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I know that we can write $(2)=(1+i)^2$, and from this question we have that $\mathbb{Z}[i]/(3)$ is a field. But perhaps I'm confused why this is so. In particular, doesn't the same computation in the selected answer to that question give us $\mathbb{Z}[i]/(2)\cong (\mathbb{Z}/2\mathbb{Z})[i]$, a field?

ponchan
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"$(\Bbb Z/2\Bbb Z)[i]$" is $(\Bbb Z/2\Bbb Z)[x]/(x^2+1) = (\Bbb Z/2\Bbb Z)[x]/((x+1)^2)$ which is not a field because e.g. $(x+1)(x+1) = 0$ in the quotient but $x+1 \ne 0$ in the quotient.

Kenny Lau
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    Note that there are a few closely related ways of thinking about this: $2$ is a sum of two squares, so it ramifies in $\mathbb{Z}[i]$ (since $a^2+b^2 = (a+bi)(a-bi)$); equivalently (though the equivalency is nontrivial), $-1$ is a square modulo $2$ so the polynomial $x^2+1$ factors mod $2$ and you get zero divisors. By contrast, $3$ is not a sum of two squares, and $-1$ isn't a square mod $3$ (since $1^2=2^2=1$ and $0^2=0$). – Steven Stadnicki Mar 03 '21 at 22:20
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    You are confusing a prime being not inert (f<n) vs a prime being ramified (e=n). These conditions are not equivalent. Indeed, $5$ (e=1, f=1, g=2) is a sum of two squares $5 = 1^2 + 2^2$ and it is not ramified. – Kenny Lau Mar 03 '21 at 22:29
  • So does the selected answer to the referenced question fail to justify that the ring is a field? Or is it considered clear that $(\Bbb Z/3\Bbb Z)[i]$ is a field even though $(\Bbb Z/2\Bbb Z)[i]$ is not? – Karl Mar 03 '21 at 22:33
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    That would depend on the context, i.e. the background of the OP of the referenced question. Technically speaking, the accepted answer of the referenced question has not justified fully why $(\Bbb Z/3\Bbb Z)[i]$ is a field. – Kenny Lau Mar 03 '21 at 22:40
  • @KennyLau D'oh. Thank you; mea culpa. – Steven Stadnicki Mar 03 '21 at 23:11