# Minimal polynomial (field theory)

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 J$_α$ 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 J$_α$
More specifically, J$_α$ 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, J$_α$ 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 J$_α$ since 0α$^i$ = 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 J$_α$, 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 J$_α$. 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 J$_α$, 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 J$_α$ 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 J$_α$ of minimal degree n > 0. We have that r := f−g ∈ J$_α$ (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 / c$_m$ (writing c$_m$ ∈ F for the non-zero coefficient of highest degree in r) is a monic polynomial of degree m < n such that r / c$_m$ ∈ J$_α$ (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 J$_α$

The minimal polynomial f of α generates the ideal J$_α$, i.e. every g in J$_α$ 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], J$_α$ 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 J$_α$, 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 $L/K$ the minimal polynomial of any $\alpha \in L$ not in $K$ can be computed as
   $f(x)=\prod _{\sigma \in {\text{Gal}}(L/K)}(x-\sigma (\alpha ))$
if $\alpha$ has no stabilizers in the Galois action. Since it is irreducible, which can be deduced by looking at the roots of $f'$, it is the minimal polynomial. Note that the same kind of formula can be found by replacing $G={\text{Gal}}(L/K)$ with $G/N$ where $N={\text{Stab}}(\alpha )$ is the stabilizer group of $\alpha$. For example, if $\alpha \in K$ then its stabilizer is $G$, hence $(x-\alpha )$ 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) = x$^2$ − 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 $d$, computing the minimal polynomial of an element $a+b{\sqrt {d\,}}$ can be found using Galois theory. Then
   ${\begin{aligned}f(x)&=(x-(a+b{\sqrt {d\,}}))(x-(a-b{\sqrt {d\,}}))\\&=x^{2}-2ax+(a^{2}-b^{2}d)\end{aligned}}$
in particular, this implies $2a\in \mathbb {Z}$ and $a^{2}-b^{2}d\in \mathbb {Z}$. This can be used to determine ${\mathcal {O}}_{\mathbb {Q} ({\sqrt {d\,}}\!\!\!\;\;)}$ 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) = x$^4$ − 10x$^2$ + 1 = (x − √2 − √3)(x + √2 − √3)(x − √2 + √3)(x + √2 + √3).
Notice if $\alpha ={\sqrt {2}}$ then the Galois action on ${\sqrt {3}}$ stabilizes $\alpha$. Hence the minimal polynomial can be found using the quotient group ${\text{Gal}}(\mathbb {Q} ({\sqrt {2}},{\sqrt {3}})/\mathbb {Q} )/{\text{Gal}}(\mathbb {Q} ({\sqrt {3}})/\mathbb {Q} )$.

### 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(2π/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.