0

I want to show that $L^+$ is non-empty where $L$ is a full-rank integer lattice and $L^+$ denotes the set of elements of $L$ having positive coordinates I have an indication that I did not understand $L \otimes \mathbb{Q} ⊆ \mathbb{Z}^n \otimes \mathbb{Q}$ is equal to $\mathbb{Q}^n$; any totally positive element of $\mathbb{Q}^n$ has hence an integer multiple in $L$ thanks

anomaly
  • 25,364
  • 5
  • 44
  • 85
mehyeddine
  • 21
  • 4

1 Answers1

1

Choose $x\in \mathbb{Q}^n$ with all $x_i > 0$. Since $L$ is a full-rank lattice, there exists some nonzero $n\in \mathbb{Z}$ with $nx\in L$. Assuming without loss of generality that $n > 0$, the point $nx\in L^+$.

hardmath
  • 37,015
anomaly
  • 25,364
  • 5
  • 44
  • 85
  • why if $L$ is a full-rank lattice, there exists some nonzero $v∈Z $ with vx∈L$ – mehyeddine Jun 30 '14 at 09:12
  • Every element of $L\otimes \mathbb{Q}$ is of the form $\lambda \otimes q$ for $\lambda\in L$ and $q\in \mathbb{Q}$. – anomaly Jun 30 '14 at 16:47
  • more details please.i don't understood – mehyeddine Jun 30 '14 at 17:42
  • If you have integer vectors that are linearly dependent over $\mathbb{Q}$, they are also linearly dependent over $\mathbb{Z}$. Conversely if you have full-rank $\mathbb{Z}^n$ lattice, there are $n$ linearly independent (over $\mathbb{Z}$) vectors, and these are linearly independent over $\mathbb{Q}$, and hence they span $\mathbb{Q}^n$. – hardmath Jul 01 '14 at 00:10
  • ok I understand that the basis of $L$ span $Q^n$. But i don't understand the role of $L⊗Q⊆Q^n$. more détaille please – mehyeddine Jul 01 '14 at 17:39