Minimal polynomial (field theory)

For the minimal polynomial of a matrix, see Minimal polynomial (linear algebra).

In field theory, given a field extension E / F and an element α of E that is an algebraic element over F, the minimal polynomial of α is the monic polynomial p, with coefficients in F, of least degree such that p(α) = 0. The minimal polynomial is irreducible over F, and any other non-zero polynomial f with f(α) = 0 is a (polynomial) multiple of p.

Proof: Let E / F be a field extension over F as above, $\alpha \in$ E, and $f \in$ F[x] a minimal polynomial. Suppose f = g * h where $g,h \in$ F[x]\F. Hence f(α) = 0. Since fields are also integral domains, we have that g(α) = 0 or h(α) = 0. As both the degrees of both g and h are smaller than the degree of f, we get a contradiction as f does not have a minimal degree. We conclude that minimal polynomials are irreducible.

For example, for $F = \mathbb{Q}, E = \mathbb{R}, \alpha = \sqrt 2$ the minimal polynomial for α is p(x) = x² − 2.

If

$\alpha = \sqrt 2 + \sqrt 3$

then

$p(x) = x^4 - 10 x^2 + 1 = (x - \sqrt 2 - \sqrt 3)(x + \sqrt 2 - \sqrt 3)(x - \sqrt 2 + \sqrt 3)(x + \sqrt 2 + \sqrt 3)$

is the minimal polynomial.

The base field F is important as it determines the possibilities for the coefficients of p(x). For instance if we take $F = \mathbb{R}$, then

$p(x) = x - \sqrt 2$

is the minimal polynomial for

$\alpha = \sqrt 2$.

References

• Weisstein, Eric W., "Algebraic Number Minimal Polynomial" from MathWorld.
• Minimal polynomial at PlanetMath.