Show that there is a disc in $\mathbb{C}$ with radius R , so that no primes of $\mathbb{Z}[i]$ are contained in the disc.

I was thinking of taking a disc which which does not touch (0,0), for example : $|z-R|+ R<|R|$

But then how does one show that this disc doesn't contain any primes in $\mathbb{Z}[i]= \mathbb{Z}+\mathbb{Z}i$.

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    $\begingroup$ I've noted a couple of times that you shouldn't "sign" your messages by putting blank lines and "V" at the bottom, because every message already comes with your signature at the bottom right. $\endgroup$ – Arturo Magidin Nov 1 '11 at 20:03
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    $\begingroup$ Your disc should be $|z-z_0|<R$, where $z_0$ is the center. |z_0| has to be large because there are lots of small primes. $\endgroup$ – Ross Millikan Nov 1 '11 at 20:05
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    $\begingroup$ The first sentence is incoherent, and its second clause is completely divorced from the clause that precedes. "$\mathbb{Z}[i]$ does not contain any prime numbers" is false; presumably, it's supposed to be somehow connected with the first clause? And is $R$ a specific number, or an arbitrary number? And are "prime numbers" rational primes, or prime elements of $\mathbb{Z}[i]$? A definite prerequisite for doing mathematics successfully is attention to detail. $\endgroup$ – Arturo Magidin Nov 1 '11 at 20:05
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    $\begingroup$ You've asked 12 questions and you haven't accepted any of the answers that were given to you: See here how you can accept answers. Was none of the answers you got helpful to you? $\endgroup$ – t.b. Nov 1 '11 at 20:11
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    $\begingroup$ @VVV: You might want to thank Thomas Andrews; but you still need to say what $R$ is. Is it arbitrary? If so, try something like the argument used to show that every for every $d\gt 0$ there are $d$ consecutive integers, none of which are primes. $\endgroup$ – Arturo Magidin Nov 1 '11 at 20:15

It suffices to show for $R$ a positive integer.

Let $P$ be the product of all $z\in \mathbb{Z}[i]$ with $0<|z|<4R$.

Let $D = P + 2R$.

Then if $|z-D|<R$, then $R<|z-P|<3R$. So $z-P$ is a divisor of $P$.

But then $z-P$ divides $z=(z-P) + P$.

On the other hand, since $R<|z-P|$, $z-P$ is not a unit of $\mathbb{Z}[i]$, so $z$ is not a prime.

You have to show additionally that $u(z-P)\neq z$ is not possible for any unit $u\in \mathbb{Z}[i]$, but that's not hard.

As noted above, $|z-P|<4R$.

On the other hand $|z| + R \geq |z| + |z-D| \geq |D| \geq |P| - 2R$. So $|z|\geq |P|-3R$.

Since it is easy to show that $|P|\geq 81R^4>7R$, we see that $|z|>4R$. So $z$ is not a unit times $z-P$.

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  • $\begingroup$ why pick $D=P+2R$ ? $\endgroup$ – VVV Nov 1 '11 at 21:21
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    $\begingroup$ Basically, you want to pick $D$ so that if $|z-D|<R$, then not only is $|z-P|<4R$ (to make $z-P$ a divisor of $P$) but also that $z-P$ is not a unit. By picking this $D$, |z-P|>R, so $z-P$ cannot be a unit. $\endgroup$ – Thomas Andrews Nov 1 '11 at 21:27
  • $\begingroup$ Note that there are no possible primes in the disk of radius $4R$ around $P$ except for $P\pm 1$ and $P\pm i$. So we are really just finding any old circle of radius $R$ inside this bigger circle around $P$ so that these anomalies are excluded. We could have chosen $D=P+R+2$. $\endgroup$ – Thomas Andrews Nov 1 '11 at 21:46
  • $\begingroup$ Which is similar to the case of $\mathbb Z$, where, to find $n$ consecutive non-primes, we choose $(n+1)!+2,...,(n+1)!+(n+1)$. We have to avoid $(n+1)!\pm 1$, because it isn't obvious that these are not prime. $\endgroup$ – Thomas Andrews Nov 1 '11 at 21:50

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