0
$\begingroup$

I am having trouble understanding the following proof relating to factor rings:

Prove that $\mathbb{Z}[i]/\langle 2+i \rangle = \{\overline{0}, \overline{1}, \overline{2}, \overline{3}, \overline{4}\}$ where $\overline{a} = a + \langle 2+i \rangle$.

I think that this implies that

$\overline{0} = 0 + \langle 2+i \rangle$.

$\overline{1} = 1 + \langle 2+i \rangle$.

$\overline{4} = 4 + \langle 2+i \rangle$.

where $\langle 2+i \rangle$ is a principal ideal.

And I am also aware that there exists some $k \in \mathbb{Z}$ such that $a+bi+k(2+i) \in \mathbb{Z}$ which is $k = -b$.

I am confused because I think that the definition of ideal implies that since $k \in \mathbb{Z}$ and therefore $k \in \mathbb{Z}[i]$, then $k(2 + i)$ where $(2+i) \in \langle 2+i \rangle$ is also in the ideal. So in this case is each $\overline{a}$ only comprised of whole integers, since we can choose some $k$ to obtain an integer for every number $0-4$? Or does each $\overline{a}$ also include some numbers which are not whole integers?

Furthermore, part of this proof will involve proving that $5 \mathbb{Z} = \langle 2+i \rangle \cap \mathbb{Z}$. I am unsure of what elements the principal ideal has in this case so I am having trouble thinking about what the intersection on the right hand side will be. Is the principal ideal only allowed to have complex numbers in this case? or can it also have integers? If so, how would we obtain an integer element?

I appreciate any help! Thank you.

$\endgroup$
5
  • $\begingroup$ What is means is that for each $m+ni\in\Bbb Z[i]$ is in exactly one of the five sets $0+\left<2+i\right>,\ldots,4+\left<2+i\right>$. $\endgroup$ – Angina Seng Apr 24 '20 at 5:13
  • $\begingroup$ @AnginaSeng Thank you, but are these sets comprised of only integers or only of complex numbers, or both? $\endgroup$ – wombatz Apr 24 '20 at 5:16
  • $\begingroup$ Note $(2+i)(2-i)=5$ $\endgroup$ – J. W. Tanner Apr 24 '20 at 5:18
  • $\begingroup$ $a+\left<2+i\right>=\{a+(m+ni)(2+i):m,n\in\Bbb Z\}$. $\endgroup$ – Angina Seng Apr 24 '20 at 5:19
  • $\begingroup$ Since $\mathbb{Z}[i]$ is a conmutative ring with identity $\langle 2+i\rangle$ consists in the set $\langle 2+i\rangle=\lbrace (a+bi)(2+i)~\mid~a, b\in\mathbb{Z}\rbrace$ so if $(a+bi)(2+i)\in\langle 2+i\rangle$ multiplying $2a-b+(a+2b)i=(a+bi)(2+i)$ whence $(a+bi)(2+i)\in\mathbb{Z}$ iff $a=-2b$ so using the last equality you'll see that $(a+bi)(2+i)\in 5\mathbb{Z}$. On the other hand $5q=(2q-qi)(2+i)\in\langle 2+i\rangle$. Hope it helps you! $\endgroup$ – Yeipi Apr 24 '20 at 5:21
0
$\begingroup$

You have several questions here. I'll look at why $\left<2+i\right>\cap\Bbb Z=5\Bbb Z$.

The elements of $\left<2+i\right>$ are $(m+ni)(2+i)$ for $m$, $n\in\Bbb Z$. Now $$(m+ni)(2+i)=(2m-n)+(m+2n)i.$$ That's only in $\Bbb Z$ when $m=-2n$. In this case, then $$(m+ni)(2+i)=(-2n+ni)(2+i)=-5n.$$ So we can get all multiples of $5$, but only them, as integer elements of the ideal $\left<2+i\right>$.

Follow-up exercise: prove $\left<3+2i\right>\cap\Bbb Z=13\Bbb Z$.

$\endgroup$
1
$\begingroup$

The easiest way to see this is using the ring isomorphism theorems. $$\frac{\mathbf Z[i]}{(2+i)}\stackrel{(1)}{\cong} \frac{\mathbf Z[X]/(X^2+1)}{(2+X)}\stackrel{(2)}{\cong} \frac{\mathbf{Z}[X]/(2+X)}{(X^2+1)}\stackrel{(3)}{\cong} \frac{\mathbf Z}{(5)}\cong \mathbf{F}_5$$

$(1)$: We can write $\mathbf{Z}[i]\cong \mathbf{Z}[X]/(X^2+1)$.

$(2)$: This is the third isomorphism theorem.

$(3)$: Here we send $X\mapsto -2$.

$\endgroup$
0
$\begingroup$

As a result of $5\mathbb{Z}=\langle 2+i\rangle\cap\mathbb{Z}$ you have $\mathbb{Z}/5\mathbb{Z}=\mathbb{Z}/\left(\langle 2+i\rangle\cap\mathbb{Z}\right)$ so using second isomorphism theorem for groups $$\mathbb{Z}/5\mathbb{Z}=\mathbb{Z}/\left(\langle 2+i\rangle\cap\mathbb{Z}\right)\cong\left(\mathbb{Z}+\langle 2+i\rangle\right)/\langle 2+i\rangle$$

But as you've said $a+bi+k(2+i)\in\mathbb{Z}$ which means that for every $a+bi\in\mathbb{Z}[i]$ then $a+bi\in\mathbb{Z}+\langle2+i\rangle$. Therefore $$\mathbb{Z}/5\mathbb{Z}\cong\mathbb{Z}[i]/\langle2+i\rangle$$

Notice that the above isomorphism is just a group isomorphism, not a ring isomorphism but it gives you a bijection between the field with 5 elements and $\mathbb{Z}[i]/\langle 2+i\rangle$.

$\endgroup$
0
$\begingroup$

In general, the number of residue classes in $\Bbb Z[i]$ modulo a Gaussian integer $a+bi$ is the norm $n(a+bi)=a^2+b^2$.See here for a proof: "Gaussian integer - Wikipedia" https://en.m.wikipedia.org/wiki/Gaussian_integer

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.