Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This exercise is from a book called "Introduction a L'Algebre et L'Analyse Modernes" de M.Zamansky, I attempted to solve. But I don't know if my solutions are correct (they seem too short to be correct). I am very grateful if somebody could take a look at it.

Let L be a subgroup of $\mathbb{Z}^{3}$. Let $ q_{1}\mathbb{Z}$, $(q_{1}\ge 0)$ be the group of all $ x_{1} \in \mathbb{Z}$ with $ (x_{1},0,0) \in L$ and let $\displaystyle u_{1}=(q_{1},0,0)$. Let $q_{2}\mathbb{Z}$, $(q_{2}\ge 0)$ be the group of all $x_{2} \in \mathbb{Z}$ so that there exists $x_{21} \in \mathbb{Z}$ with $(x_{21},x_{2},0) \in L$. If $q_{2}>0$ then $ u_{2}=(q_{21},q_{2},0) \in L$; otherwise $u_{2}=0$. Let $ q_{3}\mathbb{Z}$, $(q_{3}\ge 0)$ be the group of all $x_{3} \in \mathbb{Z}$ so that there are $x_{31},x_{32} \in \mathbb{Z}$ with $ (x_{31},x_{32},x_{3}) \in L$. If $ q_{3} > 0$ then $ u_{3} = (q_{31},q_{32},q_{3}) \in L$ otherwise $u_{3}=0$
i) It holds that: $L= \mathbb{Z}u_{1} + \mathbb{Z}u_{2}+ \mathbb{Z} u_{3}$
ii) If $L\ne \{0\}$ then the $u_{i}$ with $q_{i}>0 $ are $\mathbb{Z}$ linearly independent.
iii) There are $q_{1},q_{2},q_{3}$ for $L= \{(x_{1},x_{2},x_{3}) \in \mathbb{Z}^{3}; 2x_{1}+4x_{2} + 5x_{3} = 0 \}$

Attempt ( with Dylan Moreland's hints) :

i) Let $x=(x_{1},x_{2},x_{3})$ be an element of L, because $u_{3}=(q_{31},q_{32},q_{3})$ and the third coordinate is only found in $u_{3}$ and not in $u_{1}, u_{2}$, there $\exists a_{3} \in \mathbb{Z}$ so that $a_{3}u_{3} = x_{3}$. Similarly, there are $a_{1},a_{2} \in \mathbb{Z}$ so that $a_{1}q_{1}+a_{2}q_{21}+a_{3}q_{31} = x_{1}$ and $a_{2}q_{2}+a_{3}q_{32} = x_{2}$. But this is the same as saying $a_{1}u_{1}+a_{2}u_{2}+a_{3}u_{3} = (x_{1},x_{2},x_{3})$, so $L= \mathbb{Z}u_{1} + \mathbb{Z}u_{2} + \mathbb{Z} u_{3}$

ii) Suppose we have $a_{i} \in \mathbb{Z}$ such that:$$a_{1}u_{1}+a_{2}u_{2}+a_{3}u_{3}=0$$ If u_{3}=0, then we can ignore it, otherwise we must have $a_{3}=0$. If $u_{2}=0$, then we can ignore it, otherwise we must have $a_{2}=0$. If $u_{1}=0$, then we can ignore it, otherwise we must have $a_{1}=0$. So $u_{i}$ with $q_{i}$ must be $\mathbb{Z}$linearly independent.

Can't we just write the 3 vectors into a matrix like : $\begin{pmatrix}q_{1} & 0& 0 \\ q_{21} & q_{2} & 0 \\ q_{31} & q_{32} & q_{3} \end{pmatrix}$

This matrix is a 3x3 matrix with rank 3, so its vectors must be linearly independent!

iii) If $(x_{1},0,0) \in L$, then $2x_{1} = 0$, so we can put $u_{1}=0$. If $(x_{1},x_{2},0) \in L$,then $2x_{1}+4x_{2}=0$, so $x_{1}=-2x_{2}$ and we can put $u_{2}= (-2,1,0)$. If $(x_{1},x_{2},x_{3}) \in L$, then $2x_{1}+4x_{2}= + 5x_{3} = 0$. If we put $x_{3} = x_{1}+x_{2}$, then we get $7x_{1}+9x_{2} = 0$, and this has solutions for $x_{1}=-9$ and $x_{2}= 7$ , so $u_{3} = (-9, 7, 2)$

These are my old attempts :

i) Assume $L=(l_{1},l_{2},l_{3}) $If one can show that $\mathbb{Z}(q_{1},0,0)+\mathbb{Z}(q_{21},q_{2},0)+ \mathbb{Z}(q_{31},q_{32},q_{3}) - (l_{1},l_{2},l_{3})=0$, then $L=\mathbb{Z}u_{1} + \mathbb{Z}u_{2}+ \mathbb{Z} u_{3}$. Now one can fix $\mathbb{Z}_{1}, \mathbb{Z}_{2}, \mathbb{Z}_{3}$ so that $\mathbb{Z}_{1}u_{1} + \mathbb{Z}_{2}u_{2}+\mathbb{Z}_{3}u_{3}-L$ = 0 with $L=(\mathbb{Z}_{1}q_{1}+\mathbb{Z}_{2}q_{21}+\mathbb{Z}_{3}q_{31}, \mathbb{Z}_{2}q_{2}+\mathbb{Z}_{3}q_{32}, \mathbb{Z}_{3}q_{3})$, so $L=\mathbb{Z}u_{1} + \mathbb{Z}u_{2}+ \mathbb{Z} u_{3} $

ii) Assume $u_{i}$ with $q_{i}>0$ are $\mathbb{Z}$ linearly dependent. Then all possible combinations are: $\mathbb{Z}u_{2} = \mathbb{Z}u_{1}$ or $\mathbb{Z}u_{3}; \mathbb{Z}u_{1}=\mathbb{Z}u_{2}$or $\mathbb{Z}u_{3}, \mathbb{Z}u_{3}=\mathbb{Z}u_{1}$ or $\mathbb{Z}u_{2}$; $\mathbb{Z}u_{1} - \mathbb{Z}u_{2} = \mathbb{Z} u _{3}; \mathbb{Z}u_{1}- \mathbb{Z}u_{3}= \mathbb{Z}u_{2}; \mathbb{Z}u_{3}- \mathbb{Z}u_{2} = \mathbb{Z}u_{1}$

and one of them has to be true. Since no one is, the $u_{i}$ must be independent.

iii) One can set $x_{21},x_{32},x_{31}$ to 0 and let : $x_{3}=x_{1}+x_{2}$, which is the same as $7x_{1}+9x_{2}=0$ and this has solutions for $x_{1}=-9n, x_{2}=7n $ where $n\in \mathbb{Z}$

share|cite|improve this question
"$q_{1,2,3}\mathbb{Z}$" has no meaning in the exercise. I can guess what you mean, but I shouldn't have to. And from my guess, it would seem you aren't proving anything, you are just asserting your conclusions. – Arturo Magidin Dec 8 '11 at 2:30
I changed it, sorry. – VVV Dec 8 '11 at 18:02
@VVV: Moderators are unable to separate questions. – Zev Chonoles Dec 9 '11 at 18:37
up vote 2 down vote accepted

(i) Let $l = (l_1, l_2, l_3)$ be an element of $L$; we'd like to write this as a linear combination of $u_1, u_2, u_3$. Since $u_3$ is the only purported generator that has anything to do with the third coordinate in $\mathbf Z^3$, let's begin there. From the construction of $u_3$, we see that there exists an $a_3 \in \mathbf Z$ such that the third coordinate of $a_3q_3$ is $l_3$. In other words, $l - a_3u_3$ has a zero in the third coordinate. Let's write it out: \[ (l_1 - a_3q_{31}, l_2 - a_3q_{32}, 0). \] Note that $l - a_3u_3$ is also in $L$, so on it you could try to run this argument again, working with the second coordinate and $u_2$. Does this help?

(ii) Suppose we have $a_1, a_2, a_3 \in \mathbf Z$ such that \[ a_1u_1 + a_2u_2 + a_3u_3 = 0. \] If $u_3 = 0$, then ignore it; otherwise, we must have $a_3 = 0$. Continue in this way.

(iii) Use the given recipe. If $(l_1, 0, 0) \in L$, then $2l_1 = 0$ and hence $u_1 = 0$. If $(l_1, l_2, 0) \in L$ then $2l_1 + 4l_2 = 0$, and so $l_1 = -2l_2$. Thus $l_2$ can be anything, so $u_2 = (-2, 1, 0)$. Can you find a $u_3$?

[I'd like to add some words on matrices and Smith normal form later. Also, a similar proof shows the more general fact that a submodule of a rank $n$ free module over a PID is also free of rank $\leq n$.]

share|cite|improve this answer
Thank you very much.How do you tell coordinates? $x-a_{3}u_{3}$ is the same as : $(x_{1},x_{2},x_{3})-a_{3}(x_{31},x_{32},x_{3})$? You used x instead of q, but then why is u defined with q and not with x? – VVV Dec 9 '11 at 0:03
@VVV I'm trying to avoid the notation in the book because I think it's just confusing. Try to forget about $x_\text{double subscript}$ for now. – Dylan Moreland Dec 9 '11 at 5:24
@VVV I think I understand your question better now. The point is that there is some $a_3 \in \mathbf Z$ such that $a_3q_3 = x_3$. I didn't see the need to mention this intermediate step, but perhaps I should have. – Dylan Moreland Dec 9 '11 at 8:27
Thank you!!! I edited my answer using your hints. Please tell me if this is what you had in mind. – VVV Dec 9 '11 at 9:12
@VVV Sure, let me look it over. I will try to edit the answer to use the book's notation later on, I think. – Dylan Moreland Dec 9 '11 at 19:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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