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Find the quotient ring $\mathbb{Z}[i]/(4+i)$ by identifying elements with the lattice points in the square generated by $4+i$.

I know that $N(4+i) = 17$.

Therefore, $4+i$ is irreducible.

Now here's where I am stuck - lattice points:

$I = (4+i)$ so $Z = 4+i$

$$(m+ni)(4+i) = 4m + mi + 4ni - n = m(4+i) + n(-1+4i)$$

Would greatly appreciate any help!

Zev Chonoles
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abstract
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  • Look at the image of $\mathbb{Z}$ under the projection map $\mathbb{Z}[i] \to \mathbb{Z}[i]/(4+i)$. What is it? – Zardo Jun 21 '15 at 18:33
  • I'm afraid to say I'm lost. I realize it is supposed to be a square generated by a linear combination of m and n (as indicated in the question), but I have no idea how to plot it. – abstract Jun 21 '15 at 18:37
  • Look here, below: http://math.stackexchange.com/questions/23358/quotient-ring-of-gaussian-integers – Zardo Jun 21 '15 at 18:39
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    There are lots of threads with this very question on this site. Try taking a look at some of them: http://math.stackexchange.com/questions/23358/quotient-ring-of-gaussian-integers, http://math.stackexchange.com/questions/52537/order-of-some-quotient-ring-of-gaussian-integers, http://math.stackexchange.com/questions/373073/quotient-rings-of-gaussian-integers – Viktor Vaughn Jun 21 '15 at 19:34

1 Answers1

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As you have already noted, $4+ \Bbb i$ is irreducible; $\Bbb Z [ \Bbb i ]$ is a Euclidean domain, therefore a principal ideal domain (PID), and we know that in a PID the ideal generated by an irreducible element is maximal; hence, the ideal $(4+ \Bbb i)$ is maximal, so the quotient ring $\Bbb Z [ \Bbb i ] / (4+ \Bbb i)$ is a field.

On the other hand, every element of $\Bbb Z [ \Bbb i ]$ can be written as $q(4 + \Bbb i) + r$ with $r=0$ or $N(r) < N(4+\Bbb i) = 17$. According to this, the class modulo $(4+\Bbb i)$ of any element is the class of its remainder modulo $4+ \Bbb i$. But what are these posible remainders? Well, they are constructed from the pairs of natural numbers satisfying the inequation $a^2 + b^2 <17$, giving $\{ 0, \Bbb i, 2 \Bbb i, 3 \Bbb i, 4 \Bbb i, 1, 1+ \Bbb i, 1+2 \Bbb i, 1+3 \Bbb i, 2, 2+ \Bbb i, 2+2 \Bbb i, 2+3 \Bbb i, 3, 3+ \Bbb i, 3+2 \Bbb i, 4 \}$ - a total of $17$ elements.

How many fields with $17$ elements are there? Up to an isomorphism, the only one is $\Bbb Z _{17}$ and you are done.

Alex M.
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