Quaternion Polynomial Equation

[alexfloo]

I'm working on the following:

"Prove that x^2 - 1=0" has infinitely many solutions in the division ring Q of quaternions."

The Quaternions are presented in my book in the representation as two-by-two square matrices over ℂ. The book gives that for a quaternion


(sorry for the terrible notation, I wasn't able to figure out how to do a matrix in here and I didn't want to do the tex by hand)
\stackrel{a+bi\ \ c+di}{-c+di\ \ a-bi}

has inverse

\frac{1}{a^2+b^2+c^2+d^2}\stackrel{a-bi\ \ -c-di}{c-di\ \ a+bi}.

Now if x^2-1=0, then clearly x=x^{-1}. This means that the individual components must be equal, so a+bi is a real multiple of its conjugate, which requires that it is real, and that multiple is one, or that it is imaginary, and that multiple is -1. Since the multiple is identically positive (sum of squares) b=d=0, and since the multiple must equal 1, a^2+c^2=1. This gives us

\stackrel{a\ \ c}{-c\ \ a}=\stackrel{a\ \ -c}{c\ \ a},

so c=-c=0. Therefore, a=1, which seems to imply that 1 (the quaternion unity) is the unique solution.

What am I missing?

 

[Ben Niehoff]

Are you sure there isn't a typo in the question? I would expect x^2 + 1 = 0 to have infinitely many solutions. x^2 - 1 = 0 should have two solutions.

 

[Deveno]

if q* is the quaternial conjugate, then qq* is real. note that q^-1 = q*/|q|, so from q = q^-1, we get q = q*/|q|, implying that q is real.

but x² -1 = 0 has just two real solutions, 1 and -1.

if it IS a typo, consider (ai + bj + ck)², where a²+b²+c² = 1 (which has infinitely many solutions (a,b,c)).

 

[alexfloo]

In fact, it wasn't a typo at all. The problem just before it references the polynomial x^2-1 over an entirely different field, and I misread. Apologies, and thanks for your help!