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Show that, if $ a + bi$ is prime in $\mathbb{Z} [i]$, then $a - bi $ is prime in $\mathbb{Z}[i]$

Since $\mathbb{Z}[i]$ is $ED$, then if $a+bi$ is irreducible then $a+bi$ is prime. But now how I can relate $a+bi$ with $a-bi$?

Thanks for help.

Zev Chonoles
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Sara
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2 Answers2

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If $a-bi$ is not irreducible then take conjugate on both sides to obtain a contradiction.Conjugate of $a+bi$ is $a-bi$. So if $a-ib=(p+iq)(r+is)$ then clearly $a+ib=(p-iq)(r-is)$.

happymath
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In fact the property holds in every quadratic ring $\mathbb Z[\sqrt{d}]$ if one uses the definition of prime elements: say $a+b\sqrt{d}$ is prime, and want to show that $a-b\sqrt{d}$ is also prime.

Following a nitpick from the comments let's first show that $a-b\sqrt{d}$ is not invertible: suppose it is, and then there is $z\in\mathbb Z[\sqrt{d}]$ such that $(a-b\sqrt d)z=1$. From here we get $(a+b\sqrt d)\overline{z}=1$, so $a+b\sqrt d$ is invertible, a contradiction.

Now suppose $a-b\sqrt{d}\mid z_1z_2$ where $z_i\in\mathbb Z[\sqrt{d}]$. Then $a+b\sqrt{d}\mid\overline{z}_1\overline{z}_2$ where $\overline{z}_i$ is the conjugate of $z_i$. Since $a+b\sqrt{d}$ is prime it follows $a+b\sqrt{d}\mid\overline{z}_1$ or $a+b\sqrt{d}\mid\overline{z}_2$. Suppose $a+b\sqrt{d}\mid\overline{z}_1$. Then $a-b\sqrt{d}\mid z_1$ and we are done.

user26857
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  • This is fantastic. Thanks user 26857 – Sara Jun 04 '15 at 16:50
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    @Sara Beware, the above is not a complete proof. In addition to the above we also need to show that $,a - b i,$ is a (nonzero) nonunit in order to conclude that it is prime. Do you see how to do that? – Bill Dubuque Jun 10 '15 at 23:23
  • @user26857 I have seen many proofs fail due to that (and analogous) oversights, so I thought it worthy of explicit mention (esp. at this level). – Bill Dubuque Jun 10 '15 at 23:58
  • @BillDubuque AFAIK $(0)$ is a prime ideal in any integral domain, so $0$ is a prime element, or not? – user26857 Jun 12 '15 at 20:53
  • @user26857 Depends on convention. That's why I wrote (nonzero) in parens. In any case, the main point is that one does need to prove that it is a nonunit, since units are never considered prime. – Bill Dubuque Jun 12 '15 at 21:00