Quaternions, Taking the Square Root

Taking the Square Root

Let u be a quaternion over a ring R. Assume R is an integral domain, with characteristic other than 2. If v = u2, and we are given v, can we find u?

Apply the formula and see where it leads.

u = a + bi + cj + dk

v = a2-b2-c2-d2 + 2a(bi+cj+dk)

  1. The square root of 0 is 0. If R is the reals, or embeds in the reals, 0 has no other square root. If R is Zp, 0 may have other square roots; this is handled in case 3.

  2. The imaginary components of v are all 0, and a is nonzero. This means b c and d are 0, and v is a square in R. When R is the reals, v is a positive real and u is ± its square root.

  3. The imaginary components of v are 0, because a = 0. Thus v = -(b2+c2+d2) in R. There is a square root for every point on the sphere b2+c2+d2 = -v. In the reals, v has to be 0, which was handled in case 1, or a negative number, giving a sphere of solutions. Other rings may lead to other solutions. For instance, 2i+j is the square root of 0 over Z5.

  4. At least one of the three imaginary components of v is nonzero. This means a is nonzero. When searching for u, replace b with b′/a, c with c′/a, and d with d′/a. We can always multiply by a back again, to recover b c and d. Now u2 looks like this.

    v = a2 - (b′2+c′2+d′2)/a2 + 2(b′i+c′j+d′k)

    Well you can tell straight away what b′ c′ and d′ are. Substitute for these and set the first expression to the real component of v. This gives a quadratic in a2. Solve the quadratic, then take the square root of that to find a. Of course a belongs to R, so the quadratic has to produce a square in R. In the reals, the quadratic has to yeild a positive solution.

    Let's take the square root of 3+4i over the reals. With b′ = 2, solve a4-3a2-4 = 0. This gives a2 = 4 or -1, the latter being impossible. Thus a is ±2, and the square root is plus or minus 2+i.

    Assume the real component of v is j. Now -j becomes the middle coefficient on the quadratic formula, and the discriminant is larger than j2, whence its square root is larger than j. Exactly one of the two solutions is positive, and v has exactly two square roots.

  5. If R has characteristic 2, the imaginary components of v must be 0, and u is any point on the quadratic surface a2-b2-c2-d2 = v.