$\newcommand\L{\mathcal{L}}\newcommand\b{\mathbf{b}}$ I always thought that the dimension of a lattice is equal to the number of vectors in its basis. I. e., $$ \L(\b_1,...,\b_n)= \left\{\sum_{i=1}^n x_i\b_i:x_i \in \mathbb{Z}\right\} $$ has dimension $n$. This belief was due to,
- Hoffstein, Jeffrey, et al. An Introduction to Cryptography (pg. 388). Springer New York, 2014.
The dimension of $\L$ is the number of vectors in a basis for $\L$.
$\L$ is equal to the set $\L(\b_1,...,\b_d)= \left\{\sum_{i=1}^d x_i\b_i:x_i \in \mathbb{Z}\right\}$ of all integer linear combinations of the $\b_i$'s. The integer $d$ is the dimension of the lattice $\L$.
Note that in Gama et al. we have $d=n$.
However, recently, while reading,
- Micciancio, Daniele, and Shafi Goldwasser. Complexity of lattice problems: a cryptographic perspective. Vol. 671. Springer Science & Business Media, 2002.
page 1 has the same equation (the first one in this post) and says,
A lattice in $\mathbb{R}^m$ is the set of all integral combinations of $n$ linearly independent vectors $\b_1,...\b_n$ in $\mathbb{R}^m (m\geq n)$. The integers $n$ and $m$ are called the rank and dimension of the lattice, respectively.
So it seems that what the first two definitions call dimension, this third definition calls rank! E. g., The lattice $\L(\b_1), \b_1=[0, 1]$ is one dimensional according to the first two definitions and two dimensional according to the third while having a rank of 1.
Could someone clarify if there no standard it depends on the author or am I missing something?