Multiply dimensions with a proved basis

Try

Adjoin √2 to Q, then adjoin √3. The first step has degree two. Is the second also degree two, or did the first extension already contain √3? The answer requires a membership argument, not multiplying two familiar quadratic degrees automatically.

Assume √3=a+b√2 with rational a,b. Squaring gives 3=a²+2b²+2ab√2. Irrationality forces ab=0. Each resulting case would make either three or three halves a rational square, which prime-exponent parity rules out.

Predict

The candidate basis for Q(√2,√3) is 1,√2,√3,√6. Predict how it arises by multiplying a basis for the first step by a basis for the second. Then identify the two different independence statements needed to prove the product basis is independent over Q.

Experiment

Write a general element first using coefficients in the intermediate field, then expand those coefficients over Q.

Stage	Form
Over Q(√2)	u+v√3
Expand u,v	(a+b√2)+(c+d√2)√3
Over Q	a+b√2+c√3+d√6

If the final expression is zero, independence of 1,√3 over Q(√2) makes u=v=0. Independence of 1,√2 over Q then makes a=b=c=d=0. This proves more than a convenient notation: the coordinates are unique.

Understand

For a finite tower F⊆E⊆K, choose an F-basis u₁,…,u_m of E and an E-basis v₁,…,v_n of K. The mn products uᵢvⱼ span K over F by successive expansion. If an F-linear combination of these products vanishes, group by vⱼ. Their E-independence makes each E-coefficient zero; the uᵢ independence then kills every F-coefficient. Thus [K:F]=[K:E][E:F], the tower law.

It follows that an intermediate degree divides the total finite degree. A degree-three field over Q cannot be contained in a degree-eight extension, because eight would equal three times a positive integer. This is an obstruction to containment, not a proof that every divisor occurs as an intermediate degree.

For the biquadratic example, √3 is not in Q(√2), while it satisfies a quadratic over that field. Therefore the second degree is exactly two, giving total degree four. To justify that three halves is not a rational square, write b=u/v in lowest terms and compare prime exponents in 2u²=3v²: the exponent of three has opposite parity on the two sides. The same parity argument handles a²=3.

Degrees of generators over the original base need not multiply. If α=√2 and β=−√2, both have degree two over Q, but adjoining β after α adds nothing. The second tower degree is one. More generally, the correct factor is [F(α,β):F(α)], not automatically [F(β):F].

An element in a finite extension also has degree dividing the total degree: apply the tower to F⊆F(α)⊆E. This fact will translate geometric construction steps into arithmetic obstructions. It does not say every algebraic element of a permitted degree lies in the extension.

Apply

Given degrees three and four along an actual tower, construct the twelve product-basis symbols and prove why they are independent. Then contrast this with two degree-two generators that produce only one quadratic field.

In your proof, label the field over which each coefficient is allowed to vary. Saying that vectors are “independent” without a base field can conceal the exact step where a tower argument fails.

Continue

Straightedge-and-compass constructions next create towers of real quadratic extensions. The tower law provides rigorous impossibility results while also showing why a power-of-two degree test is not a complete construction method.