Some confusion about Gaussian ring

Question

Is $\mathbb{Z} [i] $ is field ? yes/No

yes, I thinks it will field because it is integral domain

Is its True ?

Answer 1

No, it is not a field. For instance, $2$ has no inverse in $\mathbb Z[i]$.

Answer 2

Well, ${\Bbb Z}[i] = \{a+bi\mid a,b\in{\Bbb Z}\}$ is a subset of ${\Bbb C}=\{a+bi\mid a,b\in{\Bbb R}\}$ and so an integral domain. But its clearly not a field.

Answer 3

Note that $\mathbb{Z} \subseteq \mathbb{Z}[i]$ and $\mathbb{Z}$ is not a field. To see this, we know that $1$ is the identity in $\mathbb{Z}$, so how would we invert $2$? We would need to multiply it by $\frac{1}{2}$, which is not an element of $\mathbb{Z}$. In fact, the only invertible elements in $\mathbb{Z}$ are $\pm 1$, hence the integers do not form a field.

Hence, $\mathbb{Z}[i]$ is not a field.

Answer 4

We know that $$\mathbb Z[i]\cong\mathbb Z[x]/(x^2+1).$$

And $$(x^2+1)\subsetneq (3,x^2+1)\subsetneq\mathbb Z[x]\quad \text{(why?)}$$ implies $(x^2+1)$ is not a maximal ideal of $\mathbb Z[x]$, hence $\mathbb Z[i]$ is not a field.