Minimal polynomial (field theory)

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Short description: Concept in abstract algebra


In field theory, a branch of mathematics, the minimal polynomial of an element α of an extension field of a field is, roughly speaking, the polynomial of lowest degree having coefficients in the smaller field, such that α is a root of the polynomial. If the minimal polynomial of α exists, it is unique. The coefficient of the highest-degree term in the polynomial is required to be 1.

More formally, a minimal polynomial is defined relative to a field extension E/F and an element of the extension field E/F. The minimal polynomial of an element, if it exists, is a member of F[x], the ring of polynomials in the variable x with coefficients in F. Given an element α of E, let Jα be the set of all polynomials f(x) in F[x] such that f(α) = 0. The element α is called a root or zero of each polynomial in Jα

More specifically, Jα is the kernel of the ring homomorphism from F[x] to E which sends polynomials g to their value g(α) at the element α. Because it is the kernel of a ring homomorphism, Jα is an ideal of the polynomial ring F[x]: it is closed under polynomial addition and subtraction (hence containing the zero polynomial), as well as under multiplication by elements of F (which is scalar multiplication if F[x] is regarded as a vector space over F).

The zero polynomial, all of whose coefficients are 0, is in every Jα since 0αi = 0 for all α and i. This makes the zero polynomial useless for classifying different values of α into types, so it is excepted. If there are any non-zero polynomials in Jα, i.e. if the latter is not the zero ideal, then α is called an algebraic element over F, and there exists a monic polynomial of least degree in Jα. This is the minimal polynomial of α with respect to E/F. It is unique and irreducible over F. If the zero polynomial is the only member of Jα, then α is called a transcendental element over F and has no minimal polynomial with respect to E/F.

Minimal polynomials are useful for constructing and analyzing field extensions. When α is algebraic with minimal polynomial f(x), the smallest field that contains both F and α is isomorphic to the quotient ring F[x]/⟨f(x)⟩, where ⟨f(x)⟩ is the ideal of F[x] generated by f(x). Minimal polynomials are also used to define conjugate elements.

Definition

Let E/F be a field extension, α an element of E, and F[x] the ring of polynomials in x over F. The element α has a minimal polynomial when α is algebraic over F, that is, when f(α) = 0 for some non-zero polynomial f(x) in F[x]. Then the minimal polynomial of α is defined as the monic polynomial of least degree among all polynomials in F[x] having α as a root.

Properties

Throughout this section, let E/F be a field extension over F as above, let α ∈ E be an algebraic element over F and let Jα be the ideal of polynomials vanishing on α.

Uniqueness

The minimal polynomial f of α is unique.

To prove this, suppose that f and g are monic polynomials in Jα of minimal degree n > 0. We have that r := f−g ∈ Jα (because the latter is closed under addition/subtraction) and that m := deg(r) < n (because the polynomials are monic of the same degree). If r is not zero, then r / cm (writing cm ∈ F for the non-zero coefficient of highest degree in r) is a monic polynomial of degree m < n such that r / cm ∈ Jα (because the latter is closed under multiplication/division by non-zero elements of F), which contradicts our original assumption of minimality for n. We conclude that 0 = r = f − g, i.e. that f = g.

Irreducibility

The minimal polynomial f of α is irreducible, i.e. it cannot be factorized as f = gh for two polynomials g and h of strictly lower degree.

To prove this, first observe that any factorization f = gh implies that either g(α) = 0 or h(α) = 0, because f(α) = 0 and F is a field (hence also an integral domain). Choosing both g and h to be of degree strictly lower than f would then contradict the minimality requirement on f, so f must be irreducible.

Minimal polynomial generates Jα

The minimal polynomial f of α generates the ideal Jα, i.e. every g in Jα can be factorized as g=fh for some h' in F[x].

To prove this, it suffices to observe that F[x] is a principal ideal domain, because F is a field: this means that every ideal I in F[x], Jα amongst them, is generated by a single element f. With the exception of the zero ideal I = {0}, the generator f must be non-zero and it must be the unique polynomial of minimal degree, up to a factor in F (because the degree of fg is strictly larger than that of f whenever g is of degree greater than zero). In particular, there is a unique monic generator f, and all generators must be irreducible. When I is chosen to be Jα, for α algebraic over F, then the monic generator f is the minimal polynomial of α.


Examples

Minimal polynomial of a Galois field extension

Given a Galois field extension [math]\displaystyle{ L/K }[/math] the minimal polynomial of any [math]\displaystyle{ \alpha \in L }[/math] not in [math]\displaystyle{ K }[/math] can be computed as

[math]\displaystyle{ f(x) = \prod_{\sigma \in \text{Gal}(L/K)} (x - \sigma(\alpha)) }[/math]

if [math]\displaystyle{ \alpha }[/math] has no stabilizers in the Galois action. Since it is irreducible, which can be deduced by looking at the roots of [math]\displaystyle{ f' }[/math], it is the minimal polynomial. Note that the same kind of formula can be found by replacing [math]\displaystyle{ G = \text{Gal}(L/K) }[/math] with [math]\displaystyle{ G/N }[/math] where [math]\displaystyle{ N = \text{Stab}(\alpha) }[/math] is the stabilizer group of [math]\displaystyle{ \alpha }[/math]. For example, if [math]\displaystyle{ \alpha \in K }[/math] then its stabilizer is [math]\displaystyle{ G }[/math], hence [math]\displaystyle{ (x-\alpha) }[/math] is its minimal polynomial.

Quadratic field extensions

Q(√2)

If F = Q, E = R, α = √2, then the minimal polynomial for α is a(x) = x2 − 2. The base field F is important as it determines the possibilities for the coefficients of a(x). For instance, if we take F = R, then the minimal polynomial for α = √2 is a(x) = x − √2.

Q(√d )

In general, for the quadratic extension given by a square-free [math]\displaystyle{ d }[/math], computing the minimal polynomial of an element [math]\displaystyle{ a + b\sqrt{d\,} }[/math] can be found using Galois theory. Then

[math]\displaystyle{ \begin{align} f(x) &= (x - (a + b\sqrt{d\,}))(x - (a - b\sqrt{d\,})) \\ &= x^2 - 2ax + (a^2 - b^2d) \end{align} }[/math]

in particular, this implies [math]\displaystyle{ 2a \in \mathbb{Z} }[/math] and [math]\displaystyle{ a^2 - b^2d \in \mathbb{Z} }[/math]. This can be used to determine [math]\displaystyle{ \mathcal{O}_{\mathbb{Q}(\sqrt{d\,}\!\!\!\;\;)} }[/math] through a series of relations using modular arithmetic.

Biquadratic field extensions

If α = √2 + √3, then the minimal polynomial in Q[x] is a(x) = x4 − 10x2 + 1 = (x − √2 − √3)(x + √2 − √3)(x − √2 + √3)(x + √2 + √3).

Notice if [math]\displaystyle{ \alpha = \sqrt{2} }[/math] then the Galois action on [math]\displaystyle{ \sqrt{3} }[/math] stabilizes [math]\displaystyle{ \alpha }[/math]. Hence the minimal polynomial can be found using the quotient group [math]\displaystyle{ \text{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q})/\text{Gal}(\mathbb{Q}(\sqrt{3})/\mathbb{Q}) }[/math].

Roots of unity

The minimal polynomials in Q[x] of roots of unity are the cyclotomic polynomials. The roots of the minimal polynomial of 2cos(2pi/n) are twice the real part of the primitive roots of unity.

Swinnerton-Dyer polynomials

The minimal polynomial in Q[x] of the sum of the square roots of the first n prime numbers is constructed analogously, and is called a Swinnerton-Dyer polynomial.

See also

References



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