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Prove that (3) is a maximal ideal in $\mathbb{Z}[i]$, and thus $\mathbb{Z}[i]/(3)$ is a field. (Hint: for $a+bi\not\in(3)$, show that $(a+bi,3)=(1).)$ How many elements are in this field?

We've only recently learned about maximal ideals so my understanding of them is quite poor, I'm having a tough time approaching this question. Any help is appreciated, thank you!

Bella Dohn
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    $3$ is prime in $\Bbb Z[i]$, which is a PID, so that $(3)$ is a maximal ideal. The quotient has $9$ elements, see this post. You will find all you need on this site MSE here. Start searching for it. – Dietrich Burde Nov 29 '21 at 16:06
  • If you know the concept of Euclidean Domain it should be pretty straight forward (https://en.wikipedia.org/wiki/Euclidean_domain) using the argument posted from @Dietrich Burde. – Augusto Matteini Nov 29 '21 at 16:07
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    Two tools that are useful, norm $|a+bi|=\sqrt{a^2+b^2}$ and Euclidean division. Assume that you have $(3)\subset I\subset \mathbb{Z}[i]$. Take $x\in I\setminus (3)$ and compute its remainder $a+bi$ modulo $3$. We must have $a^2+b^2=|a+bi|<9/2$. The only options for $a,b$ are $\pm1,0$. All of them give $a+bi$ either units or $0$. In the first case $I=\mathbb{Z}[i]$, since it would contain that remainder that is a unit. In the second case we get that $x\in(3)$, which contradicts that $x\notin (3)$. – plop Nov 29 '21 at 16:07
  • Hint: $\mathbb{Z}[i]/(3) \cong \mathbb{Z}[x]/(3,x^2+1) \cong (\mathbb{Z}/3\mathbb{Z})[x]/(x^2+1)$ – lhf Nov 29 '21 at 16:08
  • Well, we could get remainders $2+0i$ or $0+2i$ too. But in that case, divide $3$ by these to get remainder $1$. In this case we got $1\in I$, which implies $I=\mathbb{Z}[i]$. – plop Nov 29 '21 at 16:13

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