In field theory, given a field extension E / F and an element α of E which 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, 
E, and 
F[x] a minimal polynomial. Suppose f = g * h where 
F[x]\F. Hence f(α) = 0. Since a polynomial ring over a field is an integral domain (Proof here), we have that 
. 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 
the minimal polynomial for α is p(x) = x2 − 2. If 
then
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 
, then 
is the minimal polynomial for 
.
References
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