2
$\begingroup$

Here's my attempt as I think I vaguely remember something similar:

We have $\phi : \mathbb{Z} \to \mathbb{Z}[i]/(7+5i)$ by $\phi(n) = n + (7+5i)$. I would like now to prove that $\ker\phi = \langle 74 \rangle$ and so by the first isomorphism theorem $$\mathbb{Z}[i]/(7+5i) \cong \mathbb{Z} / \langle 74 \rangle= \mathbb{Z}_{74}.$$

Questions:

  1. How do we prove that $\ker \phi = \langle 74 \rangle $
  2. Is the last equals sign correct?
$\endgroup$
6
  • 1
    $\begingroup$ you probably mean $\langle 74 \rangle$ $\endgroup$ Commented Jun 11, 2012 at 17:43
  • $\begingroup$ Can you prove that $74$ is at least contained in the kernel? Do you know that this map is surjective? $\endgroup$ Commented Jun 11, 2012 at 17:51
  • $\begingroup$ Ah yes I do, I've edited that in $\endgroup$ Commented Jun 11, 2012 at 17:52
  • 1
    $\begingroup$ $\phi$ is surjective because $31$ is a square root of $-1$ modulo $74$, hence $\phi(31) \in i + \langle 74 \rangle \subseteq i + \langle 7 + 5i \rangle$. $\endgroup$ Commented Jun 11, 2012 at 17:56
  • 3
    $\begingroup$ Dear user, A very similar question was asked and answered here: math.stackexchange.com/q/23358/221 See this answer in particular. Regards, $\endgroup$ Commented Jun 11, 2012 at 17:59

2 Answers 2

Reset to default
7
$\begingroup$

Hint $\rm\: n\in ker\ \phi\iff 7\!+\!5{\it i}\:|\:n \iff (7\!-\!5{\it i})(7\!+\!5{\it i})\:|\:(7\!-\!5{\it i})\,n\iff 74\:|\:7n\!-\!5n{\it i}\iff 74\:|\:n$

Written in fraction form it amounts to simply rationalizing denominators, viz. for $\rm\:w\ne 0$

$$\rm\ w\:|\:n\ in\ R \iff \frac{n}w \in R\iff \frac{nw'}{ww'}\in R$$

so reducing the problem of divisibility by the complex number $\rm\:w\:$ to the much simpler problem of divisibility by the integer $\rm\:ww'\in \Bbb Z.$

See also my post here on analogous ways of determining units, i.e. divisors of $1$ (vs. $\rm\:n\in \mathbb Z).$

$\endgroup$
7
  • $\begingroup$ Could I ask why $74\:|\:7n\!-\!5n{\it i}\iff 74\:|\:n$? $\endgroup$ Commented May 18, 2021 at 19:37
  • $\begingroup$ @Chris If $\,a,b\in\Bbb Q\,$ then $\,a+b\,i\in \Bbb Z[i]\!\iff\! a,b\in \Bbb Z.\,$ So $\,7n/74 -(5n/74)\,i\in \Bbb Z[i]\,$ $\Rightarrow 7n/74\in \Bbb Z\,$ so $74\mid n$ by $\gcd(74,7)=1$ and Euclid's Lemma. The method used in the answer is a special case of the method of simpler multiples $\endgroup$ Commented May 18, 2021 at 19:59
  • $\begingroup$ Thanks for answering. I also thought that $c|a+bi \iff c|a$ and $c|b$ (in $\Bbb Z$). So, $74|7n-5ni \iff 74|7n$ and $74|5ni \iff 74|n$. What do you say? $\endgroup$ Commented May 18, 2021 at 21:05
  • $\begingroup$ @Chris But you need to justify the first $\!\iff\!$, e.g. as I did in my prior comment. $\endgroup$ Commented May 18, 2021 at 21:11
  • 1
    $\begingroup$ @Chris Yes. Of course we can also prove that without using fractions - using only divisibility language. I presumed that was the step you were asking about in your first comment. $\endgroup$ Commented May 18, 2021 at 21:46
6
$\begingroup$

One can even prove a more general result (and it's a fun exercise, too). Given any $n\in\Bbb Z$ not a perfect square, and any $a,b\in\Bbb Z$, we have that $\Bbb Z[\sqrt{n}]/\langle a+b\sqrt{n}\rangle\cong\Bbb Z/\langle a^2-b^2n\rangle$ iff $a,b$ are coprime in $\Bbb Z$. This problem, then, is just the special case of this more general result with $n=-1$. A key observation, here, is that $a,b$ are coprime in $\Bbb Z$ iff there exist $k,l\in\Bbb Z$ such that $ak+bl=-1$.

If you decide to play with this and get stuck, just let me know, and I can give you some more hints.

$\endgroup$

You must log in to answer this question.