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Transcendental extension

In field theory, a transcendental extension of a field F is a field extension E/F that is not algebraic over F, meaning it contains at least one element that is transcendental over F.^[1] An element \alpha \in E is transcendental over F if it is not algebraic over F, i.e., there exists no nonzero polynomial in F[X] with \alpha as a root.^[2] Such extensions contrast with algebraic extensions, where every element satisfies a polynomial equation over the base field, and they arise naturally in contexts like the field of rational functions F(x), where x is transcendental over F.^[1] A key subclass is the purely transcendental extension, in which there exists an algebraically independent set S \subseteq L over K such that L = K(S), the rational function field generated by S.^[3] The maximal size of such an algebraically independent set is called the transcendence degree of the extension, denoted \operatorname{tr.deg}_K L, which is an invariant measuring the "transcendental dimension" and equals the cardinality of any transcendence basis for L over K.^[4] For instance, the extension \mathbb{Q}(\pi)/\mathbb{Q} has transcendence degree 1, since \pi is transcendental over \mathbb{Q}, while the field of complex numbers \mathbb{C}/\mathbb{Q} has infinite transcendence degree, as it contains uncountably many algebraically independent transcendentals.^[2] Transcendental extensions play a fundamental role in algebraic geometry, number theory, and the study of differential fields, where they help classify infinite-degree extensions beyond purely algebraic ones.^[5]

Fundamentals

Definition

In field theory, a field extension K/F is transcendental if it contains at least one element that is transcendental over the base field F. An element \alpha \in K is said to be transcendental over F if it is not algebraic over F, meaning that \alpha is not a root of any non-zero polynomial with coefficients in F.^[6]^[7] This contrasts with an algebraic extension, in which every element of K is algebraic over F, i.e., satisfies some non-zero polynomial equation over F. Thus, transcendental extensions are exactly those field extensions that are not algebraic.^[6]^[5] A basic example of a transcendental extension is the simple extension obtained by adjoining a single transcendental element x to F, denoted F(x), which is isomorphic to the field of rational functions in one indeterminate over F. More generally, transcendental extensions arise by adjoining one or more transcendental elements to F.^[6]^[8]

Transcendental elements and algebraic independence

In field theory, an element \alpha in an extension field E of a base field F is called transcendental over F if it is not algebraic over F, meaning there is no nonzero polynomial f(X) \in F[X] such that f(\alpha) = 0.^[9] Equivalently, \alpha is transcendental if the evaluation homomorphism F[X] \to E given by g(X) \mapsto g(\alpha) has trivial kernel, so F[\alpha] \cong F[X] as rings, and thus the field of fractions satisfies F(\alpha) \cong F(X), the rational function field in one variable over F.^[9] A set \{\alpha_i\}_{i \in I} \subseteq E of elements is said to be algebraically independent over F if there exists no nontrivial polynomial relation, that is, no nonzero multivariate polynomial P(X_i)_{i \in I} \in F[(X_i)_{i \in I}] such that P(\alpha_i)_{i \in I} = 0.^[9] If such a relation does exist, the set is algebraically dependent over F. A single element \alpha is algebraically independent over F precisely when it is transcendental over F, and more generally, if a set \{\alpha_1, \dots, \alpha_n\} is algebraically independent, then F(\alpha_1, \dots, \alpha_n) \cong F(X_1, \dots, X_n), the rational function field in n variables over F.^[9] An algebraically independent set over F is maximal if it is not properly contained in any larger algebraically independent set in E; any such maximal set serves as a transcendence basis for the extension E/F, generating E as an algebraic extension over the purely transcendental extension F(\{\alpha_i\}).^[9] Prominent examples of transcendental elements over the rationals \mathbb{Q} include \pi and e, each proven individually to be transcendental but conjectured—without proof—to be algebraically independent over \mathbb{Q}.^[10]^[11]

Transcendence Basis

Construction and properties

To construct a transcendence basis for a field extension K/F, start with a generating set G \subset K such that K = F(G). Consider the partially ordered set of all algebraically independent subsets of G (ordered by inclusion); by Zorn's lemma, this set has a maximal element B, which is algebraically independent over F. Moreover, K is algebraic over F(B), so B is a transcendence basis for K/F.^[12]^[13] A transcendence basis B for K/F is algebraically independent over F by construction, and the subextension F(B)/F is purely transcendental, meaning K decomposes as an algebraic extension of the purely transcendental extension F(B)/F. Any two transcendence bases of K/F have the same cardinality, known as the transcendence degree of K/F.^[4]^[12]^[13] The field F(B) is isomorphic to the rational function field F(x_i \mid i \in I) in |B| variables, where I is an index set with |I| = |B|; this holds whether B is finite or infinite. For finite bases of cardinality n, F(B) \cong F(x_1, \dots, x_n), the field of rational functions in n indeterminates over F. In the infinite case, with |B| = \kappa a cardinal (possibly transfinite), F(B) is the fraction field of the polynomial ring F[x_i \mid i \in I] over an index set I of cardinality \kappa, preserving algebraic independence and the structure of the extension.^[4]^[13]

Existence and cardinality

Every field extension K/F admits a transcendence basis. To prove this, consider the partially ordered set of all algebraically independent subsets of K over F, ordered by inclusion. This poset satisfies the conditions of Zorn's lemma, as any chain has an upper bound given by its union, which remains algebraically independent. Thus, there exists a maximal element B, which is algebraically independent, and K is algebraic over F(B), making B a transcendence basis.$$](https://stacks.math.columbia.edu/tag/030F) All transcendence bases of K/F have the same cardinality, denoted \operatorname{tr.deg}(K/F), the transcendence degree of K over F. For finite bases, this follows by induction using the replacement property: if B' is another basis with |B'| < |B|, an element of B can be replaced by one from B' while preserving algebraic independence and maximality, leading to a contradiction. For infinite bases, suppose |B'| \leq |B|; for each \alpha \in B', there is a finite subset B_\alpha \subset B such that \alpha is algebraic over F(B_\alpha). The union B^* = \bigcup_{\alpha \in B'} B_\alpha has cardinality |B^*| \leq |B'| \cdot \aleph_0 = |B'| (since |B'| is infinite), and B^* generates an extension over which B is algebraic, implying B = B^* by maximality and thus |B| = |B'|. The symmetric argument yields equality.[$$(https://stacks.math.columbia.edu/tag/030F) If K/F is finitely generated as a field extension, then \operatorname{tr.deg}(K/F) is finite. In this case, any algebraically independent subset must be finite, as infinite independence would contradict the finite generation; transcendence bases thus achieve a maximal finite size by Zorn's lemma applied within the finitely generated context.$$](http://virtualmath1.stanford.edu/~conrad/121Page/handouts/trdeg.pdf) More generally, the transcendence degree encodes the "transcendental dimension" of the extension, analogous to vector space dimension, with K decomposing as a vector space over the purely transcendental subfield F(B) of dimension equal to the algebraic degree over that subfield.[$$

Examples and Illustrations

Single transcendental extensions

A quintessential example of a single transcendental extension is the rational function field \mathbb{Q}(x) over the base field \mathbb{Q}, where x is an indeterminate transcendental over \mathbb{Q}. The elements of \mathbb{Q}(x) are ratios of polynomials with rational coefficients, specifically quotients p(x)/q(x) where p, q \in \mathbb{Q} and q \neq 0, forming a field under the usual operations. Since x satisfies no nonzero polynomial equation over \mathbb{Q}, there are no algebraic relations among its powers, making \{x\} a transcendence basis for this extension.^[14]^[2] Another illustration arises from transcendental numbers, such as adjoining \pi to the rationals to form \mathbb{Q}(\pi) over \mathbb{Q}. Here, \pi is transcendental over \mathbb{Q}, as proven by Lindemann in 1882 using properties of the exponential function, ensuring that \mathbb{Q}(\pi) is a single transcendental extension of transcendence degree 1. Similarly, \mathbb{Q}(e) over \mathbb{Q} qualifies, with e transcendental as established by Hermite in 1873. In both cases, the extension consists of rational expressions in the transcendental element, analogous to the rational function field but embedded in the reals.^[14]^[4] Geometrically, single transcendental extensions correspond to function fields of curves, particularly the affine line over a field k, whose function field is k(t) with t transcendental over k. This field encodes rational functions on the line \mathbb{A}^1_k, providing a bridge to algebraic geometry where such extensions model the "dimension 1" behavior of varieties.^[15] In these examples, structural properties like automorphisms and derivations remain straightforward. For instance, the k-automorphisms of k(t) fixing k are precisely the fractional linear transformations t \mapsto (at + b)/(ct + d) with ad - bc \neq 0 and a, b, c, d \in k. Derivations, such as the standard d/dt on k(t) satisfying the Leibniz rule d(fg)/dt = f \, d(g)/dt + g \, df/dt, extend naturally and are determined by their action on t, where d(t)/dt = 1. These features highlight the simplicity of single generator transcendental extensions.^[16]^[17]

Extensions with multiple transcendentals

In transcendental extensions generated by multiple elements, a key distinction arises when the generators are algebraically independent over the base field, leading to a transcendence degree equal to the number of such elements. Consider the finite case of the extension \mathbb{Q}(x, y)/\mathbb{Q}, where x and y are algebraically independent transcendentals over \mathbb{Q}. This field is isomorphic to the field of rational functions in two variables over \mathbb{Q}, denoted \mathbb{Q}(x, y), and possesses transcendence degree 2.^[5] For the infinite case, the field of rational functions in countably many variables over \mathbb{Q} provides an example; it is constructed as the direct limit (union) of the fields \mathbb{Q}(x_1, \dots, x_n) over all finite n \in \mathbb{N}, with transcendence basis \{x_n \mid n \in \mathbb{N}\}.^[18] A concrete example from complex analysis is the field of all meromorphic functions on the complex plane \mathbb{C}, which has transcendence degree equal to the cardinality of the continuum over \mathbb{C}.^[19] In contrast, when the added elements are algebraically dependent, the transcendence degree does not increase accordingly; for instance, \mathbb{Q}(\pi, 2\pi)/\mathbb{Q} has transcendence degree 1, as $2\pi is algebraic over \mathbb{Q}(\pi) (satisfying the linear equation X - 2\pi = 0), and \pi is known to be transcendental over \mathbb{Q}.^[20]

Transcendence Degree

Definition and basic properties

In field theory, the transcendence degree of a field extension K/F, denoted \operatorname{tr.deg}_F(K), is defined as the cardinality of any transcendence basis for K over F. A transcendence basis is a maximal algebraically independent subset of K over F, and all such bases have the same cardinality, ensuring the definition is well-posed. If K/F is algebraic, meaning every element of K satisfies a polynomial equation over F, then \operatorname{tr.deg}_F(K) = 0; otherwise, the transcendence degree is a positive finite integer or an infinite cardinal.^[8] The transcendence degree possesses several elementary properties that establish it as a fundamental invariant of field extensions. For any tower of field extensions F \subseteq L \subseteq K, the additivity formula holds: \operatorname{tr.deg}_F(K) = \operatorname{tr.deg}_F(L) + \operatorname{tr.deg}_L(K). This relation applies uniformly to finite and infinite cases, where the sum in the infinite setting is the cardinal sum corresponding to the disjoint union of transcendence bases. Additionally, the transcendence degree is preserved under isomorphisms of field extensions, as it depends solely on the algebraic independence structure relative to the base field.^[8] When the transcendence degree is finite, say \operatorname{tr.deg}_F(K) = n < \infty, a key characterization emerges for finitely generated extensions. In this case, K is finitely generated as a field extension of F if and only if it is a finite-degree algebraic extension of a purely transcendental extension of degree n over F. A purely transcendental extension of degree n is of the form F(x_1, \dots, x_n), where the x_i are indeterminates algebraically independent over F. This property underscores the transcendence degree's role in decomposing extensions into transcendental and algebraic components.^[21]^[5]

Computation in specific cases

Computing the transcendence degree of a field extension often involves identifying a transcendence basis or leveraging structural properties of the extension. In the case of function fields, consider the function field k(X) of an affine curve X over an algebraically closed field k. Here, \operatorname{tr.deg}(k(X)/k) = 1, as k(X) is finitely generated over k with a single algebraically independent element corresponding to a separating transcendental, and the extension is algebraic over this rational function field.^[22] This holds more generally for the function field of any irreducible affine variety of dimension 1 over k, where the transcendence degree equals the geometric dimension.^[4] For extensions mixing algebraic and transcendental elements, such as \mathbb{Q}(\pi, \sqrt{2})/\mathbb{Q}, the transcendence degree is 1. The number \pi is transcendental over \mathbb{Q}, so \{\pi\} is a transcendence basis for \mathbb{Q}(\pi)/\mathbb{Q}, giving \operatorname{tr.deg}(\mathbb{Q}(\pi)/\mathbb{Q}) = 1.^[23] Adjoining \sqrt{2}, which is algebraic over \mathbb{Q} (satisfying x^2 - 2 = 0), yields an algebraic extension \mathbb{Q}(\pi, \sqrt{2})/\mathbb{Q}(\pi), preserving the transcendence degree by the additivity property over towers of extensions.^[4] In the context of integral domains, the transcendence degree of the fraction field relates to the domain's structure when integrally closed. For the domain A = \mathbb{Z}, which is integrally closed in its fraction field K = \mathbb{Q}(x), we have \operatorname{tr.deg}(K / \operatorname{Frac}(\mathbb{Z})) = \operatorname{tr.deg}(\mathbb{Q}(x)/\mathbb{Q}) = 1, since \{x\} forms a transcendence basis over \mathbb{Q} and the extension is algebraic thereafter.^[4] To compute transcendence degrees algorithmically, particularly for fraction fields of finitely generated algebras over a field, Noether normalization provides a method by embedding the algebra as a finite module over a polynomial subring in the appropriate number of variables. Specifically, for an affine domain A finitely generated over a field k with \operatorname{tr.deg}( \operatorname{Frac}(A)/k ) = d, there exist algebraically independent elements y_1, \dots, y_d \in A such that A is integral over k[y_1, \dots, y_d], confirming the degree equals d.^[24] This technique applies to affine varieties, reducing the computation to finding such parameters via linear projections or generic choices.

Advanced Properties

Relations to differentials

In the rational function field k(x) over a field k, the derivation \frac{d}{dx} satisfies the Leibniz rule \frac{d}{dx}(fg) = f \frac{d}{dx}(g) + g \frac{d}{dx}(f) for f, g \in k(x). It is defined on the subring k by \frac{d}{dx}(c) = 0 for c \in k and \frac{d}{dx}(x) = 1, extended by linearity and the Leibniz rule, and then to quotients via \frac{d}{dx}\left(\frac{f}{g}\right) = \frac{g \frac{d}{dx}(f) - f \frac{d}{dx}(g)}{g^2}.^[25] This construction generalizes to multiple variables in k(x_1, \dots, x_n), where partial derivations \frac{\partial}{\partial x_i} each satisfy the Leibniz rule and vanish on the other indeterminates x_j for j \neq i. These partial derivations span the k-vector space of all derivations from k(x_1, \dots, x_n) to itself.^[25] The module of Kähler differentials \Omega_{A/F} for an F-algebra A encodes all F-derivations from A into A-modules via a universal derivation d: A \to \Omega_{A/F} obeying the Leibniz rule d(ab) = a \, db + b \, da. For a purely transcendental algebra A = F[x_1, \dots, x_n], \Omega_{A/F} is the free A-module with basis \{dx_1, \dots, dx_n\}. Localizing to the fraction field K = F(x_1, \dots, x_n), \Omega_{K/F} \cong \bigoplus_{i=1}^n K \, dx_i as K-modules.^[26] For a separable field extension K/F, the transcendence degree equals the dimension of the K-vector space \Omega_{K/F}. If \{t_i \mid i \in I\} is a transcendence basis, then \{dt_i\} forms a basis for \Omega_{K/F}, as the separable algebraic extension over F(t_i) contributes no additional differentials. This links the geometric notion of "transcendental directions" to algebraic dimension.^[27] In characteristic p > 0, Kähler differentials detect p-independence in extensions. A set \{t_i\} is p-independent over F if the dt_i are linearly independent in the quotient \Omega_{K/F} / d(K^p), where K^p denotes the subfield of p-th powers; this distinguishes transcendental elements from those algebraic over p-th power subfields. For instance, in the purely inseparable extension K = k(\alpha) with minimal polynomial X^p - a = 0 for a \in k not a p-th power, \Omega_{K/k} is free of rank 1 on d\alpha.^[28]

Key theorems on extensions

Lüroth's theorem provides a fundamental characterization of subfields within the rational function field k(x) over a field k. Specifically, if k is a field of characteristic zero and L is a subfield such that k \subset L \subset k(x), then L = k(f) for some rational function f \in k(x), meaning L is a purely transcendental extension of transcendence degree 1 over k.^[29] This result extends to positive characteristic p > 0 under additional conditions, such as when L does not contain elements algebraic over k of degree divisible by p, ensuring the extension remains simple transcendental.^[30] The theorem relies on the concept of a transcendence basis, where a single element serves as such a basis for L/k.^[30] Zariski's lemma addresses the structure of finitely generated field extensions, particularly their algebraic nature. It states that if k is a field and K is a finitely generated extension of k that is algebraic over k, then K/k is a finite field extension. In the context of transcendental extensions, this lemma implies that the transcendence degree is preserved under algebraic extensions, including finite ones; for instance, if L/K is a finite extension and K/k has transcendence degree n, then L/k also has transcendence degree n, as any additional elements would contradict the algebraic finiteness. This preservation is crucial for understanding how algebraic parts do not alter the transcendental dimension. Every field extension E/F admits a transcendence basis, yielding a purely transcendental extension T/F (the rational function field in those indeterminates) such that E is algebraic over T, and the transcendence degree of E/F equals that of T/F. This decomposition is unique up to isomorphism, with the transcendental part generated by a transcendence basis and the algebraic part forming the algebraic closure relative to that basis. The theorem underscores that every extension can be viewed as an algebraic extension of a purely transcendental one, providing a complete structural classification.^[4] The Siegel-Shidlovsky theorem offers bounds on the transcendental degrees of fields generated by E-functions, which are entire functions satisfying certain growth and arithmetic conditions analogous to exponential functions. It asserts that if f_1(z), \dots, f_n(z) are E-functions linearly independent over \mathbb{Q}(z) and satisfying a linear differential system of order n over \mathbb{Q}(z), then the transcendence degree over \overline{\mathbb{Q}} of the field generated by f_1(\alpha), \dots, f_n(\alpha) at a nonzero algebraic point \alpha \in \overline{\mathbb{Q}} equals n.^[31] This result provides sharp algebraic independence measures, limiting dependencies in the values of these functions and extending Siegel's earlier work on individual E-functions.^[32]

Applications

In field theory and Galois theory

In field theory, the study of transcendental extensions within Galois theory often focuses on their automorphism groups rather than the classical notion of Galois extensions, which requires algebraicity. For a simple transcendental extension k(x)/k, where k is a field and x is transcendental over k, the group of k-automorphisms \mathrm{Aut}_k(k(x)) consists precisely of the fractional linear transformations x \mapsto \frac{ax + b}{cx + d} with a, b, c, d \in k and ad - bc \neq 0. This group is isomorphic to the projective general linear group \mathrm{PGL}(2, k).^[33] When considering separable closures, the absolute Galois group becomes more complex due to the infinite nature of the extension, leading to non-discrete topologies like the Krull topology on the automorphism group.^[6] Purely transcendental extensions differ fundamentally from algebraic ones in Galois theory, as they are not algebraic over the base field and thus cannot be Galois in the standard sense, which demands normality and separability for algebraic extensions. For instance, the extension \mathbb{Q}(x)/\mathbb{Q} is purely transcendental but not Galois, since x satisfies no polynomial equation over \mathbb{Q}, preventing the formation of a normal closure within an algebraic setting.^[6] Such extensions typically yield non-Galois structures, where the fixed field of the automorphism group does not align with classical Galois correspondence principles applicable to algebraic cases. In characteristic zero, Galois actions preserve the transcendence degree of extensions, a property essential for descent techniques in Galois theory. Field automorphisms, including those in Galois groups of algebraic parts over transcendental bases, map algebraically independent sets to algebraically independent sets, ensuring that the cardinality of a transcendence basis remains invariant under such actions.^[6] This preservation facilitates the descent of properties from larger fields to base fields, maintaining structural invariants like transcendence degree during Galois descent. Lüroth's theorem provides a foundational structure for Galois theory over rational function fields by asserting that intermediate fields between k and k(x) are themselves rational function fields, aiding realizations in the inverse Galois problem. However, for function fields of higher genus or extensions of higher degree, Luroth-like problems exhibit counterexamples where subfields are not purely transcendental, complicating the solvability of the inverse Galois problem and leading to unresolved cases for certain finite groups as Galois groups over such bases.^[34] Seminal work using Hurwitz families demonstrates realizations for many groups over function fields, but the intricacies increase with higher degrees, highlighting ongoing challenges in transcendental settings.^[34]

In algebraic geometry and number theory

In algebraic geometry, the function field of an irreducible affine variety V over an algebraically closed field \overline{k} is the fraction field of its coordinate ring, and the dimension of V is defined as the transcendence degree of this function field over \overline{k}.^[35] For instance, the affine space \mathbb{A}^n has dimension n, as its function field \overline{k}(x_1, \dots, x_n) has transcendence degree n. This equivalence highlights how transcendental extensions capture the geometric dimension, with the transcendence basis corresponding to a set of algebraically independent rational functions generating the field up to algebraic closure.^[35] The Noether normalization lemma further connects transcendental extensions to the structure of varieties by asserting that for a finitely generated algebra S over a field k that is an integral domain, there exists a transcendence degree r (equal to the Krull dimension of S) and elements y_1, \dots, y_r \in S that are algebraically independent over k such that S is a finite module over the polynomial ring k[y_1, \dots, y_r].^[24] This finite extension implies that the fraction field of S is a finite algebraic extension of the purely transcendental extension k(y_1, \dots, y_r), embedding the variety into a finite cover of affine space and facilitating proofs of finiteness and dimension results.^[24] In number theory and arithmetic geometry, transcendental extensions arise in the study of heights on varieties defined over finitely generated fields K/k with positive transcendence degree, where generalized Northcott properties ensure only finitely many points of bounded height and bounded degree over such bases.^[36] For example, over fields of transcendence degree r over \mathbb{Q}, vector-valued heights on subvarieties satisfy finiteness theorems analogous to the classical Northcott theorem, which bounds the number of algebraic points of bounded height in projective space over number fields, thereby controlling arithmetic complexity in transcendental settings.^[36] Transcendental extensions over \mathbb{Q}, such as \mathbb{Q}(e) and \mathbb{Q}(\pi), are central to Diophantine approximation, where numbers like e and \pi exhibit specific irrationality measures bounded using techniques inspired by Roth's theorem.^[37] Roth's theorem states that for any algebraic irrational \alpha and \epsilon > 0, there are only finitely many rationals p/q satisfying |\alpha - p/q| < q^{-2-\epsilon}, and while this applies directly to algebraics, extensions involving transcendentals like e (proven irrational via Fourier series truncations) and \pi (via continued fractions and Hermite's method) yield measures \mu(e) = 2 and improved bounds for \mu(\pi) \leq 7.103205334137\dots (as of 2020), with it conjectured that \mu(\pi) = 2 as for almost all irrationals.^[37]^[38] This links approximation quality to transcendence proofs. In model theory, transcendence bases are essential for characterizing models of the theory \mathrm{ACF}_0 of algebraically closed fields of characteristic 0, which admits quantifier elimination and is decidable.^[39] Models of \mathrm{ACF}_0 are determined up to elementary equivalence by their transcendence degree over the prime field \mathbb{Q}, with a transcendence basis allowing isomorphisms between models via algebraic independence preservation, enabling algorithmic decision procedures for sentences in the language of rings.^[39]

References

[1]
Transcendental Extension -- from Wolfram MathWorld
An extension field of a field F that is not algebraic over F, ie, an extension field that has at least one element that is transcendental over F.
[2]
[PDF] FIELD THEORY 1. Fields, Algebraic and Transcendental Elements ...
We may view E as a vector space over F, and so doing we define the degree of E over F by [E : F] = dimF E. The degree of an extension may be finite or infinite.
[3]
transcendental extension in nLab
- **Definition of Transcendental Extension**: A field extension $ K \subset L $ is transcendental if there exists an element $ \alpha \in L $ that is transcendental over $ K $, meaning every polynomial with coefficients in $ K $ having $ \alpha $ as a root is the zero polynomial.
[4]
Section 9.26 (030D): Transcendence—The Stacks project
Let K/k be a field extension. The transcendence degree of K over k is the cardinality of a transcendence basis of K over k. It is denoted \text{trdeg}_ k(K).
[5]
[PDF] Transcendental extensions - Brandeis
Because of this we can define a purely transcendental field extension to be an extension k(y1,··· ,yn) generated by a set of algebraically independent ...<|control11|><|separator|>
[6]
[PDF] Fields and Galois Theory - James Milne
extensions, and showed that every field can be obtained as an algebraic extension of a purely transcendental extension. He also proved that every field has an ...Missing: primary | Show results with:primary
[7]
[PDF] Section V.1. Field Extensions
Dec 30, 2023 · F is a transcendental extension if at least one element of F is transcendental over K. Example V.1.A. The most common example of an algebraic ...
[8]
[PDF] 18.782 Arithmetic Geometry Lecture Note 12 - MIT OpenCourseWare
Oct 17, 2013 · Definition 12.3. The transcendence degree of a field extension L/K is the cardinality of any (hence every) transcendence basis for L/k.
[9]
[PDF] arXiv:2306.14352v2 [math.NT] 17 Sep 2023
Sep 17, 2023 · After eleven years, in 1893, D. Hilbert [3] and A. Hurwitz [4] managed to give simpler proofs for the transcendence of e and π. In 1882 F.
[10]
[PDF] Fields and Galois Theory - James Milne
Sep 28, 2008 · Swinnerton-Dyer, H. P. F., A brief guide to algebraic number theory. Cambridge, 2001, p133. Page 86. 82. 7 INFINITE GALOIS EXTENSIONS.
[11]
[PDF] Transcendence of e and π - G Eric Moorhouse
ily verify its transcendence. The most important transcendental numbers in nature are e and π, which were shown to be transcenden- tal by Hermite and ...<|control11|><|separator|>
[12]
[PDF] Conjectures. Algebraic independence of transcendental numbers
Apr 23, 2021 · Introduction to Transcendental Number Theory 8. Conjectures. Algebraic independence of transcendental numbers. Michel Waldschmidt. Sorbonne ...
[13]
Lemma 9.26.3 (030F)—The Stacks project
Let E/F be a field extension. A transcendence basis of E over F exists. Any two transcendence bases have the same cardinality.
[14]
[PDF] Zorn's lemma and some applications, II - Keith Conrad
When L/K has a transcendence basis S, the intermediate field K(S) is purely transcen- dental over K, in the sense that it is generated over K by elements that ...
[15]
[PDF] EXPLORING TRANSCENDENTAL EXTENSIONS
Definition: A field E containing a field F is called an extension field of F (or simply an extension of F, denoted by E/F). Such an E is regarded as an ...
[16]
[PDF] Algebraic Geometry (Math 6130)
so the fields of rational functions in (a) and (b) are isomorphic to k(t), which is also the field of rational functions of the affine line k. On the other hand ...
[17]
[PDF] Homework 4 Solutions
Therefore, the automorphisms of the rational function field k(t) that fix k are precisely the fractional linear transformations. Problem 3 [14.2.13] Prove that ...
[18]
[PDF] 18.782 Arithmetic Geometry Problem Set 10 - MIT OpenCourseWare
A derivation on a function field F/k is a k ... that we get a derivation δx for each transcendental x in F. ... rational function field. Compute div dt and prove ...
[19]
[PDF] INFINITE EXTENSIONS 1. The Algebraic Closure Recall that a field ...
The cardinality of a transcendence basis for Ω over k is called the transcendence degree of Ω over k. It is an analogue of the concept of the dimension of a ...
[20]
[PDF] University of Minnesota Research Report - CORE
differential transcendence degree (see Definition 1.3 below),. (1.4). Diff-Trans 8°M/C = c , where c is the cardinality of the continuum, [3]. For the proof ...
[21]
[PDF] Transcendence of e and π
Apr 7, 2006 · Corollary 1 (Hermite–Lindemann). If α is algebraic (over Q) and α 6= 0, then eα is tran- scendental. Hence e and π are transcendental. Proof.
[22]
Lemma 9.26.11 (037J)—The Stacks project
### Summary of Finitely Generated Field Extensions and Transcendence Degree
[23]
Section 53.2 (0BXX): Curves and function fields—The Stacks project
A variety is a curve if and only if its function field has transcendence degree 1, see for example Varieties, Lemma 33.20.3.
[24]
proof of Lindemann-Weierstrass theorem and that e and π - π
Mar 22, 2013 · But e was not shown to be transcendental until 1873 (by Hermite, see [3] and [4] ), and Lindemann showed π to be transcendental as well in the ...
[25]
Section 10.115 (00OW): Noether normalization—The Stacks project
In this section we prove variants of the Noether normalization lemma. The key ingredient we will use is contained in the following two lemmas.
[26]
Derivatives and Derivations
A derivative is an A-linear map from A[X] to A[X], while a derivation is an A-linear map from B to M that satisfies the product rule.Missing: dx | Show results with:dx
[27]
[PDF] Contents - UChicago Math
We have shown that separable algebraic extensions have no Kähler differentials, but that purely transcendental extensions have a free module of rank equal to ...
[28]
[PDF] Samit Dasgupta Duke University Math 790, 3/2/2021 Introduction to ...
Mar 2, 2021 · KÄHLER DIFFERENTIALS AND TRANSCENDENCE DEGREE. Corollary. Let K C L be fields of characteristic 0. Then. dimL ΩL/K = trdK L. More generally ...
[29]
[PDF] Math990, HW1
Let K/k be a finitely generated extension of transcendence degree n and let d be the dimension of the K-vector space of Kahler differentials Ω1. K/k. Prove ...
[30]
[PDF] Lüroth's Theorem, and some related results
Lüroth's Theorem, and some related results, developed as a series of exercises. A simple transcendental extension of a field k means an extension of the ...Missing: paper source
[31]
Lüroth field extensions - ScienceDirect.com
Theorem 1. If F is a Lüroth field extension of field K of characteristic zero that coincides with it's algebraic closure in F then so is a purely transcendental ...Missing: original | Show results with:original
[32]
Lüroth's Theorem -- from Wolfram MathWorld
A theorem that can be stated either in the language of abstract algebraic curves or transcendental extensions.
[33]
[PDF] In Memoriam Ernst Steinitz (1871-1928) - Institute for Mathematics
Sep 12, 2010 · Steinitz defined the notion of transcendence degree and he showed that every field can be obtained as an algebraic extension of a purely ...
[34]
[PDF] A refined version of the Siegel-Shidlovskii theorem
Using Y. André's result on differential equations satisfied by E-functions, we derive an improved version of the Siegel-Shidlovskii theorem. It gives a.Missing: source | Show results with:source
[35]
[PDF] ON p-ADlC G-FUNCTIONS - OSU Math
Siegel's work established the algebraic independence of the values at algebraic points of jE-functions satisfying linear differential equations of the second ...Missing: source | Show results with:source
[36]
[PDF] FIELD AUTOMORPHISMS OF R AND Qp - Keith Conrad
Automorphisms of a rational function field. For a field K, we write K(t) for the field of rational functions in one indeterminate with coefficients in K. A ...
[37]
Fields of definition of function fields and hurwitz families — groups ...
groups as galois groups. M. Fried University of California at Irvine, Irvine, California. Pages ...
[38]
[PDF] 18.782 Introduction to Arithmetic Geometry Fall 2013 Lecture #13 10 ...
Oct 22, 2013 · Let V/k be an affine variety defined over k. The function field k(V ) of V is the fraction field of the coordinate ring k[V ]. We similarly ...
[39]
[PDF] arXiv:2010.07200v1 [math.NT] 14 Oct 2020
Oct 14, 2020 · Moriwaki [28, 29] develops a theory of heights and proves a Northcott property over finitely generated extensions K of Q with transcendence ...
[40]
[PDF] Diophantine approximation, irrationality and transcendence
Jun 2, 2010 · Theorem on the transcendence of e follows. Exercise 8. Using Hermite's method as explained in § 8.3, prove that for any non-zero r ∈ Q(i) ...
[41]
[PDF] Introduction to Model theory Zoé Chatzidakis – CNRS (Paris 7 ...
Theorem. Let T = ACF be the theory of algebraically closed fields. Then T eliminates quantifiers. Moreover, any two models of T of the same characteristic ...Missing: ACF_0 | Show results with:ACF_0