Field Theory

Introduction

Field theory is the theory of fields.

The field theory covered in this article mainly concerns field extensions. A field is an algebraic structure closed under addition, subtraction, multiplication, and division. In competitive programming, operations modulo a prime \(p\) are often required, which is equivalent to performing operations in the finite field \(\mathbf F_p\). Similar to the case of the real field \(\mathbf R\), some problems are more conveniently solved by computing in a larger field (i.e., the complex field \(\mathbf C\)). A common example is using the Fast Fourier Transform to accelerate the multiplication of real-coefficient polynomials. Similar operations can be performed for finite fields. Most readers may be relatively unfamiliar with extensions of finite fields, so understanding the theory of extensions over general fields is beneficial. At the end of this article, some algorithmic applications requiring finite field extensions are presented, along with a brief discussion of extensions over integer rings that may be needed in some applications.

Closely related to field theory is Galois theory. It connects field extensions with their automorphism groups, allowing the properties of field extensions to be understood through group-theoretic tools. Although this theory is often a core part of related algebra courses, it is far removed from the content of competitive programming, so this article will not cover it in detail. Interested readers should refer to relevant textbooks.

Notation

When no ambiguity arises, this article may omit the multiplication symbol for rings and fields and write the ring \((R,+,\cdot)\) as \(R\) and the field \((F,+,\cdot)\) as \(F\). The additive identity of a ring or field is called the zero element, and the multiplicative identity is called the unit element. Moreover, in this article, \(p\) always denotes a prime, and \(q\) always denotes a prime power, which can be written as \(p^n\), where \(n\) is a positive integer.

Field Extensions

Similar to groups and rings, the concepts of subfields and field homomorphisms can be defined.

Subfield

For a field \(F\), if its subring \(E\) is also a field, then \(E\) is called a subfield of \(F\).

Regardless of how the definitions of rings and subrings handle the unit element, a subfield \(E\) must contain the unit element of \(F\)[^subfield-one].

Field Homomorphism

A ring homomorphism \(\varphi:F\rightarrow E\) from field \(F\) to field \(E\) is also called a field homomorphism from \(F\) to \(E\).

Field Homomorphisms and the Unit Element

If, unlike the definition in this article, ring homomorphisms are required to map the unit element to the unit element, then field homomorphisms naturally also require the unit element to map to the unit element. Otherwise, the unit element might also map to the zero element.

Since fields have only trivial ideals, a field homomorphism either maps the entire field to the zero element or must be an embedding. This means that the discussion of field homomorphisms can be reduced to the discussion of subfields.

In the context of fields, the smaller field is often the more familiar one, so it is common to consider larger fields from the perspective of subfields. This leads to the concept of field extensions.

Field Extension

For a field \(F\), if \(F\) is a subfield of \(E\), then \(E\) is called an extension of \(F\), or an extension field, denoted \(E/F\).

Notation for Field Extensions

Although the notation is similar, the concept of a field extension is unrelated to quotient rings and should not be confused.

Examples

The complex field \(\mathbf C\) is an extension of the real field \(\mathbf R\), and the real field \(\mathbf R\) is an extension of the rational field \(\mathbf Q\).

Degree of a Field Extension

For a field extension \(E/F\), \(E\) is always a vector space over \(F\). The dimension of this vector space is the degree of the field extension.

Degree of a Field Extension

The degree of a field extension \(E/F\) is the dimension of \(E\) as a vector space over \(F\), i.e., \(\dim_F(E)\), denoted \([E:F]\). If the degree is finite, the extension is called a finite extension; otherwise, it is called an infinite extension.

Examples

The degree \([\mathbf C:\mathbf R]\) of the extension \(\mathbf C/\mathbf R\) is \(2\), so it is a finite extension. The extension \(\mathbf R/\mathbf Q\) is an infinite extension.

The degree of field extensions satisfies a multiplicative property.

Theorem

Let \(F\subseteq K\subseteq E\) be fields. Then the degrees of the extensions satisfy \([E:F]=[E:K][K:F]\).

Proof

For infinite degrees, this is obvious. Otherwise, if \(\{\alpha_i\}\) is a basis of \(E\) as a vector space over \(K\), and \(\{\beta_j\}\) is a basis of \(K\) as a vector space over \(F\), then it can be verified that \(\{\alpha_i\beta_j\}\) is a basis of \(E\) as a vector space over \(F\).

This article mainly discusses finite field extensions.

Characteristic of a Field

A natural starting point for studying field extensions is the smallest subfield containing the unit element of \(F\). This field is called the prime subfield of \(F\).

The structure of the prime subfield is uniquely determined by the properties of the field's unit element. The characteristic of the field summarizes these properties.

Characteristic of a Field

The characteristic of a field \(F\) is the smallest positive integer \(n\) such that \(n\cdot 1=0\); if no such \(n\) exists, the characteristic is said to be \(0\). Here, \(n\cdot 1\) denotes the sum of \(n\) unit elements \(1\). If the characteristic of \(F\) is not \(0\), then \(F\) is said to be of finite characteristic.

The characteristic of a field can be understood through ring homomorphisms. The integer ring \(\mathbf Z\) is the closure under addition, subtraction, and multiplication starting from \(0\) and \(1\). It can be seen as a "prototype": all rings containing a unit element should "inherit" part of the structure of \(\mathbf Z\)[^initial-object-ring]. Thus, for a field \(F\), consider the ring homomorphism \(\varphi:\mathbf Z\rightarrow F\) with \(\varphi(1)=1\). Such a ring homomorphism is uniquely determined: it maps \(n\in\mathbf N_+\) to \(n\cdot 1\), the sum of \(n\) unit elements \(1\). The image \(\varphi(\mathbf Z)\) embeds into \(F\) and must be unitary, commutative, and without zero divisors, hence an integral domain. Therefore, the kernel \(\ker\varphi\) must be a prime ideal. The prime ideals of \(\mathbf Z\) are of the form \((n)\), where \(n=0\) or \(n\) is a prime. The resulting \(n\) is the characteristic of the field.

The characteristic determines the structure of the prime subfield:

When the characteristic is \(0\), the homomorphism \(\varphi\) is injective, and \(\mathbf Z\) embeds into \(F\). The rational field \(\mathbf Q\), as the smallest field containing \(\mathbf Z\), must also embed into \(F\), and it is the prime subfield of \(F\).
When the characteristic is a prime \(p\), the image \(\varphi(\mathbf Z)=\mathbf Z/p\mathbf Z\) embeds into \(F\). Since \(\mathbf Z/p\mathbf Z\) is already a field, denoted \(\mathbf F_p\), it is the prime subfield of \(F\).

These discussions essentially prove the following:

Theorem

The characteristic of a field \(F\) can only be \(0\) or a prime \(p\). Fields of characteristic \(0\) have prime subfield \(\mathbf Q\), and fields of characteristic \(p\) have prime subfield \(\mathbf F_p\).

The \(\mathbf Q\) and \(\mathbf F_p\) in the theorem are also called prime fields, i.e., fields whose only subfield is themselves. Finite fields must have finite characteristic because fields of characteristic \(0\) contain at least the subfield \(\mathbf Q\).

Fields of finite characteristic and characteristic \(0\) often have different properties. For example, fields of finite characteristic have the following property:

Theorem

Let \(F\) have characteristic \(p\). Then:

In the additive group of \(F\), the order of every nonzero element is \(p\), i.e., \(px=0\) for all \(x\in F\).
"Freshman's dream": \((x+y)^p=x^p+y^p\) for all \(x,y\in F\). Consequently, the map \(x\mapsto x^p\) is an injective endomorphism of \(F\), called the Frobenius endomorphism.

Proof

For the first property, note that \(px=(p1)x=0x=0\). For the second property, note that in the binomial expansion of \((x+y)^p\), all coefficients except those of \(x^p\) and \(y^p\) are multiples of \(p\), so by the first property, \((x+y)^p=x^p+y^p\). To verify that \(x\mapsto x^p\) is an endomorphism, also verify \((xy)^p=x^py^p\), which holds because multiplication in a field is commutative. Finally, a ring homomorphism between fields mapping the unit to the unit must be injective.

For finite fields, the Frobenius endomorphism is also surjective, hence an automorphism.

Simple Extensions

Similar to the extension from the real field to the complex field, many extensions can be constructed by adding extra elements to a field and specifying their algebraic properties. In general, to avoid complications from specifying properties, consider the case where elements from \(E\setminus F\) in an extension \(E/F\) are adjoined to \(F\). The rules for operations between these extra elements and elements of \(F\) are already determined in the larger field \(E\).

Extension Generated by a Subset

Let \(E/F\) be a field extension and \(S\subseteq E\). The extension generated by \(S\) over \(F\) is the smallest subfield of \(E\) containing both \(F\) and \(S\), denoted \(F(S)\).

The simplest case is when the set \(S\) has few elements.

Finitely Generated Extension

Let \(E/F\) be a field extension. If there exists a finite set \(S=\{\alpha_1,\cdots,\alpha_n\}\subseteq E\) such that \(E=F(S)\), then \(E\) is called a finitely generated extension of \(F\), also denoted \(F(\alpha_1,\cdots,\alpha_n)\).

Simple Extension

Let \(E/F\) be a field extension. If there exists \(\alpha\in E\) such that \(E=F(\alpha)\), then \(E\) is called a simple extension of \(F\). The element \(\alpha\) is called a primitive element of this simple extension.

Examples

These examples are obtained by adding elements from \(\mathbf C\) to \(\mathbf Q\).

For a square-free integer \(D\neq 0,1\), the quadratic field \(\mathbf Q(\sqrt D)\) is a simple extension of \(\mathbf Q\) by adding \(\sqrt D\in\mathbf C\setminus\mathbf Q\). Its degree is \(2\), since \(\{1,\sqrt D\}\) forms a basis.
The field \(\mathbf Q(\sqrt 2,\sqrt 3)\) is an extension of \(\mathbf Q\) by adding \(\sqrt 2\) and \(\sqrt 3\). Of course, \(\mathbf Q(\sqrt 2,\sqrt 3)=\mathbf Q(\sqrt 2)(\sqrt 3)=\mathbf Q(\sqrt 3)(\sqrt 2)\), meaning the order of addition does not matter. This is also a simple extension because \(\mathbf Q(\sqrt 2,\sqrt 3)=\mathbf Q(\sqrt 2+\sqrt 3)\). Its degree is \(4\), since \(\{1,\sqrt 2,\sqrt 3,\sqrt 6\}\) forms a basis.
The field \(\mathbf Q(\pi)\) is a simple extension, where \(\pi\) is the circle constant. It is an infinite extension because \(\mathbf Q[\pi]\subseteq \mathbf Q(\pi)\) already has a basis \(\{1,\pi,\pi^2,\cdots\}\).
The field \(\mathbf Q(\pi,\mathrm e)\) is a finitely generated extension but not a simple extension. Here, \(\pi\) is the circle constant, and \(\mathrm e\) is the base of the natural logarithm.

These examples show that simple extensions can vary greatly in nature, depending on the properties of the added element.

Algebraic Extensions

To analyze all possible scenarios when adding an element to a field, consider the ring homomorphism from the polynomial ring \(F[x]\) to the extension \(E/F\), analogous to the earlier discussion of the characteristic. Here, \(F[x]\) plays the role of \(\mathbf Z\): it is the "prototype" of the structure obtained by adjoining an indeterminate \(x\) to \(F\) and closing under addition, subtraction, and multiplication[^polynomial-universal].

Let the ring homomorphism \(\varphi:F[x]\rightarrow E\) satisfy that \(\varphi\) restricted to \(F\) is the identity and \(\varphi(x)=\alpha\), i.e., mapping the indeterminate to some element in \(E\). Since the image \(\varphi(F[x])=F[\alpha]\) is necessarily an integral domain, the kernel \(\ker\varphi\) must be a prime ideal of \(F[x]\). The polynomial ring over a field is a principal ideal domain, so the kernel must be of the form \((f(x))\), where \(f(x)=0\) or \(f(x)\) is irreducible in \(F[x]\). This leads to the following discussion:

When \(\ker\varphi=\{0\}\), the polynomial ring \(F[x]\) embeds into \(E\), and its image \(F[\alpha]\) is an integral domain. Thus, the smallest field in \(E\) containing both \(F\) and \(\alpha\) is the field of fractions of \(F[\alpha]\), i.e., \(F(\alpha)\). This notation can be interpreted either as substituting \(\alpha\) for the indeterminate in the rational function field \(F(x)\), or as the simple extension of \(F\) generated by \(\alpha\): both interpretations yield the same result in this context.
When \(\ker\varphi=(f(x))\) with \(f(x)\) irreducible, we have \(\varphi(f(x))=f(\alpha)=0\), meaning \(\alpha\in E\) is a root of the polynomial \(f(x)\) over \(F\). Since \(F\) is a field, assume \(f(x)\) is monic. The image of \(\varphi\) is the field \(F(\alpha)\), so

\[ F[x]/(f(x))\cong F(\alpha). \]

Two subcases arise:
1. If \(f(x)\) is linear, i.e., \(f(x)=x-\alpha\), then \(\alpha\in F\), so the extension \(F(\alpha)=F\) is trivial.
2. Otherwise, \(f(x)\) is an irreducible polynomial of degree greater than one, and \(\alpha\in E\setminus F\). The image \(F[\alpha]\) is already a field containing \(F\) and \(\alpha\), so it is \(F(\alpha)\), the extension of \(F\) generated by \(\alpha\), and \(F(\alpha)\supset F\) is nontrivial.

These discussions motivate the following definitions:

Algebraic and Transcendental Elements

For an extension \(E/F\), if an element \(\alpha\in E\) is a root of some nonzero polynomial \(f(x)\) over \(F\), then \(\alpha\) is called an algebraic element over \(F\); otherwise, \(\alpha\) is called a transcendental element over \(F\).

Minimal Polynomial

For an algebraic element \(\alpha\) over a field \(F\), the monic polynomial \(f(x)\) of smallest degree having \(\alpha\) as a root is called its minimal polynomial.

The minimal polynomial here is the irreducible polynomial \(f(x)\) from the earlier analysis. Of course, it can also be directly proven that minimal polynomials are irreducible. The minimality of \(f(x)\) implies that any polynomial over \(F\) having \(\alpha\) as a root must be divisible by \(f(x)\).

Examples

\(\sqrt 2\) is algebraic over \(\mathbf Q\), with minimal polynomial \(x^2-2\).
\(\sqrt 2\) is algebraic over \(\mathbf R\), with minimal polynomial \(x-\sqrt 2\).
\(\pi\) is transcendental over \(\mathbf Q\).
In general, algebraic elements over \(\mathbf Q\) are called algebraic numbers, and transcendental elements are called transcendental numbers. In particular, if the minimal polynomial of an algebraic number is monic, it is called an algebraic integer. The algebraic integers in an algebraic extension form a ring. For example, the algebraic integers in the quadratic field \(\mathbf Q(\sqrt{D})\) form the quadratic integer ring \(\mathbf Z[\omega]\). See the quadratic integer ring page for notation.

Algebraic and Transcendental Extensions

For an extension \(E/F\), if every element of \(E\) is algebraic over \(F\), then \(E\) is called an algebraic extension of \(F\); otherwise, it is called a transcendental extension of \(F\).

The outcome of a simple extension depends on the nature of the added element. When the added element is transcendental, the simple extension is always isomorphic to the field of rational functions. In this case, no further simplification is possible. However, when the added element is algebraic, the simple extension is essentially \(F[\alpha]\), obtained by replacing the indeterminate \(x\) in the polynomial ring \(F[x]\) with \(\alpha\). From an elementary perspective, compared to the transcendental case, elements in the extension field can have no denominators; this means that a process analogous to "rationalizing the denominator" in elementary arithmetic is always possible in simple algebraic extensions. Since the extensions involved in competitive programming are mostly simple algebraic extensions, the next section discusses their computation in more detail.

The importance of simple algebraic extensions is also reflected in the following theorem:

Theorem

A field extension is finite if and only if it is a finitely generated algebraic extension.

Proof

Let \(F\) be a field, and \(E=F(\alpha_1,\cdots,\alpha_n)\) a finitely generated algebraic extension, i.e., each \(\alpha_i\) is algebraic over \(F\). Let \(E_i=F(\alpha_1,\cdots,\alpha_i)\), so \(E_0=F\) and \(E_n=E\). Note that \(\alpha_i\) must be algebraic over \(E_{i-1}\), since the minimal polynomial of \(\alpha_i\) over \(F\) is also a polynomial over \(E_{i-1}\); moreover, the degree of the minimal polynomial of \(\alpha_i\) over \(E_{i-1}\) cannot exceed that over \(F\). Hence, \([E_i:E_{i-1}]\) is finite, and by the multiplicative property of extension degrees, \([E:F]=\prod_{i=1}^n[E_i:E_{i-1}]\) is finite. Conversely, starting from \(E_0=F\), for each constructed \(E_i\), choose an element \(\alpha_{i+1}\in E\setminus E_i\) and adjoin it to \(E_i\) to obtain \(E_{i+1}=E_i(\alpha_{i+1})\), until \(E_n=E\). Since the degree decreases at each step, this process must terminate in finitely many steps. Therefore, a finite extension must be a finitely generated algebraic extension.

This means that to understand finite extensions, it suffices to understand simple algebraic extensions, as any finite extension can be obtained through finitely many simple algebraic extensions.

Structure and Computation of Simple Algebraic Extensions

In this section, let \(F\) be a number field, \(E\) an extension of \(F\), and \(\alpha\in E\setminus F\) an algebraic element over \(F\). Let the minimal polynomial of \(\alpha\) be \(f(x)\), a monic polynomial of degree \(n\), i.e.,

\[ f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0, \]

where \(a_0,a_1,\cdots,a_{n-1}\in F\) and \(f(x)\) is irreducible over \(F\).

The isomorphism \(F(\alpha)\cong F[x]/(f(x))\) indicates that operations in \(F(\alpha)\) are computations modulo \(f(x)\). By polynomial division with remainder, it suffices to consider congruence classes of polynomials of degree less than \(n=\deg f(x)\). For these polynomials, a natural basis is \(\{1,\alpha,\cdots,\alpha^{n-1}\}\). Thus, we have:

Theorem

Under the assumptions of this section, the extension field \(F(\alpha)\) can be written as

\[ F(\alpha)=\{\lambda(\alpha)=\lambda_0+\lambda_1\alpha+\cdots+\lambda_{n-1}\alpha^{n-1}:\lambda_0,\lambda_1,\cdots,\lambda_{n-1}\in F\}. \]

Here, \(\lambda(x)\) ranges over all polynomials of degree less than \(n\). Therefore, the extension degree \([F(\alpha):F]=n\), i.e., the degree of the minimal polynomial of \(\alpha\). In the extension field, addition of elements \(\lambda(\alpha)\) and \(\mu(\alpha)\) is polynomial addition, i.e., adding corresponding coefficients. Multiplication of \(\lambda(\alpha)\) and \(\mu(\alpha)\) yields \(\rho(\alpha)\), where \(\rho(x)\) is the remainder of \(\lambda(x)\mu(x)\) divided by \(f(x)\).

As a field, division in \(F(\alpha)\) can also be computed. According to the multiplication process described in the theorem, this is equivalent to solving a linear congruence equation over the polynomial ring. Analogous to integers, to compute the quotient \(\lambda(\alpha)/\mu(\alpha)\), first determine the multiplicative inverse of \(\mu(\alpha)\), then multiply by \(\lambda(\alpha)\). To compute the inverse of \(\mu(\alpha)\), solve the congruence \(\mu(x)\xi(x)\equiv 1\pmod{f(x)}\). This can be done using the extended Euclidean algorithm.

Below are a few concrete examples to illustrate the computational details.

Example

Consider the extension \(\mathbf Q(\alpha)\), where \(\alpha\) is a root of \(x^3-2x-2=0\). Compute

\[ \frac{1+\alpha}{1+\alpha+\alpha^2}. \]

First, compute the inverse of \(1+\alpha+\alpha^2\), i.e., solve the congruence

\[ (x^2+x+1)\xi(x)+(x^3-2x-2)\nu(x)=1. \]

Apply the extended Euclidean algorithm. Perform the Euclidean algorithm:

\[ \begin{aligned} x^3-2x-2 &= (x-1)(x^2+x+1)+(-2x-1),\\ x^2+x+1 &= \left(-\frac12x-\frac14\right)(-2x-1)+\frac34,\\ -2x-1 &= \left(-\frac{8}{3}x-\frac{4}{3}\right)\frac{3}{4}. \end{aligned} \]

Then compute the coefficients for the congruence:

\[ \begin{aligned} \frac{3}{4} &=(x^2+x+1)+\left(\frac12x+\frac14\right)(-2x-1)\\ &=(x^2+x+1)+\left(\frac12x+\frac14\right)\left((x^3-2x-2)-(x-1)(x^2+x+1)\right)\\ &=\left(-\frac12x^2+\frac14x+\frac54\right)(x^2+x+1)+\left(\frac12x+\frac14\right)(x^3-2x-2). \end{aligned} \]

Thus, the solution is

\[ \xi(x)=-\frac23x^2+\frac13x+\frac53,\ \nu(x)=\frac23x+\frac13. \]

This means the inverse of \(1+\alpha+\alpha^2\) is

\[ -\frac23\alpha^2+\frac13\alpha+\frac53. \]

Second, multiply the inverse by \(1+\alpha\):

\[ \begin{aligned} (1+\alpha)\left(-\frac23\alpha^2+\frac13\alpha+\frac53\right) &=-\frac23\alpha^3-\frac13\alpha^2+2\alpha+\frac53\\ &=-\frac23(2\alpha+2)-\frac13\alpha^2+2\alpha+\frac53\\ &=-\frac13\alpha^2+\frac23\alpha+\frac13. \end{aligned} \]

This is the final answer.

The example only used the condition that \(\alpha\) is a root of the equation, without specifying which root. The polynomial \(x^3-2x-2=0\) has one real root and a pair of complex conjugate roots in \(\mathbf C\). Adjoining any of them to \(\mathbf Q\) yields isomorphic extension fields. That is, these three distinct roots are algebraically indistinguishable.

In general, for an irreducible polynomial \(f(x)\) over a field \(F\), if there are distinct roots \(\alpha\neq\beta\) in an extension, these roots exhibit the same algebraic properties when used to form simple extensions of \(F\). Such roots are called conjugates. The usual notion of complex conjugation is a special case of this concept for the extension \(\mathbf C/\mathbf R\).

Example

Consider the extension \(\mathbf F_2(\alpha)\), where \(\alpha\) is a root of \(x^2+x+1=0\). For \(a+b\alpha\) and \(c+d\alpha\), the operations are

\[ \begin{aligned} (a+b\alpha)+(c+d\alpha)&=(a+c)+(b+d)\alpha,\\ (a+b\alpha)(c+d\alpha)&=ac+(ad+bc)\alpha+bd\alpha^2\\ &=(ac+bd)+(ad+bc+bd)\alpha. \end{aligned} \]

This provides rules similar to those for complex numbers. Most readers may be unfamiliar with such a root \(\alpha\), but this does not hinder performing operations in this field. In fact, \([\mathbf F_2(\alpha):\mathbf F_2]=2\), so as a vector space, \(|\mathbf F_2(\alpha)|=4\), yielding a finite field of size \(4\). Later we will see that all finite fields are constructed this way.

For small-scale computations, modulo operations with a monic polynomial \(f(x)\) can often be performed by substituting

\[ x^n=-a_{n-1}x^{n-1}-\cdots-a_1x-a_0 \]

to reduce the degree of the target polynomial. Moreover, for low-degree extensions, coefficient operation rules can often be computed directly, allowing implementations similar to complex number classes without needing to compute modulo each time.

As an example of simple algebraic extensions, refer to the reference implementation of finite fields below.

The algorithms described here can only handle extensions of relatively low degree in practice, which is sufficient for most competitive programming applications. For cases where the extension degree becomes a complexity bottleneck, appropriate polynomial techniques (Fast Fourier Transform, Fast Number Theoretic Transform, Polynomial Fast Division/Modulo, Polynomial Euclidean Algorithm, etc.) should be used to accelerate computations.

Splitting Fields

The previous section provided a detailed discussion of the structure of simple algebraic extensions. However, such extensions are often insufficient:

Example

Consider the extension \(\mathbf Q(\sqrt[3]{2})/\mathbf Q\). The algebraic element \(\sqrt[3]{2}\) has minimal polynomial \(x^3-2\) over \(\mathbf Q\). In \(\mathbf C\), \(x^3-2\) has three roots: \(\sqrt[3]{2},\sqrt[3]{2}\omega,\sqrt[3]{2}\omega^2\), where \(\omega=\mathrm{e}^{2\pi\mathrm{i}/3}\) is a primitive cube root of unity. Although \(\mathbf Q(\sqrt[3]{2})\cong\mathbf Q(\sqrt[3]{2}\omega)\cong\mathbf Q(\sqrt[3]{2}\omega^2)\), \(\mathbf Q(\sqrt[3]{2})\) does not contain the other two roots, making operations like \(\sqrt[3]{2}+\sqrt[3]{2}\omega\) impossible. To fully consider all three roots, further extension of \(\mathbf Q(\sqrt[3]{2})\) is needed, i.e., to \(\mathbf Q(\sqrt[3]{2},\sqrt[3]{2}\omega,\sqrt[3]{2}\omega^2)\).

As mentioned earlier, such an extension can be achieved by successive simple extensions. Note that the minimal polynomial of \(\sqrt[3]{2}\omega\) over \(\mathbf Q\) and over \(\mathbf Q(\sqrt[3]{2})\) are different: the former is \(x^3-2\in\mathbf Q[x]\), while the latter is \(x^2+\sqrt[3]{2}x+\sqrt[3]{4}\in \mathbf Q(\sqrt[3]{2})[x]\), since

\[ x^3-2 = (x-\sqrt[3]{2})(x^2+\sqrt[3]{2}x+\sqrt[3]{4}). \]

The original minimal polynomial splits off a linear factor after extension, so the remaining roots have minimal polynomials of lower degree over the extended field. The process of successive extensions is a process of "splitting" the polynomial. Therefore, at each simple extension, the minimal polynomial must be recomputed.

Adjoining all roots of a polynomial to a field yields the splitting field of the polynomial.

Split

Let \(F\) be a field. A polynomial \(f(x)\) is said to split over \(F\) if it can be factored into linear factors in \(F[x]\).

Splitting Field

For a polynomial \(f(x)\) over a field \(F\), an extension \(E/F\) is called a splitting field of \(f(x)\) if \(f(x)\) splits over \(E\) but not over any proper subfield of \(E\).

It can be shown that, like simple extensions, the splitting field of a given polynomial is unique up to isomorphism, independent of the construction method. Splitting fields are always finite extensions.

Normal Extension

An algebraic extension \(E/F\) is called a normal extension if for every \(\alpha\in E\), the minimal polynomial of \(\alpha\) splits over \(E\).

Normal extensions play a fundamental role in Galois theory.

Algebraically Closed Fields

Most extension concepts discussed earlier require working within a larger field. Although for simple extensions, polynomial rings allow constructing extensions without relying on a larger field, no such method exists in general. For the rational field \(\mathbf Q\) and real field \(\mathbf R\), algebraic extensions can always be assumed to lie inside the complex field \(\mathbf C\). For finite fields, there is no similar known field. In fact, for any field, an algebraic closure exists, allowing all algebraic extensions to be assumed within it. This completely resolves the issue.

Algebraic Closure

For a field \(F\), an extension \(\overline F\) of \(F\) is called an algebraic closure of \(F\) if \(\overline F\) is algebraic over \(F\) and every \(f(x)\in F[x]\) splits over \(\overline F\).

An algebraic closure is a normal extension of the field. Its construction essentially involves adjoining all possible roots of polynomials to the field. Like splitting fields, the algebraic closure of a field is unique up to isomorphism.

Theorem

Every field \(F\) has an algebraic closure.

Proof

The difficulty lies in set theory. Here we cite a proof by Artin.

The first part constructs an extension \(K_1/F\) such that every polynomial over \(F\) has at least one root in \(K_1\). For \(F\), consider the multivariate polynomial ring[^multi-poly-ring] \(R=F[\cdots,x_f,\cdots]\), where the indeterminates \(x_f\) are indexed by all monic polynomials over \(F\). Let \(I\) be the ideal generated by all \(f(x_f)\). First, \(I\neq R\), so a maximal ideal \(M\supseteq I\) exists. Otherwise, if \(1\in I\), there exist finitely many monic polynomials \(f_i\) over \(F\) and corresponding elements \(g_i\in R\) such that \(g_1f_1(x_{f_1})+\cdots+g_kf_k(x_{f_k})=1\). Let \(F(\alpha_1,\cdots,\alpha_k)\) be the algebraic extension of \(F\) obtained by adjoining roots \(\alpha_i\) of \(f_i(x)\). Substituting \(x_{f_i}=\alpha_i\) and all other indeterminates \(x_f\) appearing in \(g_i\) as \(0\) in the identity yields \(0=1\) in \(F(\alpha_1,\cdots,\alpha_k)\), a contradiction. Hence, \(I\neq R\), and the maximal ideal \(M\) is valid. The quotient \(R/M\) is a field, denoted \(K_1\), and every monic polynomial \(f(x)\) over \(F\) has a root \(\overline{x_f}\) in \(K_1\).

The second part inductively obtains an algebraically closed field \(K\) containing \(F\) (definition below). Repeat the construction: from \(K_i\), construct \(K_{i+1}\) such that every polynomial over \(K_i\) has at least one root in \(K_{i+1}\). Moreover, \(K_i\) naturally embeds into \(K_{i+1}\), so define their union \(K=\bigcup_{i=1}^\infty K_i\). It is easy to verify that \(K\) is also a field, and any polynomial over \(K\) has coefficients entirely in some \(K_i\), so one of its roots appears in \(K_{i+1}\subseteq K\). This shows that every polynomial over \(K\) has at least one root in \(K\), so \(K\) is algebraically closed.

Finally, let \(\overline F\) be the set of all elements in \(K\) that are algebraic over \(F\). It is clearly a field; for any \(\alpha,\beta\in\overline F\), we have \(\alpha\pm\beta,\alpha\beta,\alpha/\beta\in F(\alpha,\beta)\subseteq\overline F\). It is also an algebraic extension of \(F\), since its elements are algebraic over \(F\). For any polynomial \(f(x)\) over \(F\), all its roots are algebraic over \(F\), hence in \(\overline F\), so \(f(x)\) splits into linear factors. Thus, \(\overline F\) is an algebraic closure of \(F\).

Examples

The algebraic closure of \(\mathbf R\) is \(\mathbf C\).
The algebraic closure of \(\mathbf Q\) is the set of all algebraic numbers (i.e., algebraic elements in the extension \(\mathbf C/\mathbf Q\)), denoted \(\overline{\mathbf Q}\).

All algebraic extensions of an algebraic closure are trivial. Such fields are called algebraically closed fields.

Algebraically Closed Field

A field \(F\) is called algebraically closed if every non-constant polynomial \(f(x)\) over \(F\) has at least one root \(\alpha\in F\).

In fact, it has the following equivalent definitions:

Theorem

For a field \(F\), the following are equivalent:

\(F\) is algebraically closed.
Every polynomial \(f(x)\) over \(F\) splits.
The only irreducible polynomials over \(F\) are linear.
\(F\) has no nontrivial algebraic extensions.
\(F\) has no nontrivial finite extensions.
\(F\) is the algebraic closure of some field.

Proof

The equivalence of the first five is clear from definitions. For the sixth, an algebraically closed field is clearly its own algebraic closure, as it has no nontrivial algebraic extensions. Conversely, to show that the algebraic closure of a field is algebraically closed, let \(\overline F\) be an algebraic closure of \(F\), and let \(f(x)\) be a polynomial over \(\overline F\). Let \(\alpha\) be a root of \(f(x)\) in its splitting field, and let \(S\subseteq\overline F\) be the set of nonzero coefficients of \(f(x)\). Then \(F(S)(\alpha)=F(S\cup\{\alpha\})\) is a finite extension, so \(\alpha\) must be algebraic over \(F\). By definition of algebraic closure, the minimal polynomial of \(\alpha\) over \(F\) splits over \(\overline F\), so \(\alpha\in\overline F\). This shows that every polynomial

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