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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.

Shoutre
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    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. – Thomas Andrews Oct 13 '15 at 18:39
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    So, tile the plane with squares where one side of the square is the segment between $0$ and $2+i$. – Thomas Andrews Oct 13 '15 at 18:41
  • Not sure I understand your first comment. Like a square sub-lattice that has been rotated? – Shoutre Oct 13 '15 at 18:43
  • Oh, okay, I think I get it. Well, but how can I be sure that all the elements of this ideal are in this sub-lattice? – Shoutre Oct 13 '15 at 18:46
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    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.) – Thomas Andrews Oct 13 '15 at 18:46
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    Because every element is of the form $a(2+i)+b(-1+2i)=(a+bi)(2+i)$ – Thomas Andrews Oct 13 '15 at 18:47
  • Related: https://math.stackexchange.com/questions/1586353 (and http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/modulesoverPID.pdf, page 4) – Watson Nov 22 '18 at 09:47

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

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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

cactus314
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