Treat a larger field as a vector space

Try

Take numbers of the form a+b√2 with a,b rational. Adding and multiplying two such numbers stays in the same form because (√2)²=2. Predict whether every nonzero expression also has an inverse of that form. Multiply by a−b√2 and inspect the rational denominator.

This set is a field containing Q, denoted Q(√2). It is also a vector space over Q, with basis one and √2. Field multiplication and vector-space coordinates describe complementary parts of its structure.

Predict

Could a²−2b² vanish for rational a,b not both zero? If b is nonzero, this would make √2 rational; if b is zero, it forces a=0. Predict why this denominator argument proves field closure and why irrationality also proves independence of the proposed basis.

Experiment

Compute in the two-coordinate basis.

Operation	Expanded expression	Reduced coordinates
(1+√2)+(3−2√2)	4−√2	(4,−1)
(1+√2)(3−2√2)	3+√2−4	(−1,1)
(1+√2)⁻¹	(1−√2)/(1−2)	(−1,1)

The last two rows happen to have the same result, but for different reasons. Verify the inverse by multiplying it by 1+√2 and obtaining one. Coordinates alone do not define multiplication unless the relation between basis elements is also specified.

Understand

An extension E/F means F is a subfield of E, with the same identity. Its degree [E:F] is the dimension of E as an F-vector space, possibly infinite. The notation F(α) means the smallest subfield containing F and α; F[α] means all polynomial expressions in α with coefficients in F. These can differ.

An element α is algebraic over F if some nonzero polynomial in F[x] vanishes at α. Otherwise it is transcendental over F. A finite-dimensional extension is algebraic: if its degree is n, the n+1 vectors 1,α,…,αⁿ are linearly dependent, giving a nonzero annihilating polynomial. The converse requires care: an algebraic extension can have infinite degree when it contains infinitely many independent adjunctions.

For √2, every polynomial expression reduces to a+b√2, and the inverse calculation shows this set is already a field. Independence follows because a+b√2=0 with b nonzero would make √2 rational. Thus [Q(√2):Q]=2. The same reasoning applies to a nonsquare rational d with an appropriate square root.

If t is transcendental over F, evaluation F[x]→F[t] has zero kernel, so F[t] is a polynomial ring. It is not a field: a polynomial inverse of t would give a nonzero polynomial relation t p(t)−1=0. Its fraction field F(t) adds rational expressions and is strictly larger.

Algebraicity depends on the base field. Every element a of a field is algebraic over that field, since x−a vanishes there. Thus an element transcendental over Q becomes algebraic over Q(a). Likewise algebraic does not mean rational or already in the base field; √2 is the basic counterexample.

Apply

Prove that expressions a+b√5 form a field and compute its degree over Q. State separately the spanning argument, independence argument, and inverse formula. These are distinct proof obligations even though they use the same quadratic relation.

Then contrast a finite-dimensional extension with an arbitrary algebraic extension. The power-dependence proof starts from a finite basis; it does not prove every algebraic extension has one. Keep the quantifiers on elements and on the whole extension separate.

Continue

Minimal polynomials next give the canonical relation for one algebraic element. Their degree determines the basis length and explains why every simple algebraic ring F[α] is already a field.