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For a bounded operator on a complex Banach space, the spectrum is a nonempty compact subset of the complex plane contained in the closed disk of radius equal to the operator norm.

If $C$ is a commutative algebra of finite dimension over a field $K$, every prime ideal is maximal and $|\mathrm{Spec}C|\le {\dim }_{K}C$. Indeed, the quotient by a prime is a finite-dimensional integral domain. Multiplication by any nonzero element is an injective linear endomorphism, hence surjective, so the quotient is a field. Distinct maximal ideals are pairwise comaximal. For any finite list ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$, construct $e_i\equiv 1(\mathrm{mod}{\mathfrak{m}}_i)$ and $e_i\equiv 0(\mathrm{mod}{\mathfrak{m}}_j)$ for $j\ne i$ by multiplying elements in ${\mathfrak{m}}_j$ congruent to one modulo ${\mathfrak{m}}_i$. These give a surjection $C\to {\prod }_{i}C/{\mathfrak{m}}_i$. Each factor has positive $K$-dimension, so $r\le {\dim }_{K}C$, excluding an infinite list.

With $m$ the characteristic-prime-free part of $n$, all roots of unity over a finite field lie in ${\mathbb{F}}_{q^s}$ exactly when $m\mid q^s-1$. This follows from the cyclicity of the multiplicative group of a finite field. For $m=1$ the least positive degree is one. A primitive root of order $m$ has the same field degree, because its Frobenius orbit closes exactly when $m\mid q^s-1$.

A field $k$ is a splitting field for a finite group $G$ if every simple module over its group algebra $kG$ is absolutely simple: it remains simple over every extension field. Equivalently, the quotient by the Jacobson radical is a product of full matrix algebras over $k$. In characteristic $p$, containing all $m$th roots of unity, where $|G|=p^{a}m$ and $p\nmid m$, is sufficient. This is the modular splitting-field theorem; it is substantially stronger than merely having eigenvalues for one chosen group element.

The square-class group identifies two nonzero elements of a field when their quotient is a square. Multiplication makes it an abelian group of exponent two, hence a vector space over ${\mathbb{F}}_2$. It records quadratic extensions and the images of Kummer theory for $n=2$.

A square root of unity is an element $x$ of a ring with identity satisfying $x^2=1$. In a field of characteristic different from two, the only possibilities are $1$ and $-1$: the difference of squares identity $(x-1)(x+1)=0$ has no other roots because a field has no zero divisors. In a general ring more solutions may occur; modulo eight the four residue classes $1,3,5,7$ all have square one.

The image of the linear map $A+B$ is contained in the sum of the images of $A$ and $B$. Since the dimension of a vector space of this sum is at most the sum of their dimensions, matrix rank is subadditive. Iteration shows that a sum of $r$ matrices of matrix rank at most one has matrix rank at most $r$, over any field.

A subfield of a field $K$ is a subset containing the additive and multiplicative identities and closed under addition, subtraction, multiplication, and inversion of nonzero elements. It is itself a field under the operations inherited from $K$.

The sum-product phenomenon says that a finite subset of a field cannot usually have both a small sumset and a small product set unless it has additional subfield-like structure.

Over a field whose characteristic is not two, a bilinear form has a unique decomposition into a symmetric bilinear form $S$ and an alternating bilinear form $K$:Adding and subtracting the equations $B(v,w)=S(v,w)+K(v,w)$ and $B(w,v)=S(v,w)-K(v,w)$ both proves uniqueness and gives these formulas. In matrix notation, $S$ has matrix $(A+A^T)/2$ and $K$ has skew-symmetric matrix $(A-A^T)/2$.^[ref]

The diagonal values $q(v)=B(v,v)$ alone do not determine a general bilinear form, because $K(v,v)=0$ for every vector. For the matrix of a bilinear form in the example above,Thus two different bilinear forms, represented by $A$ and its symmetric matrix part, have exactly the same quadratic form. To recover a bilinear form from its diagonal, one needs extra information such as the hypothesis that it is a symmetric bilinear form.

Over every field, including characteristic two, there is a natural isomorphismThe middle summand maps $v\otimes w$ to the mixed product $vw$. Monomial bases prove bijectivity, and the Leibniz rule makes the map equivariant for Lie algebra representations.

A symplectic basis $(e_1,\dots ,e_n,f_1,\dots ,f_n)$ has pairings $\omega (e_i,f_j)={\delta }_{ij}$ and all pairings between two $e_i$ or two $f_i$ equal to zero.

A tensor product turns a bilinear map into a linear map. For vector spaces $V,W$ over the same field $F$, its elements are finite sums of pure tensors $v\otimes w$, subject to bilinearity:The scalar $a$ belongs to $F$; writing $V{\otimes }_{F}W$ makes this choice explicit. These rules preserve separate linearity while allowing arbitrary linear combinations of products.^[ref]

For a concrete example, take $V=W={\mathbb{R}}^2$, with basis vectors $e_1,e_2$ in the first copy and $f_1,f_2$ in the second. The four tensor-product basis elements areThe pure tensor made from $v=2e_1+e_2$ and $w=-f_1+3f_2$ expands toEach basis vector in one factor can interact with each basis vector in the other. This is why dimensions multiply; the tensor-product coefficient matrix displays these four interactions as a two-by-two grid.

The tensor product is different from the Cartesian product $V\times W$. With componentwise vector addition and scalar multiplication, the latter contains pairs and has dimension $\dim V+\dim W$. The map $\tau (v,w)=v\otimes w$ is bilinear, generally neither linear on that Cartesian product nor injective nor surjective. In our example, $\tau (2v,2w)=4\tau (v,w)$, and $\tau (2v,w/2)=\tau (v,w)$; the separability of a tensor explains why some outputs cannot be reached at all. Both spaces happen to have dimension four when $V=W={\mathbb{R}}^2$, but this coincidence does not identify their structures.

The theory of formally real fields consists of the field axioms and all sentences asserting that $-1$ is not a sum of $m$ squares, one for each positive integer $m$. The real closure theorem and model completeness make the Theory of real closed fields its model companion.

A module $M$ over an integral domain $R$ is torsion-free when $rm=0$ with $0\ne r\in R$ implies $m=0$, equivalently when its torsion submodule is zero. Every free module over an integral domain is torsion-free, as one can check in each coefficient. Every vector space is torsion-free over its scalar field.

The converse requires hypotheses. A finitely generated module over a principal ideal domain is free if it is torsion-free, but the module $\mathbb{Q}$ over $\mathbb{Z}$ is torsion-free and not free. Any two nonzero rational numbers have a nontrivial integer relation, so a possible basis of a module could have at most one element; no single $q$ generates all rational numbers, since $q/2\notin \mathbb{Z}q$ for $q\ne 0$. The finite-generation hypothesis cannot be dropped. See Keith Conrad's Modules over a PID, Corollary 2.6.