Minimal polynomial (field theory)

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.

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.

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.

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.

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.

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.

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.

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 = (x − √2 − √3)(x + √2 − √3)(x − √2 + √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 ${\displaystyle {\text{Gal}}(\mathbb {Q} ({\sqrt {2}},{\sqrt {3}})/\mathbb {Q} )/{\text{Gal}}(\mathbb {Q} ({\sqrt {3}})/\mathbb {Q} )}$ .

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.

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.

• Ring of integers
• Algebraic number field

• Weisstein, Eric W. "Algebraic Number Minimal Polynomial". MathWorld.
• Minimal polynomial at PlanetMath.
• Pinter, Charles C. A Book of Abstract Algebra. Dover Books on Mathematics Series. Dover Publications, 2010, p. 270–273. ISBN 978-0-486-47417-5