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If $a+bi$ is not a unit of $\mathbb{Z}[i]$ prove that $a^2+b^2>1$.

Definition: Let $R$ be commutative ring with unit element. An element $u\in R$ we call unit if it's inverse $u^{-1}$ also lies in $R$.

Proof: Suppose by contradiction that $a^2+b^2\leq 1$ then $a^2+b^2\in \{0,1\}$.

If $a^2+b^2=1$ then we have four cases $a+bi\in \{\pm i, \pm 1\}$. But each of them is unit of $\mathbb{Z}[i]$. In this case we get contradiction.

If $a^2+b^2=0$ then $a+bi=0$. Here we have no contradiction, since by definition $0$ is NOT unit.

Can anyone explain this to me, please?

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    $\begingroup$ Explanation: The problem poser accidentally forgot to exclude zero. $\endgroup$
    – Hagen von Eitzen
    May 26, 2018 at 19:33
  • $\begingroup$ @HagenvonEitzen, Do you mean that we should add a nonzero element in the definition of unit element? $\endgroup$
    – RFZ
    May 26, 2018 at 19:35
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    $\begingroup$ $0$ is an element of $\mathbb{Z}[i]$ that fails $a^2+b^2>1$. Therefore, the problem should state that "If $a+bi\not=0$ is not a unit of $\mathbb{Z}[i]$, then $a^2+b^2>1$." Alternately, "If $a+bi$ is not a unit of $\mathbb{Z}[i]$, then $a+bi=0$ or $a^2+b^2>1$." $\endgroup$
    – Michael Burr
    May 26, 2018 at 19:36
  • $\begingroup$ @MichaelBurr, Indeed, but I am proving via contradiction. I am supposing that $a^2+b^2\leq 1$. And the case when it is equal to zero confuses me. Because in this case we get $a=b=0$. $\endgroup$
    – RFZ
    May 26, 2018 at 19:38
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    $\begingroup$ The original problem is wrong - another condition is needed. $\endgroup$
    – Michael Burr
    May 26, 2018 at 19:39

2 Answers 2

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In general, for a number field $K$ we have that $|Nm(\alpha)|>1$ for every (nonzero) non-unit $\alpha\in \mathcal{O}_K$, see this duplicate. Now $Nm(\alpha)=a^2+b^2$ for $K=\Bbb{Q}[i]$, and $\mathcal{O}_K=\Bbb{Z}[i]$.

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If $a+ib$ is unit so there exists $c+id\in \mathbb{Z}[i]$ such that $(a+ib)(c+id)=1$ hence $(a^2+b^2)(c^2+d^2)=1$ then $a^2+b^2=1$. Since $a,b\in \mathbb{Z}$, $(a,b)=(\pm 1,0)$ or $(a,b)=(0,\pm 1)$ thus $a+ib\in \{\pm 1,\pm i\}$. Conversely, it is clear that $\pm 1, \pm i$ are units therefore the units of $\mathbb{Z}[i]$ are $\pm 1, \pm i$.

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  • $\begingroup$ Your reasoning for what the units of Gaussian integers $\mathbb Z[i]$ is correct, but the Question asks us to prove something about the nonunits. Perhaps you misread the Question? Some further reasoning would be necessary to draw the requested conclusion. $\endgroup$
    – hardmath
    May 28, 2018 at 2:07
  • $\begingroup$ @hardmath The present answer completely classifies units as those elements $a+bi$ such that $a^2+b^2=1$. Hence it also classifies nonunits as those elements such that $a^2+b^2\ne1$. $\endgroup$
    – egreg
    May 28, 2018 at 8:49
  • $\begingroup$ @egreg: Although the OP does have tag "proof-verification", the structure of the Question is of that form. So I feel the contribution you see in the above Answer is somewhat ephemeral and would be better stated as already found in the Question. $\endgroup$
    – hardmath
    May 28, 2018 at 16:20

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