Ring of quaternions over $\mathbb{Z_3}$ 
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,
  
  
*
  
*$R$ is a field.
  
*$R$ is a division ring.
  
*$R$ has zero divisors.
  
*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.
 A: 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.
A: Wedderburn's theorem says that finite division rings are commutative. Is $R$ commutative? Can it have no zero divisors?
A: Since Wedderburn's Theorem is mentioned, let me also use another big name to give a short generalisation of lisyarus's answer in the case of $\mathbb Z_p$ instead of $\mathbb Z_3$ for some odd prime $p$. This generalization doesn't seem to be on MSE, and this is a good place to post it.
In this case, we can use the very beautiful Lagrange's four-square theorem to claim that $\exists \;a,b,c,d\in \mathbb Z_p$ such that $a^2+b^2+c^2+d^2=p$.
Now, all of $a,b,c,d$ can't be $0$ since that would imply $p=0$. Also, all of $a,b,c,d$ can't be $0\;(\operatorname{mod }p)$ as that would imply $p^2|p$ which is absurd. So, at least one of is not $0$ mod $p$. So, if we define
$$u=a+bi+cj+dk\\v=a-bi-cj-dk$$
then
$$uv=a^2+b^2+c^2+d^2=p=0\;(\operatorname{mod }p)$$
although both $u$ and $v$ are not $0$ mod $p$.
This completes the proof.
