1
$\begingroup$

Define all points in the affine integral lattice $\mathcal{L}=\{(x,y,z,t) : x+y+z+t=5$ and $x-z \equiv 0$ (mod $12$)$\} \subset \mathbb{Z}^4$.

This is a question from a practice exam I have with no solutions. I'm not certain what to do but this is what I have so far. It could be completely wrong so take it with a pinch of salt. I will be using an algorithm that our lecturer described in one of our lectures.

First define $H =\{(x,y,z,t) : x+y+z+t=0$ and $x-z \equiv 0$ (mod $12$)$\}$.

We will be considering $\pi_i:\mathbb{Z}^4\rightarrow \mathbb{Z}$.

Consider $\pi_4:\mathbb{Z}^4\rightarrow \mathbb{Z}$.

Then ker($\pi_4$) we choose a vector satisfying the conditions stated in $H$. i.e. $(1,9,1,1) \in H$

Next we consider ker($\pi_4)\cap H=\{(x,y,z,0) : x+y+z=0$ and $x-z \equiv 0$ (mod $12$)$\}$.

This time I choose the vector $(1,10,1,0) \in$ ker($\pi_4)\cap H$.

I then go to do this again and find that the next two vectors must be $0$, as in the first case I consider $\{(x,y,0,0) : x+y=0$ and $x \equiv 0$ (mod $12$)$\}$ so $x$ must be zero and hence so must $y$, and similarly for the next case.

So I therefore think that I have a basis for $H$, i.e.

$$ H= \begin{bmatrix} 1 \\ 9 \\ 1 \\ 1 \\ \end{bmatrix} \cdot \mathbb{Z} + \begin{bmatrix} 1 \\ 10 \\ 1 \\ 0 \\ \end{bmatrix}\cdot \mathbb{Z} $$

And then to rewrite to include the $5$ I write: $$ \mathcal{L}= \begin{bmatrix} 0 \\ 5 \\ 0 \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 1 \\ 9 \\ 1 \\ 1 \\ \end{bmatrix} \cdot \mathbb{Z} + \begin{bmatrix} 1 \\ 10 \\ 1 \\ 0 \\ \end{bmatrix}\cdot \mathbb{Z} $$

So basically my question is: have I answered the question and/or have I found a basis for $\mathcal{L}$?

EDIT: I edited the

$$ \begin{bmatrix} 0 \\ 5 \\ 0 \\ 0 \\ \end{bmatrix} $$

from a $$ \begin{bmatrix} 5 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix} $$ so that it wouldn't effect the $x-z$.

$\endgroup$
1
$\begingroup$

Your second vector, $$ \begin{bmatrix} 1 \\ 10 \\ 1 \\ 0 \end{bmatrix}, $$ is not a member of $H$.

Specifically, $$1+10+1 = 12 \ne 0.$$ Notice that the first equality is not modulo 12.

This means that your basis is not a basis for $H$.

$\endgroup$
0
$\begingroup$

I realised that I made a mistake in my answer.

The third and fourth vector should respectively be defined as:

$ \begin{bmatrix} 12 \\ -12 \\ 0 \\ 0 \\ \end{bmatrix} $ and $ \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix} $

Therefore the basis is as follows:

$$ \mathcal{L}= \begin{bmatrix} 0 \\ 5 \\ 0 \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 1 \\ 9 \\ 1 \\ 1 \\ \end{bmatrix} \cdot \mathbb{Z} + \begin{bmatrix} 1 \\ 10 \\ 1 \\ 0 \\ \end{bmatrix}\cdot \mathbb{Z} + \begin{bmatrix} 12 \\ -12 \\ 0 \\ 0 \\ \end{bmatrix}\cdot \mathbb{Z} $$

$\endgroup$

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.