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 αE 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 := fgJα (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 cmF for the non-zero coefficient of highest degree in r) is a monic polynomial of degree m < n such that r / cmJα (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 = fg, 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 ${\displaystyle L/K}$ the minimal polynomial of any ${\displaystyle \alpha \in L}$ not in ${\displaystyle K}$ can be computed as

${\displaystyle f(x)=\prod _{\sigma \in {\text{Gal))(L/K)}(x-\sigma (\alpha ))}$

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

#### 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) = x2.

#### Q(√d )

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

{\displaystyle {\begin{aligned}f(x)&=(x-(a+b{\sqrt {d\,))))(x-(a-b{\sqrt {d\,))))\\&=x^{2}-2ax+(a^{2}-b^{2}d)\end{aligned))}

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

If α = 2 + 3, then the minimal polynomial in Q[x] is a(x) = x4 − 10x2 + 1 = (x23)(x + 23)(x2 + 3)(x + 2 + 3).

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

### 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.