Taking Square Roots in quadratic extension Fields

Given a positive number R in a quadratic extension field F of a field B, find its square root or proove that R has no square root. We assume that taking square roots in the field B is given.

Algorithm

Let B be a field, let r ∈ B a non-square element of B (i.e. ∄ x ∈ B: x² = r)

Let F = B[√r] a the quadratic extension of B containing the square root of r.

Let R ∈ F with R = a + b√r, a ∈ B, b ∈ B.

Let m = a² - b²r. Note that m ∈ B.

If m has no square root in B, then R has no square root in F.

Otherwise let n = √m

If (n+a)/2 has a square root u in B then X = u + b/(2u)√r is the square root of R.

If (n+a)/(2r) has a square root v in B then X = b/(2v) + v√r is the square root of R.

Otherwise R has no square root in B.

Correctness

Claim 1

If m has no square root in B, then R has no square root in F.

Proof 1

We proove the converse: if R has a square root X in F then m has a square root in B. Assume X = u + v√r and R = a + b√r = X².

(u + v√r)² = u² + v²r + 2uv√r

therefore

a = u² + v²r and

b = 2uv

m = a² - b²r = (u² + v²r)² - (2uv)² = (u⁴ + v⁴r² - 2u²v²r) = (u² - v²r)². ∎

Claim 2

If neither (n+a)/2 nor (n+a)/(2r) has a square root in B then R has no square root in F.

Proof 2

We proove the converse: if R has a square root in F (i.e. X as above) then at least one of (n+a)/2 or (n+a)/(2r) has a square root in B.

From the proof 1 we can see that n = |u² - v²r|.

Case 1: u² > v²r

In this case n = u² - v²r and (n+a)/2 = u² and (n+a)/(2r) has no square root in B.

We see further that due to the condition u² > v²r, taking the (positive) square root u of (n+a)/2 implies X > 0, no matter the sign of v = b/2u.

Case 2: u² < v²r

In this case n = v²r - u² and (n+a)/(2r) = v² and (n+a)/2 has no square root in B.

We see further that due to the condition u² < v²r, taking the (positive) square root v of (n+a)/(2r) implies X > 0, no matter the sign of u = b/2v.

Thereby it is proven that at least on of (n+a)/2 or (n+a)/(2r) has a square root in B. Further it is shown, that the square root X is found by the algorithm. ∎

Example

Take the square root of 3 - 2√2 in ℚ[√2].

Identify B = ℚ, r = 2, a = 3, b = -2.

m = 9 - 4⋅2 = 1 has a square root n=1 in ℚ.

(n+a)/2 = 2 does not have a square root in ℚ.

(n+a)/(2r) = 1 does have a square root v = 1 in ℚ.

We therefore set X = b/(2v) + v√r = -1 + 1√2 = √2 - 1 and verify that indeed (√2 - 1) > 0 and (√2 - 1)² = 3 - 2√2.

References

1. Anders Kaseorg: constructible, an implementation of constructible numbers in Haskell, including this algorithm.