Minimal Polynomials and Simple Extensions

The polynomial that defines α

Fix α algebraic over K. Among all nonzero polynomials in K[x] that vanish at α, there is a unique monic one of smallest degree. This is the minimal polynomial m_α(x). The structure here is an evaluation homomorphism: the map K[x] → K(α) sending x ↦ α has a kernel, and because K[x] is a principal ideal domain that kernel is a single principal ideal (m_α). Three properties pin it down, and you should be able to recite them.

1. m_α is irreducible over K. (If it factored, one factor would already vanish at α, contradicting minimality.)
2. m_α divides every polynomial in K[x] that vanishes at α. (It generates the whole kernel.)
3. deg m_α = [K(α):K]. The degree of the polynomial equals the degree of the extension.

Finding it in practice

Producing a polynomial that kills α is easy; proving it is minimal means proving irreducibility. Your toolkit from Volume I is exactly what you reach for: the Eisenstein criterion, reduction mod p, the rational root test, and degree-counting. A polynomial of degree 2 or 3 over a field is irreducible iff it has no root in that field — but at degree 4 and up that shortcut fails, since a quartic can factor into two irreducible quadratics with no roots at all.

Find the minimal polynomial of  a = 2^(1/2) + 3^(1/2)  over Q.

Let a = 2^(1/2) + 3^(1/2).
  a^2 = 2 + 2*6^(1/2) + 3 = 5 + 2*6^(1/2)
  a^2 - 5 = 2*6^(1/2)
  (a^2 - 5)^2 = 4 * 6 = 24
  a^4 - 10a^2 + 25 = 24
  a^4 - 10a^2 + 1 = 0

Candidate:  f(x) = x^4 - 10x^2 + 1.

Irreducible over Q?  Rational Root Test: possible roots +-1, neither works -> no linear factors.
No factorization into two rational quadratics x^2+bx+c, x^2-bx+d works out
(matching coefficients forces b,c,d off the rationals).  So f is irreducible.

Therefore m_a(x) = x^4 - 10x^2 + 1  and  [Q(a):Q] = 4.
Note: Q(a) = Q(2^(1/2), 3^(1/2)), the degree-4 field from Guide 1 — a single element a
generates it. (That is the primitive element phenomenon, Guide 4.)

Computing inside K(α)

Once you know m_α has degree n, every element of K(α) is written uniquely as c₀ + c₁α + … + c_{n-1}α^{n-1} with cᵢ ∈ K — the powers 1, α, …, α^{n-1} are a basis. Multiplication is just polynomial multiplication followed by reducing modulo m_α. Inversion looks scary but is mechanical: run the extended Euclidean algorithm on the polynomials.

Invert  (1 + a)  in  K = Q[a]/(a^3 - 2),  where a = 2^(1/3).

We want u(x) with  (1 + x) u(x) = 1  mod (x^3 - 2).
Extended Euclid on  x^3 - 2  and  x + 1:
  x^3 - 2 = (x + 1)(x^2 - x + 1) - 3
  => 3 = (x + 1)(x^2 - x + 1) - (x^3 - 2)
  => 1 = (x + 1) * [ (x^2 - x + 1)/3 ]  -  (x^3 - 2)/3

Reduce mod (x^3 - 2):  (1 + a)^(-1) = (a^2 - a + 1)/3.

Check:  (1 + a)(a^2 - a + 1) = a^3 + 1 = 2 + 1 = 3.  Divide by 3 -> 1.  Correct.