Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am stuck with these hard-star exercises for some time now (they are from a book called "Introduction a L'Algebre et L'Analyse Modernes" de M.Zamansky"), if somebody sees the right way, I will be very grateful to know 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 $u_{3}=0$

1.** If $q_{1}>0, q_{2}> 0 , q_{3}>0$, then $u_{1},u_{2},u_{3}$ can be chosen so that $0\le q_{21}, q_{31}< q_{1}, 0 \le q_{32} < q_{2}$
2.** It is possible to find $q_{1},q_{2},q_{3},q_{21},q_{31},q_{32}$ for $L=\{(x_{1},x_{2},x_{3}) \in \mathbb{Z}^{3}; 2x_{1}+4x_{2}+5x_{3} = 0 mod 7 \}$

I have come this far:

  1. Assume one can not choose $u_{1}, u_{2}, u_{3}$ with $q_{1}, q_{2}, q_{3} > 0$ so that $0\le q_{21}, q_{31}< q_{1} , 0\le q_{32} < q_{2}$. I thought about giving a counterexample with numbers but since one doesn't know what L is that is not really possible…

  2. One looks at $2x_{1}+4x_{2}+5x_{3} = 7$. $2x_{1}$ and $4x_{2}$ are always even. I don't see how to continue. But why is the book asking for $q_{21}, q_{31}, q_{32}$, too, if they can be set to 0?

As you see it is not very much… Maybe for solving this it helps to know the solution of the previous exercise: If L is a subgroup of $\mathbb{Z}^{3}$, linearly independent, linear equations .

share|improve this question
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.