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Find the elements of $\mathbb Z[i]/\langle 3+2i\rangle$

It is a theorem that $\mathbb Z[i]/(a+bi)$ is isomorphic to $\mathbb Z/(a^2 + b^2)$ if $a,b$ are coprime, so we know $\mathbb Z[i]/\langle 3+2i\rangle$ will have 13 elements.

A reasonable first hypothesis would be to suggest all elements that are in a sense "smaller" than $3+2i$ will not be divisible by it, so we first posit that the coset representatives for $0, 1, 2, 3, i, 2i, 1+i, 2+i, 3+i, 1+2i, 2+2i$ will be 11 of the 13 elements we are looking for.

For the last two elements, another reasonable hypothesis is that they will be elements that fail to be divisible by $3+2i$ on account of a difference of 3 or less between their real part and that of $3+2i$, or similarly wrt. a difference of 2 or less in the imaginary part, because with bigger differences, we're likely to encounter some equivalent coset representative instead. So we have as candidates $3i, 1+3i, 2+3i, 4i, 1+4i, 2+4i$, and maybe a few more, and we find by painstaking trial and error, using $i(3+2i) \equiv 0$, that $2+4i, 2+3i$ seem to represent cosets distinct from all others.

Two questions:

  1. Is there a more efficient way to go about this procedure of finding elements through exhaustion?
  2. How can I prove that these coset representatives are, without a doubt, representatives of nonidentical cosets, i.e., how can I prove that, out of the list of 13 coset representatives I've picked, I've listed exactly 13 cosets and no less?

Edit: thanks to users Dietrich Burde and Arthur, we know that from $-3 \equiv 2i$, we can obtain $5 \equiv i$, from which we can simplify our above set to the seemingly simpler set of coset representatives $\{-6, -5, \dots, 0, 1, \dots, 6\} + \langle 3 + 2i \rangle$. But it is still left to show, i.e., give a proof, that these must be distinct cosets.

shintuku
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  • Compare with this post. There the elements were counted similarly. It is easy to check the list using $2i=-3$. – Dietrich Burde Jun 13 '23 at 18:30
  • Are we assuming $i^2 = -1$ here? – Arthur Jun 13 '23 at 18:36
  • @DietrichBurde thanks for your comment! we can indeed use $2i = -3$ to check some elements of the list when we at least have $2i$, but what about elements like $2+3i, 1+i, 2+i, 3+i$? I will begin studying the post you linked, thanks again – shintuku Jun 13 '23 at 18:37
  • @Arthur yes that's right – shintuku Jun 13 '23 at 18:37
  • @DietrichBurde oh! we obtain $2i = -3$ implies $i = 5$! thank you very much – shintuku Jun 13 '23 at 18:47
  • sorry, I meant $2i \equiv -3$ implies $i \equiv 5$: $2i \equiv -3 \implies -2 \equiv -3i \implies -2 + 2i \equiv -i \implies -2 + -3 \equiv -i$, but it is still left to show ${-6, \dots, 0, \dots, 6} + I$ is a set of distinct cosets – shintuku Jun 13 '23 at 18:54
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    Doesn't the theorem that $\mathbb{Z}[i] / \langle 2 + 3i \rangle \simeq \mathbb{Z} / \langle 13 \rangle$ as rings pretty much immediately imply that ${ -6 + I, \ldots, 6 + I }$ enumerates the elements of the ring? – Daniel Schepler Jun 13 '23 at 19:25
  • @DanielSchepler that makes a lot of sense, but I'm having a hard time trying to figure out the way in which it would imply that the elements of that set are distinct since we aren't given an explicit isomorphism. But this is probably the way to go – shintuku Jun 13 '23 at 19:34
  • Whatever the isomorphism is, since it's a ring homomorphism it must send $-6+I \mapsto -6+\langle 13 \rangle, \ldots, 6+I \mapsto 6+\langle 13 \rangle$ which is all the elements of $\mathbb{Z} / \langle 13 \rangle$. – Daniel Schepler Jun 13 '23 at 19:38
  • @DanielSchepler ah! yes, that's it. thank you very much! – shintuku Jun 13 '23 at 19:39
  • Though one possible description of the isomorphism is $(a+bi) + \langle 3+2i \rangle \mapsto (a+5b) + \langle 13 \rangle$. – Daniel Schepler Jun 13 '23 at 19:39
  • See as well here. – Jean Marie Jun 13 '23 at 20:13
  • Same methods in the linked dupes apply here. Let $,w = 3+2i,$ so $w\bar w = 13 = 0,$ and $\bmod 13!:\ i\equiv -3/2\equiv 10/2\equiv 5\ \ $ – Bill Dubuque Jun 13 '23 at 20:52

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