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The Gaussian integer $\mathbb{Z}[i]$ is an Euclidean domain that is not a field, since there is no inverse of $2$. So, where is wrong with the following proof?
Fake proof
First, note that $\mathbb{Z}[X]$ is a integral domain. Since $x^2+1$ is an irreducible element in $\mathbb{Z}[X]$, the ideal $(x^2+1)$ is maximal, which implies $\mathbb{Z}[i]\simeq\mathbb{Z}[X]/(x^2+1)$ is a field.
"Since $x^2+1$ is an irreducible element, the ideal $(x^2+1)$ is maximal"
Is this true in a generic integral domain? Consider the ring $Z[x,y].$ We have that $x$ is an irreducible element, but $(x)$ is not a maximal ideal, as it is contained in the ideal $(x,y)$ which is still not the entire ring.
$(x^2+1)$ is a prime ideal but not maximal.
it happens in a ring of Krull-dimension $\geq 2$. $\dim \mathbb{Z}[X] = 2$.
The statement that $(x^2+1)$ is maximal is false.
The maximal ideals of $\mathbb Z[x]$ are of the form $(p, x)$ where $p$ is a prime.
