Convert algebraic relations into a quotient field

Try

Let α be a real root of x³−2. Every higher power can be reduced using α³=2: α⁴=2α, α⁵=2α², and α⁶=4. Predict a spanning set for all polynomial expressions in α. Then ask whether a shorter rational relation might make that spanning set dependent.

Eisenstein at two proves x³−2 irreducible over Q. This supplies the missing independence: no nonzero rational polynomial of degree below three can vanish at α.

Predict

Choose a monic annihilating polynomial of least positive degree for an algebraic element. If it factored into two smaller positive-degree polynomials, evaluating at the element would give a zero product in a field. Predict why one smaller factor would also vanish and contradict minimality.

Experiment

Reduce a product in the basis 1,α,α² with α³=2.

Step	Expression
Expand	(1+α)(1−α+α²)=1+α³
Apply relation	1+2=3
Divide by scalar	(1+α)⁻¹=(1−α+α²)/3

Verify the inverse by expanding again. A quotient field calculation is not merely reducing powers; nonzero polynomial expressions must also have inverses. Bézout identities provide a systematic way to find them.

Understand

For an algebraic α over F, choose the monic polynomial mα of least degree with mα(α)=0. It is irreducible by the zero-product argument in the prediction. If any f vanishes at α, divide f=qmα+r with deg r below deg mα. Evaluation gives r(α)=0, forcing r=0 by minimality. Thus mα divides every annihilating polynomial. Two monic least-degree choices divide each other and therefore agree, proving uniqueness.

Evaluation F[x]→F[α] has kernel (mα), so F[α]≅F[x]/(mα). An irreducible polynomial generates a maximal ideal in F[x]: if a polynomial g is not divisible by mα, their gcd is one, and a Bézout identity ug+vmα=1 gives an inverse to g modulo mα. Consequently the quotient is a field and F[α]=F(α).

Division supplies a basis 1,α,…,α^(d−1), where d=deg mα. These elements span because remainders have degree below d. A nontrivial linear dependence would be a lower-degree annihilating polynomial, so they are independent. Hence [F(α):F]=deg mα.

The coefficient field is essential. The minimal polynomial of √2 over Q is x²−2, but over Q(√2) it is x−√2. A polynomial that vanishes need not be minimal: x⁴−4 vanishes at √2 but factors and has unnecessarily high degree. Always prove irreducibility or minimality before reading off an extension degree.

The quotient construction also creates an algebraic root when none is already available. For irreducible m∈F[x], the field F[x]/(m) contains the class α of x, and m(α)=0. The constants embed because a nonzero constant cannot lie in a positive-degree principal ideal. This is a concrete existence argument for adjoining a root.

Apply

For α³=3 with irreducible x³−3, reduce α⁵ and explain why the three basis coordinates are unique. Then find an inverse of a simple expression by either an identity or the polynomial Euclidean algorithm.

Audit a claimed basis by checking both spanning and independence. A reduction relation proves spanning, but only minimality excludes a hidden shorter relation. Keeping these arguments separate prevents a reducible quotient from being mistaken for a field extension of the claimed degree.

Continue

The tower law next combines bases through successive adjunctions. It uses the degree over the current intermediate field, which can be smaller than the degree over the original base.