Is $R[i,j,k]$ a division ring whenever $R$ is a field?

Question

Let $R$ be a field , let us adjoin $i,j,k$ to $R$ and write $$R[i,j,k]:=\{a+bi+cj+dk :a,b,c,d \in R \}$$ where $i,j,k$ satisfies $a+bi+cj+dk=0 $ iff $a=b=c=d=0$, $i,j,k$ commutes with every element of $R$, $i^2=j^2=k^2=-1$ and $ij=k , jk=i , ki=j$.

Then is it true that $R[i,j,k]$ is a division ring whenever $R$ is a field?

Answer 1

Suppose $-1 = x^2$ in $R$, then $(x+i)(x-i)=x^2-i^2 = 0$, so $R[i,j,k]$ is not zero-divisor free. In general, we have $$R[i,j,k] \text{ is a division ring }\Leftrightarrow \forall a,b,c,d \in R: a^2+b^2+c^2+d^2= 0 \Rightarrow a=b=c=d=0.$$ For the one direction, suppose there are some $a,b,c,d$ not all zero, such that $a^2+b^2+c^2+d^2=0$ Note that in $R[i,j,k]$ we can factor the LHS as $(a+bi+cj+dk)(a-bi-cj-dk)$, so $a+bi+ci+dj$ is a zero divisor.
If we have $\forall a,b,c,d \in R: a^2+b^2+c^2+d^2= 0 \Rightarrow a=b=c=d=0$, then we can compute the inverse as $(a+bi+cj+dk)^{-1}= \frac{a-bi-cj-dk}{a^2+b^2+c^2+d^2}$.

Answer 2

Not true, You can see if $R=\mathbb{Z}_3$, which is a field, but $R[i,j,k]$ is not division ring.