If we consider the ring homomorphism $\varphi :Z[i]\rightarrow Z_3[i]$ by $a+bi \rightarrow (a\text{ mod 3})+(b\text{ mod }3)i$ then we have $\varphi$ is surjective. If we assume ker $\varphi =\langle3\rangle$ then by first isomorphism theorem, we have $Z[i]/\langle3\rangle \cong Z_3[i]$. Since $Z_3[i]$ is a field then $\langle3\rangle$ is maximal. However, I am having trouble seeing $\ker \varphi =\langle 3\rangle$. Is $\ker \varphi = \langle3\rangle+\langle3\rangle i$? If so, then I don't know what is the right answer.
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2$3$ is irreducible in this PID, so that $(3)$ is a maximal ideal. – Dietrich Burde May 24 '21 at 16:05
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1Iirc the primes congruent to $3 \pmod{4}$ are irreducible in this ring – Evariste May 24 '21 at 16:07
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1"Since $Z_3[i]$ is a field" - why is it a field? – Dietrich Burde May 24 '21 at 16:10
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When is $(a\bmod 3)+(b\bmod 3)I=(0\bmod 3)+(0\bmod 3)I?$ – Thomas Andrews May 24 '21 at 16:10
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Note, you are using $\langle 3\rangle$ to mean two things here, one as an ideal in $\mathbb Z$ and the other as an ideal in $\mathbb Z.$ – Thomas Andrews May 24 '21 at 16:14
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1@ThomasAndrews One in $\Bbb Z[i]$, you mean. – Dietrich Burde May 24 '21 at 16:15
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So how is ideal <3> in $Z[i]$ represented? – John May 24 '21 at 16:26
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1The ideal <3> of Z[i] is all a+bi where both a and b are multiples of 3. If we let <3>Z mean the ideal of Z generated by 3, then <3>Z = { 3a : a in Z } = { 3b: b in Z } ... no i anywhere. If we let <3>Z[i] mean the ideal of Z[i] generated by 3, then <3>Z[i] = { 3(a+bi) : a+bi in Z[i] }... 3 multiplies a+bi, not just a or b. As you said, that is the kernel you want <3>Z + (<3>Z)i = <3>(Z[i]) – Jack Schmidt May 24 '21 at 16:30
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Yes, Jack Schmidt this makes sense. It was tricky, getting confused by <3> in $Z[i]$ but your explanation makes sense. – John May 24 '21 at 17:37
2 Answers
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To answer the question that was actually asked, yes $\ker(\phi) = \langle 3 \rangle +\langle 3 \rangle i = \{ 3a + 3bi : a,b \in \mathbb{Z} \} = \{ 3(a+bi) : a+bi \in \mathbb{Z}[i] \} = \langle3\rangle$ is the ideal of $\mathbb{Z}[i]$ generated by 3.
Jack Schmidt
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Yes, since $i$ is a unit of $\mathbb Z[i]$, $\langle 3 \rangle = \langle 3 \rangle i$. – hardmath May 24 '21 at 16:37
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But what about $3a+6b$? That is also in the $ker \varphi$, but it is not in <3>? – John May 24 '21 at 17:43
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It is in both. 3a+6b and 3a+6bi are patterns, not specific numbers. More specific versions would be 3+6=9 and 3+6i. Both of those are in $\langle 3\rangle$, the ideal of $\mathbb{Z}[i]$ generated by 3, because both are $\mathbb{Z}[i]$ multiples of 3. $9=3\cdot 3$ and $3+6i = 3(1+2i)$. – Jack Schmidt May 24 '21 at 17:49
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The element $3$ is irreducible in $\Bbb Z[i]$, see here:
All irreducible elements in Gaussian integers
Hence the principal ideal $(3)$ is maximal, since $\Bbb Z[i]$ is a PID. This answers the title question.
Dietrich Burde
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Good, I wanted to give another argument, since there were already answers above. – Dietrich Burde May 24 '21 at 16:41