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I am trying to compute the integral closure of $\mathbb{Z}$ in $\mathbb{Q}[i].$ I have managed to show that $\mathbb{Z}[i]$ is inside the integral closure, and I suspect it is the entire thing. Can someone give me a nudge in the right direction?

Katie Dobbs
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  • Actually I just had an idea looking at the proof that $1/2$ is not in the integral closure of $\mathbb{Z}$ over $\mathbb{Q}.$ Since the integral closure is a ring, if $a+bi$ was in it (a,b rational but not integers) then $a'+b'i$ is in it as well (where a', b' are the nearest integers to a, b). But $|a'+b'i| < (1/2)^2 + (1/2)^2 = 1/2$ and the integral closure would contain powers of this, so it would contain elements of modulus $1/2^n$ for all $n,$ which makes it impossible for $\mathbb{Z}[a'+b'i]$ to be finitely generated as a $\mathbb{Z}$-module. Can anyone verify this approach is correct? – Katie Dobbs Jun 25 '13 at 12:04
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    If the core of your question is in your comment, move it to the body of the question. – lhf Jun 25 '13 at 12:18
  • @lhf I don't really care how I do the problem, so the core is just what is currently in the body. I've accepted Cocopuffs's answer. – Katie Dobbs Jun 25 '13 at 12:24
  • Doesn't $sqrt(2)$ lie in the closure. All the ring of integers (don't confuse with Z) lies. – Lada Dudnikova Aug 30 '19 at 14:18

1 Answers1

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$\mathbb{Z}[i]$ is a Euclidean ring with respect to the norm $N(x+iy) = x^2 + y^2$, so it's a unique factorization domain, and therefore integrally closed.

Cocopuffs
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  • Ahh yes. I just did an exercise that proves that a UFD is integrally closed in its field of fractions, I should have seen this. Thank you! – Katie Dobbs Jun 25 '13 at 12:06