Minimal Polynomial (Field Theory)
Minimal polynomials are particularly useful because they help calculate the degree of algebraic extensions and the fact that they divide any polynomial with that element as a root.
Any minimal polynomial must be irreducible.
See computing minimal polynomials in number fields for specific details on how one goes about calculating the minimal polynomial.
The minimal polynomial of \(\sqrt{d}\) for \(d\) a squarefree integer is \(X^2 - d\).
The minimal polynomial of a primitive \(n^{\text{th}}\) root of unit is the cyclotomic polynomial \(\Phi_n\).
Above we call this the minimal polynomial of \(\alpha\), so it is useful to establish uniqueness and existence results. These are very straightforward proofs.
Let \(\alpha\) be algebraic over a field \(\mathbb{F}\). Then there is a monic irreducible polynomial for which \(\alpha\) is a root.
Proof
By definition of being algebraic, \(\alpha\) is the root of some polynomial \(f(X) = a_n X^n + \dots + a_1X + a_0 \in \mathbb{F}[X]\) where \(a_n \neq 0\). Since \(\mathbb{F}\) is a field, we can multiply by \(a_n^{-1}\) to get
and clearly \(f(\alpha) = 0 \implies g(\alpha) = 0\).
Let \(\alpha \in \mathbb{K}\) where \(\mathbb{K}\) is an algebraic extension of \(\mathbb{F}\). Then the minimal polynomial of \(\alpha\) is unique.
Proof
Suppose that \(f, g \in \mathbb{F}[X]\) are both minimal polynomials for \(\alpha\). That is, they are irreducible monic polynomials of minimal degree for which \(\alpha\) is a root. Then \((f - g)(\alpha) = f(\alpha) - g(\alpha) = 0 - 0 = 0\). However then either \(f - g = 0\), i.e. \(f = g\) proving uniqueness, or \(f - g\) is a polynomial of degree less than that of \(f\) and \(g\), since both polynomials have matching leading coefficients. This contradicts the minimality of the degree of \(f\) and \(g\), as \(f - g\) then can be used to construct an equivalent degree monic polynomial as per the existence result above.