Why this Tensor product of fields is a field

Question

Is it true that the ring $\mathbb{Q}[i]\otimes_\mathbb{Q}\mathbb{Q}[i]$ is a field ?

Answer 1

We have that $\mathbb{Q}(i)\cong\mathbb{Q}[x]/(x^2+1)$, therefore $$\begin{align*} \mathbb{Q}(i)\otimes_\mathbb{Q}\mathbb{Q}(i)&\cong \mathbb{Q}(i)\otimes_\mathbb{Q}\mathbb{Q}[x]/(x^2+1)\\ &\cong\mathbb{Q}(i)[x]/(x^2+1)\\ &=\mathbb{Q}(i)[x]/(x-i)(x+i)\\ &\cong\mathbb{Q}(i)[x]/(x-i)\times\mathbb{Q}(i)[x]/(x+i)\\ &\cong \mathbb{Q}(i)\times \mathbb{Q}(i) \end{align*}$$ which is not a field because it has zero-divisors, e.g., $(1,0)\cdot(0,1)=(0,0)$.