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A free module is a module with a basis of a module $(e_i{)}_{i\in I}$: every element has exactly one expression $m={\sum }_{i\in I}{r}_i{e}_i$ with $r_i\in R$ and only finitely many $r_i$ nonzero. Equivalently, the coefficient map ${⨁}_{i\in I}R\to M$ is a module isomorphism. Existence means generation; uniqueness means linear independence in a module. For finite $I$ of size $n$, this says $M\cong R^n$, a finite free module. See Neil Strickland's Rings, Modules and Linear Algebra, section 11.

The integer lattice ${\mathbb{Z}}^2$ in the opening diagram is a free module over $\mathbb{Z}$ with basis of a module $e_1=(1,0)$, $e_2=(0,1)$. For example, $(3,2)=3e_1+2e_2$, and comparing the two coordinates makes the coefficients unique. A different basis of a module is $u=(1,0)$, $v=(1,1)$, because $(a,b)=(a-b)u+bv$; in particular $(3,2)=u+2v$.

By contrast, $w_1=(2,0)$ and $w_2=(0,1)$ are linearly independent in a module, but generate only the submodule $2\mathbb{Z}\times \mathbb{Z}$. They miss $(1,0)$ because $2a=1$ has no integer solution. They would form a basis over the rational numbers. Even generation need not let one extract a basis of a module: the elements $2,3$ generate $\mathbb{Z}$, since $1=3-2$, but neither singleton generates it, and $3\cdot 2-2\cdot 3=0$ is a nontrivial linear relation between them. These are concrete failures of the usual vector space argument that divides by any nonzero coefficient.

The cyclic module $\mathbb{Z}/6\mathbb{Z}$ has no basis of a module over $\mathbb{Z}$: every element has the nontrivial relation $6m=0$, so no nonempty subset is linearly independent in a module, while the empty subset cannot generate this nonzero module. Specifying the scalar ring is essential: over $R=\mathbb{Z}/6\mathbb{Z}$, the same additive object is a free module with the one-element basis of a module $\{\overline{1}\}$, since each coefficient is itself a unique class in $R$.

A translation and positive uniform scaling of a finite configuration $F$ in a real vector space. It preserves the ratios and directions of differences between its points. In an integer lattice, the Gallai theorem for an integer lattice guarantees a monochromatic homothetic copy of every finite pattern in any finite colouring, with positive integer scaling factor.

For a square integer matrix $A$ with nonzero determinant, the image $A{\mathbb{Z}}^n$ is a subgroup of the integer lattice. Its index of a subgroup isIn particular, $A=\mathrm{diag}(2,1)$ gives all points with even first coordinate. The quotient has two cosets, represented by $(0,0)$ and $(1,0)$, and the fundamental cell has area two. A determinant of two therefore counts how much integer information is missing from the image.

The integrality hypothesis is essential. The rational matrix $R=\mathrm{diag}(2,1/2)$ has determinant one but sends $(0,1)$ to $(0,1/2)$. Its generated Euclidean lattice has the same covolume as ${\mathbb{Z}}^2$ without being the same integer lattice. The unimodular matrix criterion combines an arithmetic condition with a geometric one.

A short proof of the index formula uses Smith normal form. Write $PAQ=\mathrm{diag}(d_1,\dots ,d_n)$, with $P,Q$ unimodular matrices and positive integers $d_i$. They preserve the relevant integer structure, so the quotient has the same size as ${\prod }_i\mathbb{Z}/d_i\mathbb{Z}$, namely ${\prod }_i{d}_i=|\det A|$. No assumption that the matrix is orthogonal or its basis vectors are short is needed.

Since $2\not\equiv 1(\mathrm{mod}4)$, the ring of integers of $\mathbb{Q}(\sqrt{2})$ is $\mathbb{Z}[\sqrt{2}]$, with integral basis $1,\sqrt{2}$ and field discriminant eight.

For a finite pattern $\{v_1,\dots ,v_k\}$ in a real vector space, pull a finite colouring back to words through the displayed sum map. A combinatorial line with active coordinates J maps to $a+|J|\{v_1,\dots ,v_k\}$. The Hales-Jewett theorem therefore forces a monochromatic homothetic copy. Integer patterns remain in the integer lattice; choosing b sufficiently large gives the positive-lattice version.

A unimodular matrix is a square integer matrix whose determinant is $1$ or $-1$. The decisive property is that its inverse matrix also has integer entries: multiplication by $U$ gives a bijection of the integer lattice ${\mathbb{Z}}^n$ onto itself. Thus an invertible linear map can preserve not only real coordinates but the requirement that every coordinate is an integer. These matrices form the integral general linear group.^[ref]

For an original two-dimensional example, takeIts determinant is one, and it maps $(a,b)$ to $(a+b,b)$. Conversely, the integer point $(x,y)$ comes from $(x-y,y)$. For instance, $(3,2)$ has new coordinates $(1,2)$ relative to the columns $(1,0)$ and $(1,1)$. No integer point disappears, although the basis vectors and their fundamental cell change shape.

This is useful whenever coordinates must remain integral. A unimodular change of lattice basis can give a shorter or more convenient description without changing the objects represented. Smith normal form and Hermite normal form use controlled integer operations to simplify integer matrices; ordinary Gaussian elimination may divide by numbers that destroy integrality. The geometric freedom is substantial: a sheared basis may look completely different while describing exactly the same infinite set of integer points.

A determinant of $-1$ is allowed too. Swapping two basis columns reverses orientation while preserving the integer lattice. The condition is therefore preservation of the complete integer structure, rather than a requirement to preserve orientation.