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!

 

[blue_raver22]

Your approach is correct, but there is a key point missing in your proof. While it is true that x=x^{-1} implies that the individual components must be equal, it does not necessarily mean that the components must be real or imaginary. In fact, in the division ring Q of quaternions, the components can be any combination of real and imaginary numbers.

To prove that x^2-1=0 has infinitely many solutions in Q, we can use the fact that Q is a non-commutative division ring. This means that there exist elements in Q that do not commute, i.e. ab\neq ba for some a,b\in Q. In the case of x^2-1=0, we can choose x to be any quaternion that does not commute with itself, such as i+j or 2i-k. These are all solutions to the equation since (i+j)^2-1=-1-1=2 and (2i-k)^2-1=-3+4i.

Moreover, since Q is a division ring, every nonzero element has an inverse. This means that for every solution x to x^2-1=0, there exists another solution x^{-1} that is its inverse. Therefore, there are infinitely many solutions to this equation in Q.

In conclusion, your proof is correct in showing that the only solution in Q is the quaternion unity, but it does not account for the infinitely many solutions that exist in Q. By considering the non-commutative and division properties of Q, we can see that there are infinitely many solutions to x^2-1=0 in Q.