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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.

A lattice path is a sequence of points in an integer lattice whose consecutive differences belong to a prescribed finite set of steps.

For instance, a north-east lattice path uses only unit steps east or north in the integer lattice ${\mathbb{Z}}^2$. Restricting the allowed steps makes the counting question precise: the endpoint fixes how many of each step must occur, but their order remains free. Pascal's rule counts these lattice paths by their last step; the same binomial coefficient counts selections of step positions.

For positive integers $a,b$, the north-east lattice path recurrence isThe boundaries have only straight north-east lattice paths. The origin counts the zero-step north-east lattice path once.^[ref]

Proof outline. A north-east lattice path ending at $(a,b)$ has a final step either from $(a-1,b)$ or from $(a,b-1)$. Deleting that step gives a shorter north-east lattice path; appending it reverses the operation. The two classes are disjoint and exhaust all possibilities, so the rule of sum adds their counts. At $(4,3)$, the arrivals from $(3,3)$ contribute $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$, while those from $(4,2)$ contribute $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$: altogether $20+15=35$.

Set $P(a,b)=0$ if either integer coordinate is negative. Then the same recurrence relation applies on the nonnegative axes away from $(0,0)$: one predecessor contributes one, and the other contributes zero. Initialize $P(0,0)=1$ separately; applying the recurrence relation there would incorrectly give zero.

Computing the grid once in increasing order of $a+b$ is dynamic programming: every entry uses two earlier entries. If a specified point is forbidden, set its count to zero, then continue the same recurrence relation at allowed points. This counts north-east lattice paths avoiding the forbidden points, provided the origin and endpoint are allowed. The last-step argument remains valid because appending a step to an allowed endpoint cannot visit any new intermediate integer lattice point.

Use the same triangle and let $G$ consist of translations $x\mapsto x+m$ for $m\in {\mathbb{Z}}^2$. All its vertices have integer coordinates, and one vertex is the origin, soThe rotation through a quarter-turn $(x,y)\mapsto (-y,x)$ preserves this integer lattice, and hence belongs to $H$. It fixes the origin and is not the identity, whereas the only translation fixing the origin is the identity. It is therefore outside $G$, proving $G⊊H$. This is symmetry enlargement under group translates even though $T$ itself has trivial stabilizer subgroup.

Since $2\not\equiv 1(\mathrm{mod}4)$, the ring of integers of $\mathbb{Q}(\sqrt{2})$ iswith integral basis $1,\sqrt{2}$ and field discriminant eight. Multiplication is ordinary polynomial multiplication followed by $(\sqrt{2}{)}^2=2$:For example, $(1+\sqrt{2}{)}^2=3+2\sqrt{2}$ and $(1+\sqrt{2}{)}^3=7+5\sqrt{2}$. The integer coefficients stay integral, as required for a ring.

Under the two field embeddings, the element $a+b\sqrt{2}$ becomes the point $(a+b\sqrt{2},a-b\sqrt{2})$. The integral basis vectors become $(1,1)$ and $(\sqrt{2},-\sqrt{2})$, giving a Minkowski embedding of a real quadratic field. Their determinant has absolute value $2\sqrt{2}=\sqrt{8}$, consistent with the field discriminant.

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.

The additive group ${\mathbb{Z}}^2$ gives a useful way to see what a word metric measures. Let $e_1=(1,0)$ and $e_2=(0,1)$, and choose the four-move generating set of a group $S=\{\pm e_1,\pm e_2\}$. Its Cayley graph is the square lattice. Each move changes exactly one coordinate by one, so the word length is the L1 norm, and the word metric is Manhattan distance:At least this many moves are necessary, and moving horizontally then vertically attains the bound. Bridson gives this square lattice as a basic Cayley graph example.^[ref]

Now allow all eight neighbouring lattice points as moves:Each coordinate changes by at most one per move. Use diagonal moves until the smaller displacement is exhausted, then use axis moves: the resulting word length is the maximum norm, and the word metric is Chebyshev distance. From $(0,0)$ to $(2,3)$, an $S$-route has five steps, while the $T$-route $(0,0),(1,1),(2,2),(2,3)$ has three.

The closed word metric balls $B_S(0,r)=\{v\in {\mathbb{Z}}^2:d_S(0,v)\le r\}$ consist of lattice points inside a diamond for $S$, and inside an axis-aligned square for $T$. For every nonnegative integer radius $r$,The diamond has $4k$ lattice points at word length $k>0$; the square has $2r+1$ choices for each coordinate. At radius three the counts are $25$ and $49$. These are discrete closed balls for Manhattan distance and Chebyshev distance; an Euclidean distance bound would instead select lattice points inside a Euclidean disk.