It is pretty easy to visualize the ideals of $\mathbb{Z}$ in the "integer line".

Let's go up to $\mathbb{Z}[i]$ and consider the ideal $3\cdot\mathbb{Z}[i]$.

We can visualize it as a "sub-lattice" of the gaussian integers that contains the vertex $0+0i$ and has edges of lenght 3. This is because

\begin{equation} 3\cdot\mathbb{Z}[i]=\{(3a)+(3b)i)\:|\:a,b\in\mathbb{Z}\} \end{equation}

and so this is kind of obvious and intuitive.

What about, for example, the ideal $(2+i)\cdot\mathbb{Z}[i]$?

We can see a copy of $5\cdot\mathbb{Z}$ inside it and also the elements of $\mathbb{Z}[i]$ lying over the line $y=\frac{1}{2}x$. Are there any others points I am missing? I think that yes, but I'm not sure how to find those.

Every element of $(2+i)\cdot\mathbb{Z}[i]$ is of the form $(2a+b)+(a+2b)i$, but is not clear to me which geometric figure we get from this. I'd appreciate if someone could explain it to me.

Thank you in advance.

Also, a small extra question: Is there any online website where I can draw things in the plane described by 2 parameters? That would answer my question, even though an algebraic explanation would be great.

  • 2
    $\begingroup$ It is still a square sub-lattice, but it is skew - the sides of the squares are not parallel to the axes. - for example, with basis elements are $2+i$ and $i(2+i)=-1+2i$. Multiplication by $i$ is tantamount to rotating by 90 degrees. $\endgroup$ – Thomas Andrews Oct 13 '15 at 18:39
  • 1
    $\begingroup$ So, tile the plane with squares where one side of the square is the segment between $0$ and $2+i$. $\endgroup$ – Thomas Andrews Oct 13 '15 at 18:41
  • $\begingroup$ Not sure I understand your first comment. Like a square sub-lattice that has been rotated? $\endgroup$ – Shoutre Oct 13 '15 at 18:43
  • 1
    $\begingroup$ Draw a picture of $2+1$ and $-1+2i$ and a few somes of $a(2+i)+b(-1+2i)$ and you'll see the pattern (with $a,b$ normal integers.) $\endgroup$ – Thomas Andrews Oct 13 '15 at 18:46
  • 1
    $\begingroup$ Because every element is of the form $a(2+i)+b(-1+2i)=(a+bi)(2+i)$ $\endgroup$ – Thomas Andrews Oct 13 '15 at 18:47

It's just a square lattice enlarged by $\sqrt{2^2 + 1^2 } = \sqrt{5}$ and rotated by $\theta = \tan^{-1}(\frac{1}{2})$

enter image description here

Another way to think of it is there are 5 possible translates of this subgroup: $\mathbb{Z}[i]/(2+1j)\mathbb{Z}[i] \simeq \mathbb{Z}/5\mathbb{Z}$

enter image description here

  • $\begingroup$ Can I say that the quotient $\mathbb{Z}[i]/(2+i)\mathbb{Z}[i]$ has 5 elements because there are 5 "possible translations" somewhat like we say that $\mathbb{Z}/\mathbb{3}\mathbb{Z}$ has 3 elements because we can "translate" the ideal 3 ways? $\endgroup$ – Shoutre Oct 13 '15 at 18:56
  • $\begingroup$ Beautiful pictures! $\endgroup$ – Viktor Vaughn Oct 13 '15 at 20:51
  • $\begingroup$ @Shoutre: yes, that's right. $\endgroup$ – Ravi Fernando Oct 18 '15 at 6:43

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.