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A group representation over a field is absolutely irreducible if it remains irreducible after every field extension. For a finite-dimensional representation it is enough to check an algebraic closure. Irreducibility over the original field alone is weaker: the two-dimensional real rotation representation of the cyclic group of order three splits into two lines over the complex numbers.

The James submodule theorem follows from ${\kappa }_{t}v=⟨v,e_t⟩e_t$. A nonzero restricted form therefore gives a simple radical quotient. If $m_j$ counts equal-length rows, all integral polytabloid pairings are divisible by ${\prod }_j{m}_j!$, while pairing with row reversal gives ${\prod }_j(m_j!{)}^j$. Thus the form is nonzero modulo $p$ exactly for regular partitions. The construction and the proof commute with every field extension, so the quotient is absolutely irreducible, including over the prime field.

A field extension $L/K$ is algebraic if every element of $L$ is an algebraic element over $K$. Every finite field extension is algebraic, but algebraic extensions can have infinite degree.

An automorphism of a field extension $K/F$ is a field automorphism $\sigma :K\to K$ fixing every element of $F$. Such maps form the field automorphism group ${\mathrm{Aut}}_F(K)$ under composition. For a Finite Galois extension this is the Galois group $\mathrm{Gal}(K/F)$. Specifying the fixed base field distinguishes these maps from arbitrary field automorphisms of $K$.

Fix a number field $K$, a positive integer $d$, and a finite set $S$ of its finite places. There are finitely many field extensions $L/K$ of degree at most $d$ unramified outside $S$, inside a fixed algebraic closure. To see this, view them as extensions of the rational numbers of bounded degree. Their ramified rational primes belong to a fixed finite set. At each of these primes, the different exponent is bounded in terms of the local degree and ramification index, giving a bound for the absolute field discriminant. The Hermite–Minkowski theorem gives finitely many fields up to isomorphism, and each has only finitely many embeddings into the fixed closure. This useful form of arithmetic compactness is stated and proved for $K=\mathbb{Q}$ in Andrew Sutherland's number-theory notes, Theorem 14.25; the preceding reduction gives the general case.

A field is $C_i$ if every homogeneous polynomial of degree $d\ge 1$ in more than $d^i$ variables has a nonzero zero. Algebraically closed fields are $C_0$, finite fields are $C_1$, and the Lang-Nagata theorem for Ci fields controls function-field extensions.

If $\alpha$ and $\beta$ are separable algebraic elements, put $K=F(\alpha )$ and $L=K(\beta )$. The minimal polynomial of $\beta$ over $K$ divides its separable polynomial over $F$, and is therefore separable. The extension count for field embeddings gives $[K:F]$ embeddings of $K$ into an algebraic closure, each with $[L:K]$ extensions to $L$. Hence $L$ has $[L:F]$ embeddings. For $E=F(\alpha +\beta )$, each embedding of $E$ has at most $[L:E]$ extensions, so there are at least $[E:F]$ embeddings of $E$ and hence exactly that number. Since $E$ is a simple field extension, its minimal polynomial has $[E:F]$ distinct roots. Thus $\alpha +\beta$ is a separable algebraic element.

For a scheme over $\mathbb{R}$, its base change of a morphism of schemes to $\mathbb{C}$ has scheme-theoretic fibre over a point $p$ equal to $\mathrm{Spec}(\kappa (p){\otimes }_{\mathbb{R}}\mathbb{C})$. This is the spectrum of $\kappa (p)[s]/(s^2+1)$. It has two points if $-1$ is a square in the residue field, and otherwise one point with a quadratic field extension. In both cases it is reduced, because the quadratic polynomial has no repeated root in characteristic zero. The distinction applies to generic points as well as closed points.

A constructible circle has a constructible point as its centre and radius equal to the distance between two constructible points. When those coordinates lie in a real field $F$, its equation $(x-a{)}^2+(y-b{)}^2=r^2$ has coefficients in $F$ because the squared radius is a sum of squared coordinate differences. Intersecting it with a constructible line reduces to a quadratic equation. Intersecting two distinct circles reduces to the same situation by subtracting their normalized equations to obtain the radical axis. Thus each isolated real intersection requires at most a real quadratic field extension.

A point of the Euclidean plane is constructible if it can be obtained from $(0,0)$ and $(1,0)$ by finitely many straightedge-and-compass operations: draw a line through two obtained distinct points, draw a circle with obtained centre and radius equal to an obtained segment length, and retain isolated intersection points. Its two coordinates are constructible numbers. Conversely, constructible real coordinates can be placed on perpendicular axes and their coordinate lines intersected, giving the point. Coordinates created at each isolated intersection require only solving linear or quadratic equations over the current coordinate field; hence they lie in a real field extension obtained by a finite tower of quadratic extensions. Coincident curves supply no isolated new intersection points.

For a finitely generated field extension $L/k$ in characteristic zero, a tuple $t_i$ is algebraically independent exactly when its differentials $d{t}_i$ are linearly independent in the Kähler differentials. A relation of minimum total degree gives a nontrivial differential relation; an independent tuple extends to a transcendence basis. In positive characteristic, derivatives of nonconstant relations can vanish identically.

Suppose a Non-Archimedean absolute value on $K$ has a unique extension to every finite field extension. Any K-isomorphism $K({\alpha }_i)\to K({\alpha }_j)$ between conjugate roots pulls the absolute value back to an extension on the source field, which must be the given one. Thus $|h({\alpha }_i)|=|h({\alpha }_j)|$ for every $h\in K[X]$. This argument includes inseparable irreducible polynomials: no transitive Galois action or completeness assumption is needed.

Apply Zorn's lemma to subsets of a field extension with algebraic independence, ordered by inclusion. The union of a chain is independent because any polynomial relation involves only finitely many distinct elements, all of which lie in one member of the chain. A maximal independent set $B$ is a transcendence basis: if any element were transcendental over $K(B)$, adjoining it would preserve independence, by clearing denominators, contradicting maximality. Tuple entries must be distinct in the set-based independence definition; otherwise the nonzero polynomial $X_1-X_2$ vanishes on a repeated pair and excludes every nonempty set.

For a field extension $L/K$, an extension of an absolute value $|\cdot |$ on $K$ is an absolute value on $L$ whose restriction to $K$ is $|\cdot |$.

The complex algebraic numbers form a field, the algebraic closure of $\mathbb{Q}$ inside $\mathbb{C}$. To see closure under the field operations, if $\alpha ,\beta$ are algebraic numbers, the tower law makes $\mathbb{Q}(\alpha ,\beta )$ a finite-dimensional field extension of $\mathbb{Q}$. Every element of this extension is an algebraic element, because its successive powers are linearly dependent. In particular it contains $\alpha \pm \beta$, $\alpha \beta$ and $\alpha /\beta$ when $\beta \ne 0$.

It is an infinite-degree algebraic extension. A primitive $p$th root of unity has minimal polynomial of an algebraic element $1+X+\cdots +X^{p-1}$ of degree $p-1$ for every prime $p$, so the degrees of its finite subextensions are unbounded. Finally a complex root of a polynomial whose coefficients are algebraic numbers is itself algebraic over their finite-degree coefficient field and hence over $\mathbb{Q}$, again by the tower law. Thus every such polynomial splits in this field, establishing the algebraic closure assertion.

Let $L/k$ be a field extension. If $A{\otimes }_{k}L$ is a finitely generated algebra over $L$, then $A$ is finitely generated over $k$. Choose finitely many generators of $A{\otimes }_{k}L$ and collect the finitely many coefficients from $A$ occurring in them. The $k$-subalgebra $A_0$ generated by those coefficients satisfies $(A/A_0){\otimes }_{k}L=0$, and faithful flatness of $L/k$ gives $A=A_0$.

For an algebraically closed field $k$ and a finitely generated field extension $K/k$ of transcendence degree one, choose $t$ with $K/k(t)$ finite. The integral closures of the two standard coordinate rings of the projective line in $K$ are finite modules and glue to a finite normalization of an algebraic curve of the projective line. A one-dimensional normal local ring is a discrete valuation ring; over the perfect field $k$ the resulting projective curve is smooth. Extension of rational maps gives uniqueness of this smooth projective model.

A unital commutative ring generated by finitely many elements under addition and multiplication, starting from the image of the integers. Equivalently it is a quotient of a finite-variable polynomial ring over $\mathbb{Z}$. localization of a ring at one element remains finite type, since $R[1/r]\cong R[T]/(rT-1)$. Generation here is as an algebra, not merely as a field extension when the ring happens to be a field.

A finitely generated field extension of transcendence degree $s$ of a $C_i$ field is $C_{i+s}$. In particular a function field of a curve over a C1 field is $C_2$. The theorem includes stability under finite algebraic extensions when $s=0$.

Matrix similarity invariants are quantities unchanged by replacing $A$ with $S^{-1}AS$. Important examples are the characteristic polynomial, trace, determinant, matrix rank, and eigenvalues with their algebraic multiplicities. The characteristic polynomial convention here is ${\chi }_A(t)=\det (tI-A)$, where $I$ is the identity matrix.^[ref]

The short determinant proof works over any field: because $tI-B=S^{-1}(tI-A)S$, multiplicativity of the determinant givesThus its roots, the eigenvalues, retain their algebraic multiplicities in any field extension where the polynomial splits. Also $\det B=\det A$. The cyclic property of the trace gives $\mathrm{tr}(S^{-1}AS)=\mathrm{tr}(AS{S}^{-1})=\mathrm{tr}A$; equivalently, the trace is determined by the coefficient of $t^{n-1}$ in the characteristic polynomial for an $n$-by-$n$ matrix.^[ref]

For the two matrices in the opening example,Their entries differ, yet these quantities agree. Agreement is a useful necessary test for matrix similarity; it is not sufficient in general, as same eigenvalues do not imply similarity illustrates.

The relation also preserves repeated application. Adjacent factors $S{S}^{-1}$ cancel to giveFor the opening example, matrix powers therefore have the explicit formAt $k=2$, this is $\left(\begin{array}{cc}4& -21\\ 0& 25\end{array}\right)$. Acting on $y=(1,1{)}^T$ gives $(-17,25{)}^T$, which reconstructs as $S(-17,25{)}^T=(8,25{)}^T=A^{2}x$. More generally $q(B)=S^{-1}q(A)S$ for every polynomial $q$, so the minimal polynomial is also unchanged. Multiplication by invertible matrices preserves matrix rank, giving another invariant without finding eigenvalues.