Is $\mathbb Z[i/2]:=\{f(i/2) : f(x) \in \mathbb Z[x] \}$ (smallest subring containing $\mathbb Z$ and $i/2$ ) dense in $\mathbb C$ ?
NOTE : $\mathbb Z [i/2] \ne \mathbb Z + \dfrac i2 \mathbb Z $
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${,f(i):f(x)\in{\bf Z}[x],}$ does not contain $i/2$. – Gerry Myerson Feb 06 '17 at 11:42
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If $f(x)=x$, then $f(i)=i\ne i/2$. – Gerry Myerson Feb 06 '17 at 11:50
2 Answers
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Yes. You have $-2(i/2)^2=1/2$ so you can do nested intervals. Just show that $\Bbb Z[1/2]$ is dense in $\Bbb R$ and then proceed to $\Bbb C$.
Edit: We know that a subgroup of $\Bbb R$ has a least positive element or is dense. Hence as $\Bbb Z[1/2]$ has no minimal positive element it is dense in the reals.
Further if $D\subset k$ is dense, then also $D[x] \subset k[x]$ is dense.
Now if we apply this to $$ \Bbb Z[i/2] = (\Bbb Z [1/2]) [i] $$ we know $\Bbb Z[i/2]$ is dense in the complex numbers.
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As $\{m/2^n : m \in \mathbb Z , n \in \mathbb N\} \subseteq \mathbb Z[1/2]$ , so $\mathbb Z[1/2]$ is dense in $\mathbb R$ .
Then since $\mathbb Z[1/2]+i\mathbb Z[1/2] \subseteq \mathbb Z[i/2] \subseteq \mathbb C$ , so $\mathbb Z[i/2]$ is dense in $\mathbb C$