Minimal polynomial (field theory)

In field theory, a branch of mathematics, the minimal polynomial of an element $&alpha;$ of an extension field of a field is, roughly speaking, the polynomial of lowest degree having coefficients in the smaller field, such that $&alpha;$ is a root of the polynomial. If the minimal polynomial of $&alpha;$ 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 $&alpha;$ of $E$, let $J_{&alpha;}$ be the set of all polynomials $f(x)$ in $F[x]$ such that $f(&alpha;) = 0$. The element $&alpha;$ is called a root or zero of each polynomial in $J_{&alpha;}$

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 $J_{&alpha;}$ since $0&alpha;^{i} = 0$ for all $&alpha;$ and $i$. This makes the zero polynomial useless for classifying different values of $&alpha;$ into types, so it is excepted. If there are any non-zero polynomials in $J_{&alpha;}$, i.e. if the latter is not the zero ideal, then $&alpha;$ is called an algebraic element over $F$, and there exists a monic polynomial of least degree in $J_{&alpha;}$. This is the minimal polynomial of $&alpha;$ with respect to $E/F$. It is unique and irreducible over $F$. If the zero polynomial is the only member of $J_{&alpha;}$, then $&alpha;$ 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 $&alpha;$ is algebraic with minimal polynomial $f(x)$, the smallest field that contains both $F$ and $&alpha;$ 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 &alpha; is defined as the monic polynomial of least degree among all polynomials in F[x] having &alpha; 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.

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.

Q($\sqrt{2}$)
If F = Q, E = R, α = $\sqrt{2}$, then the minimal polynomial for α is a(x) = x2 &minus; 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 = $\sqrt{2}$ is a(x) = x &minus; $\sqrt{2}$.

Q($\sqrt{d}$&hairsp;)
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 $$\begin{align} f(x) &= (x - (a + b\sqrt{d\,}))(x - (a - b\sqrt{d\,})) \\ &= x^2 - 2ax + (a^2 - b^2d) \end{align}$$ in particular, this implies $$2a \in \mathbb{Z}$$ and $$a^2 - b^2d \in \mathbb{Z}$$. 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 = $\sqrt{2}$ + $\sqrt{3}$, then the minimal polynomial in Q[x] is a(x) = x4 &minus; 10x2 + 1 = (x &minus; $\sqrt{2}$ &minus;  $\sqrt{3}$)(x + $\sqrt{2}$ &minus; $\sqrt{3}$)(x &minus; $\sqrt{2}$ + $\sqrt{3}$)(x + $\sqrt{2}$ + $\sqrt{3}$).

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(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.