Try
Can one number encode two independent square roots? Put θ=√2+√3. Because (√3−√2)(√3+√2)=1, we have θ⁻¹=√3−√2. Therefore √2=(θ−θ⁻¹)/2 and √3=(θ+θ⁻¹)/2. Both original generators belong to Q(θ), so Q(√2,√3)=Q(θ).
A primitive element for E/F is an element θ with E=F(θ). This means every element of E is a rational expression in θ with coefficients in F. It does not mean θ generates the multiplicative group E×. The two meanings coincide only in the sufficient construction used later for finite fields.
Predict
Try θ_c=√2+c√3 with rational c. Predict whether c=0, c=1 and c=2 work. The value zero loses √3. Are there other rational exceptions? Rather than checking infinitely many coefficients numerically, compare the four possible images under embeddings.
We use the preceding embedding theorem: a finite separable extension E/F has exactly [E:F] embeddings into an algebraic closure fixing F. Review the embedding proof if this count is unfamiliar. Embeddings may land outside E; normality is not assumed in this lesson.
Experiment
The four embeddings of Q(√2,√3) give the following images. Let a=√2 and b=√3, both positive real numbers.
| Image of a | Image of b | Image of θ_c |
|---|
| a | b | a+cb |
| −a | b | −a+cb |
| a | −b | a−cb |
| −a | −b | −a−cb |
Compare pairs. Changing only the sign of a never causes equality. Changing only the sign of b causes equality exactly at c=0. Changing both signs would require c=±√(2/3), which is not rational: 2/3 is not a rational square, as prime exponents in a squared fraction are even. Thus all four images are distinct for every nonzero rational c.
For c=2, θ²=14+4√6, so (θ²−14)²=96. The polynomial X⁴−28X²+100 vanishes at θ. Recover the roots directly: √2=(θ²−10)/(2θ) and √3=(θ²+10)/(4θ). Verify both by substitution. The denominator is nonzero since θ is positive. These identities show the generated field has degree four, proving the displayed monic quartic is minimal; merely finding a vanishing quartic would not prove irreducibility.
Understand
Primitive-element theorem. Every finite separable extension E/F is generated by one element. We treat infinite and finite base fields separately because avoiding finitely many coefficients requires infinitely many available choices.
Infinite base field: prevent embedding collisions
First suppose E=F(α,β), and write n=[E:F]. List its n distinct F-embeddings σ₁,…,σₙ into an algebraic closure Ω. For a pair i≠j, equality of σᵢ(α+cβ) and σⱼ(α+cβ) means
σᵢ(α)−σⱼ(α)=c(σⱼ(β)−σᵢ(β)).
If the β images differ, at most one c in Ω solves this equation; discard it only if it belongs to F. If the β images agree, the α images must differ because α,β generate E and the embeddings are distinct. Then no c solves the equation. There are at most n(n−1)/2 bad coefficients in F, one per pair. An infinite F contains a coefficient outside this finite set.
Choose such a c and put θ=α+cβ. All n images σᵢ(θ) are distinct roots of the minimal polynomial of θ over F. Hence its degree is at least n. Since F(θ)⊆E, the tower law bounds that degree by n. Equality and the tower law force E=F(θ). This argument uses all embeddings into Ω, not just internal automorphisms, so it also works for nonnormal separable extensions.
For general finite E/F, choose finitely many generators, for example a finite F-basis. Combine the first two by the preceding argument, then combine the resulting generator with the third, and continue. Each intermediate field is finite separable over F because its elements lie in the separable extension E. Induction ends with one generator. If E=F, choose θ=0; a one-generator extension needs no combining step.
Finite base field: use a multiplicative generator
If F has q elements and [E:F]=n, then E has qⁿ elements by counting basis coordinates. The multiplicative group E× is cyclic, as proved in the Frobenius and finite-subfields lesson. Choose a generator g. The subfield F(g) contains every nonzero power of g, hence all nonzero elements of E, and also zero. Therefore F(g)=E. This handles finite base fields without assuming a good coefficient remains after excluding a finite set.
Do not infer that every nonzero element works. In F₄ over F₂, the element 1 generates only F₂ as a field. A multiplicative generator is sufficient, not required: field generation and multiplicative order are different questions.
Why separability cannot simply be deleted
Fix a prime p. Let u,v be algebraically independent indeterminates over F_p, put E=F_p(u,v), and put F=F_p(uᵖ,vᵖ). The p² monomials uⁱvʲ with 0≤i,j<p form an F-basis. They span because adjoining u and then v has degree at most p at each step. For independence, clear denominators in a proposed F-linear relation. Every resulting polynomial term has exponents congruent to its own pair (i,j) modulo p. Different pairs cannot cancel in the polynomial ring F_p[u,v], so every coefficient is zero. Thus [E:F]=p².
Every rational function z in u,v satisfies zᵖ∈F, by the characteristic-p power rule. Hence the minimal polynomial of z over F divides Xᵖ−zᵖ, giving [F(z):F]≤p. No single z generates the degree-p² extension E. This is a finite inseparable extension with no primitive element. Some inseparable extensions are simple; this counterexample disproves the universal conclusion without separability, not simplicity in every inseparable case.
Apply
Reconstruct the collision proof without looking at the explanation. Identify where each assumption is used: finiteness gives a finite embedding list and generator list; separability gives n embeddings; infinitude supplies a coefficient outside the forbidden set. Explain why an automorphism-only proof would fail already for the real field Q(∛2), even though that field is simple.
For transfer, let η=√3+√5. Find η⁻¹, recover both roots, and derive its minimal polynomial. Before checking, explain why the field degree is four rather than assuming any pair of square roots is independent.
Check the changed generator
The reciprocal is (√5−√3)/2. Hence √3=(η−2η⁻¹)/2 and √5=(η+2η⁻¹)/2. The field has degree four: if √5=a+b√3 with rational a,b, squaring forces ab=0, and neither a²=5 nor b²=5/3 has a rational solution. Thus η generates that degree-four field. Since η²=8+2√15, its minimal polynomial is X⁴−16X²+4. Its four conjugate values are ±√3±√5. The recovery formulas and degree proof, rather than approximate root values, certify generation and minimality.
Continue
Return to the complete subgroup–field examples to distinguish a field generator from a list of intermediate fields. The theorem gives existence and the finite forbidden-coefficient method; it does not promise that every chosen linear combination works or that large minimal polynomials are cheap to compute. Radical solvability, cyclotomic irreducibility and sustained field investigations remain further work. Use the saved proof tasks to record your hypotheses and repair missing steps.
