A [vector space](/page/Vector%20Space) can have many different coordinate systems, and a module can have no honest coordinate system at all. Dimension is the invariant that survives the choice of coordinates: it counts how many independent directions are needed, when such a count makes sense, and it measures how far algebraic objects are from behaving like finite-dimensional Euclidean space.

The first surprise is that a basis is not part of the structure. The plane $\mathbb{R}^2$ has the standard basis, but it also has infinitely many rotated or sheared bases. If every basis had a different number of vectors, coordinates would be bookkeeping rather than mathematics. The central theorem behind dimension says that this failure does not occur for vector spaces: all bases have the same cardinality.

[example: Two Coordinates for the Same Plane] Let $V=\mathbb{R}^2$ over $\mathbb{R}$, and set $e_1=(1,0)$, $e_2=(0,1)$, $f_1=(1,1)$, and $f_2=(1,-1)$. For any $(a,b)\in \mathbb{R}^2$, scalar multiplication and coordinatewise addition give \begin{align*} a e_1+b e_2=a(1,0)+b(0,1)=(a,0)+(0,b)=(a,b). \end{align*} Thus $e_1,e_2$ span $\mathbb{R}^2$. If \begin{align*} \alpha e_1+\beta e_2=(0,0), \end{align*} then \begin{align*} \alpha(1,0)+\beta(0,1)=(\alpha,0)+(0,\beta)=(\alpha,\beta)=(0,0). \end{align*} Equality of coordinates gives $\alpha=0$ and $\beta=0$, so $e_1,e_2$ are linearly independent. Hence $(e_1,e_2)$ is a basis of $\mathbb{R}^2$. For the second coordinate system, choose \begin{align*} c=\frac{a+b}{2} \end{align*} and \begin{align*} d=\frac{a-b}{2}. \end{align*} Then \begin{align*} c f_1+d f_2=\frac{a+b}{2}(1,1)+\frac{a-b}{2}(1,-1). \end{align*} Expanding the scalar multiples gives \begin{align*} c f_1+d f_2=\left(\frac{a+b}{2},\frac{a+b}{2}\right)+\left(\frac{a-b}{2},-\frac{a-b}{2}\right). \end{align*} Coordinatewise addition gives \begin{align*} c f_1+d f_2=\left(\frac{a+b+a-b}{2},\frac{a+b-a+b}{2}\right). \end{align*} In the first coordinate, $a+b+a-b=2a$, so $\frac{a+b+a-b}{2}=a$. In the second coordinate, $a+b-a+b=2b$, so $\frac{a+b-a+b}{2}=b$. Therefore \begin{align*} c f_1+d f_2=(a,b). \end{align*} Thus $f_1,f_2$ span $\mathbb{R}^2$. It remains to check that no redundancy has been introduced. If \begin{align*} \alpha f_1+\beta f_2=(0,0), \end{align*} then \begin{align*} \alpha(1,1)+\beta(1,-1)=(\alpha,\alpha)+(\beta,-\beta)=(\alpha+\beta,\alpha-\beta)=(0,0). \end{align*} Equality of coordinates gives \begin{align*} \alpha+\beta=0. \end{align*} It also gives \begin{align*} \alpha-\beta=0. \end{align*} Adding these two equations gives \begin{align*} 2\alpha=0. \end{align*} Since the scalars are [real numbers](/page/Real%20Numbers), $2\ne 0$, so $\alpha=0$. Substituting $\alpha=0$ into $\alpha+\beta=0$ gives $\beta=0$. Hence $f_1,f_2$ are linearly independent, so $(f_1,f_2)$ is also a basis of $\mathbb{R}^2$. Every vector $(a,b)$ has coordinates $(a,b)$ in the basis $(e_1,e_2)$ and coordinates $\left(\frac{a+b}{2},\frac{a-b}{2}\right)$ in the basis $(f_1,f_2)$. The coordinates change, but both coordinate systems use exactly two basis vectors, which is the invariant that dimension records. [/example]

Dimension begins as a counting problem, but it quickly becomes a structural language. In linear algebra it classifies finite-dimensional vector spaces over a fixed field. In module theory it becomes rank, which behaves well only under hypotheses on the ring and the module. In commutative algebra it becomes Krull dimension, which measures chains of prime ideals rather than coordinates. Here an ideal $I \trianglelefteq R$ is an additive subgroup of a ring $R$ that is closed under multiplication by elements of $R$, and $I \subseteq J$ means that every element of $I$ also lies in $J$. A chain of prime ideals is therefore a nested sequence of such ideals inside one ring. In homological algebra it becomes projective or global dimension, measuring the length of resolutions. This chapter develops vector-space dimension in detail and then uses the later notions as signposts for how the same word changes meaning in algebraic settings. These versions are related by a shared theme: dimension is the length or size of the independent data needed to build an object.

## Definition

The word dimension is used in several related settings, so the first task is to say what kind of object a dimension is meant to be. At its most general, a dimension is not just any number attached to an object; it is a size invariant designed to ignore irrelevant choices such as bases, coordinates, presentations, or resolutions.

[definition: Dimension] Let $\mathcal{C}$ be a class of mathematical objects equipped with an [equivalence relation](/page/Equivalence%20Relation) $\sim$. A dimension on $\mathcal{C}$ is a function $d: \mathcal{C} \to A$ into an ordered set $A$ such that, whenever $X,Y \in \mathcal{C}$ and $X \sim Y$, one has \begin{align*} d(X) = d(Y). \end{align*} [/definition]

This definition is a guiding abstraction rather than a complete axiom system for dimension. By itself it would allow uninformative invariants, such as constant functions, so the mathematical content in each setting comes from additional behavior: invariance under the right equivalence relation, compatibility with subobjects or quotients, additivity, or a meaningful notion of length. The chapter now specializes this broad idea to the setting where dimension first becomes a theorem rather than a convention: vector spaces over a field.

Before defining vector-space dimension, we need the objects whose size dimension measures. In algebra, linear combinations make sense only after choosing a field of scalars and an abelian group of vectors on which those scalars act. The next definition packages exactly the rules needed for addition, scalar multiplication, and coordinate expansions.

[definition: Vector Space] Let $k$ be a field. A vector space over $k$ is an abelian group $(V,+)$ together with a scalar multiplication map $\cdot: k \times V \to V$, written $(a,v) \mapsto av$, such that for all $a,b \in k$ and $u,v \in V$, \begin{align*} a(u+v) = au + av. \end{align*} \begin{align*} (a+b)v = av + bv. \end{align*} \begin{align*} (ab)v = a(bv). \end{align*} \begin{align*} 1_k v = v. \end{align*} [/definition]

A vector space may be large, but a useful coordinate system should describe every vector using a controlled list of directions. The next definition is needed because dimension will not count all vectors; it will count a set whose linear combinations reach the whole space.

[definition: Span] Let $V$ be a vector space over a field $k$, and let $\mathcal{P}(V)$ denote the power set of $V$, the set of all subsets of $V$. The span operation is the map $\operatorname{span}_k: \mathcal{P}(V) \to \mathcal{P}(V)$ defined as follows: for each subset $S \subset V$, \begin{align*} \operatorname{span}_k(S) = \left\{\sum_{i=1}^m a_i v_i : m \ge 0,\ a_i \in k,\ v_i \in S\right\}, \end{align*} where the sum with $m=0$ is the empty sum, equal to $0$. [/definition]

Span tells us when a set is large enough, but size alone does not prevent waste. If one chosen direction can already be built from the others, then the same vector may admit different coefficient descriptions. To make coordinates rigid, one has to rule out every nontrivial linear relation among the chosen vectors.

[definition: Linear Independence] Let $V$ be a vector space over a field $k$. A subset $S \subset V$ is linearly independent if, for every integer $n \ge 0$, every choice of distinct vectors $v_1, \dots, v_n \in S$, and every choice of scalars $a_1, \dots, a_n \in k$, \begin{align*} \sum_{i=1}^n a_i v_i = 0 \end{align*} implies $a_i = 0$ for all $i \in \{1, \dots, n\}$. [/definition]

Span supplies reach, while [linear independence](/page/Linear%20Independence) removes redundancy. Either condition alone is insufficient for dimension: a spanning set may contain unnecessary directions, and an independent set may fail to reach the whole space.

The obstruction is that coordinates require both existence and uniqueness at the same time. The formal notion that captures this balance is a basis: a set large enough to express every vector, but rigid enough that no vector has two different finite expressions.

[definition: Basis] Let $V$ be a vector space over a field $k$. A subset $B \subset V$ is a basis of $V$ over $k$ if $B$ is linearly independent and $\operatorname{span}_k(B) = V$. [/definition]

A basis is a candidate coordinate system, but a space can have many different bases. The obstruction to defining dimension is that two unrelated-looking bases might appear to have different sizes. Before basis size can be used as an invariant, one needs a theorem showing that changing coordinates does not change the number of basis vectors.

[quotetheorem:7838]

Once basis size is independent of the chosen basis, it becomes an invariant of the vector space itself. Later comparisons of spaces and conservation laws for linear maps need a single piece of notation for this invariant, rather than repeatedly referring to the cardinality of an arbitrary basis.