers multiplying to zero, so...?

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    $\begingroup$ I didn't even think about the quotient approach. Very nice! $\endgroup$ – Alex Wertheim Nov 17 '14 at 3:35
  • $\begingroup$ how can you use derek suggestion to show that $xy = 0$ i still don't get it $\endgroup$ – alkabary Nov 17 '14 at 3:43
  • $\begingroup$ @Alkabary I added another hint. How about now? $\endgroup$ – rschwieb Nov 17 '14 at 4:16

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