Is Julia lazy?

New to Julia
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You can use AbsteactAlgebra.jl to deal with elementary abstract algebra:

using AbstractAlgebra

@show GF7 = GF(7)
x, y = GF7(10), GF7(-2)
@show(x, y, zero(x), one(x), -x, x + y, x - y, x * y, x^3, x^-5, x == GF7(3), x == 3)
println()

@show C = GF7[1 2; 3 4]
@show P, x = PolynomialRing(GF7, "x")
@show(det(C), inv(C), lu(C), charpoly(P, C))
println()

squares(Fp) = [x^2 for x in Fp]
squareroots(k, Fp) = [x for x in Fp if x^2 == k]
@show(collect(GF7), squares(GF7), squareroots.(0:6, Ref(GF7)));

Output:

GF7 = GF(7) = Finite field F_7
x = 3
y = 5
zero(x) = 0
one(x) = 1
-x = 4
x + y = 1
x - y = 5
x * y = 1
x ^ 3 = 6
x ^ -5 = 3
x == GF7(3) = true
x == 3 = true

C = GF7[1 2; 3 4] = [1 2; 3 4]
(P, x) = PolynomialRing(GF7, "x") = (Univariate Polynomial Ring in x over Finite field F_7, x)
det(C) = 5
inv(C) = [5 1; 5 3]
lu(C) = (2, (), [1 0; 3 1], [1 2; 0 5])
charpoly(P, C) = x^2 + 2*x + 5

collect(GF7) = AbstractAlgebra.GFElem{Int64}[0, 1, 2, 3, 4, 5, 6]
squares(GF7) = AbstractAlgebra.GFElem{Int64}[0, 1, 4, 2, 2, 4, 1]
squareroots.(0:6, Ref(GF7)) = Vector{AbstractAlgebra.GFElem{Int64}}[[0], [1, 6], [3, 4], [], [2, 5], [], []]
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