2
$\begingroup$

Let $R={\alpha_0+\alpha_1i+\alpha_2j+\alpha_3k: \alpha_0, \alpha_1, \alpha_2, \alpha_3\in\mathbb{Z_3}}$ be the ring of quaternions over $\mathbb{Z_3}$. Then,

  1. $R$ is a field.
  2. $R$ is a division ring.
  3. $R$ has zero divisors.
  4. None of the above.

I don't know how to proceed basically due to the fact that I am not getting clear picture about the elements. Can anyone help me? some hints or help would be great. Thanks.

  • 0
    Have you tried obtaining inverses in the same way as you would when the coefficients are real? – 2017-01-08
  • 1
    What is $\mathbb Z_3$? Do you mean 3-adic numbers or integers mod 3? – 2017-01-09

2 Answers 2

3

First, quaternions cannot be a field, since the multiplication is not commutative.

Now, use the fact that for quaternions over any commutative ring

$$qq^*=(a+bi+cj+dk)(a-bi-cj-dk)=a^2+b^2+c^2+d^2$$

Real quaternions are a division ring due to the following fact:

$$a^2+b^2+c^2+d^2=0 \Leftrightarrow a=0, b=0, c=0, d=0$$

This becomes false once we move from $\mathbb{R}$ to $\mathbb{Z}_3$. In the latter, we can find $a,b,c,d$ (with at least one not zero) such that $a^2+b^2+c^2+d^2=0$. For example, take $a=b=c=1, d=0$. Now

$$(1+i+j)(1-i-j)=1+1+1=0$$

Thus, it contains zero divisors.

  • 0
    so it is not a division ring right? – 2017-01-08
  • 0
    @KushalBhuyan Yes, since division rings have no zero divisors. – 2017-01-09
4

Wedderburn's theorem says that finite division rings are commutative. Is $R$ commutative? Can it have no zero divisors?