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Dirichlet's Unit Theorem and a Fermat-Style Application

The units of O_K are not just roots of unity — Dirichlet's theorem gives their full structure: a finite torsion part times a free abelian group of explicit rank r₁+r₂-1. We then assemble class group, units, and splitting into a clean attack on x²+5=y³.

The structure of the unit group

An element u in O_K is a unit iff its norm N(u) = ±1. In Z the only units are ±1, but in larger fields there can be infinitely many — e.g. in Z[√2] the element 1+√2 has norm -1, so all its powers (1+√2)^k are distinct units. [[dirichlet-unit-theorem|Dirichlet's unit theorem]] describes the unit group O_K* completely as a finitely generated abelian group.

Let r₁ be the number of real embeddings K → R and r₂ the number of conjugate pairs of complex embeddings, so r₁ + 2r₂ = n. Then O_K* ≅ μ_K × Z^r where μ_K is the finite cyclic group of roots of unity in K and the rank is r = r₁ + r₂ − 1. The proof embeds units via logarithms of their archimedean absolute values into a hyperplane in R^(r₁+r₂); the image is a full-rank lattice, again by the geometry of numbers.

Unit rank  r = r1 + r2 - 1.   Examples:

 Q             n=1, r1=1, r2=0  ->  r = 0.  Units = {±1}.
 Q(i)          n=2, r1=0, r2=1  ->  r = 0.  Units = {±1,±i} (finite!)
 Q(sqrt(-5))   n=2, r1=0, r2=1  ->  r = 0.  Units = {±1}.
 Q(sqrt(2))    n=2, r1=2, r2=0  ->  r = 1.  Units = ±(1+sqrt(2))^Z.
 Q(sqrt(5))    n=2, r1=2, r2=0  ->  r = 1.  fund. unit = (1+sqrt(5))/2.
 Q(cbrt(2))    n=3, r1=1, r2=1  ->  r = 1.  one fundamental unit.

Imaginary quadratic (r2=1, r1=0): rank 0, units are JUST roots of
unity -- finite.  Real quadratic (r1=2): rank 1, one fundamental
unit u, every unit is ±u^k  (this is the Pell equation engine).
Imaginary quadratic fields have finite unit group (rank 0); real quadratic fields have rank 1 — a single fundamental unit drives Pell's equation.

Putting it together: x² + 5 = y³ has no integer solutions

Here is the payoff — a genuine Fermat-style Diophantine equation cracked by working in O_K = Z[√-5]. We use everything: factor in O_K, control ideals via the class group (h_K = 2), use the finite unit group (just ±1), and read off splitting. Watch how the class number 2 does the decisive work.

Solve  x^2 + 5 = y^3  in integers.   Work in O_K = Z[sqrt(-5)].

Factor LHS:   (x + sqrt(-5))(x - sqrt(-5)) = y^3.   (*)

Step 1 (coprimality of ideals).  Let p | gcd of the two factors.
Then p | their difference 2 sqrt(-5) and p | their sum 2x.
If y is even then x^2 ≡ -5 ≡ 3 (mod 4): impossible (squares are 0,1).
So y is odd, x is even -> x^2+5 odd, and one checks the ideals
  (x+sqrt(-5)) and (x-sqrt(-5)) are COPRIME in O_K.

Step 2 (use unique factorization of IDEALS).  Their product is
the cube (y)^3, and they are coprime, so each is itself a cube:
  (x + sqrt(-5)) = a^3   for some ideal a.

Step 3 (use the class group, h_K = 2).  In Cl(K) ≅ Z/2Z,
  [a]^3 = [(x+sqrt(-5))] = [principal] = 1,  so  [a]^3 = 1.
But |Cl(K)| = 2, so [a]^2 = 1 too; gcd(3,2)=1 forces [a] = 1.
  ==>  a IS principal:  a = (a + b sqrt(-5)).

Step 4 (use the unit group, O_K* = {±1}).  Then
  x + sqrt(-5) = (unit)(a + b sqrt(-5))^3,  unit = ±1 = (∓1)^3,
so WLOG  x + sqrt(-5) = (a + b sqrt(-5))^3.
Expand:  (a + b sqrt(-5))^3 = a(a^2 - 15 b^2) + b(3a^2 - 5 b^2) sqrt(-5).
Match sqrt(-5)-coefficients:  b(3a^2 - 5b^2) = 1.
  ==> b = ±1 and 3a^2 - 5b^2 = ±1  ==>  3a^2 = 5 ± 1 = 6 or 4.
Neither gives integer a (a^2 = 2 or 4/3).  CONTRADICTION.

Conclusion:  x^2 + 5 = y^3  has NO integer solutions.
Every tool in the track appears: ideal factorization, the class number 2 (via gcd(3,2)=1), and the trivial unit group — combining to kill the equation.