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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.

Math.StackExchange
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3 Answers3

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"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.

JoshuaZ
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    I believe a counterexample in this particular case is $(x^2+1) \subsetneq (x^2+3,2) \subsetneq \mathbb Z[x].$ – ktoi Aug 29 '18 at 02:58
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    More generally, the ideal $(x^2+1,n)$ is an ideal strictly containing your ideal for any integer n. "Most" of the time, this isn't even a maximal ideal., (unless n is a prime congruent to one mod 4). This ties in totally with the fact that Z[X] is a two dimensional ring. – meh Aug 29 '18 at 15:27
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$(x^2+1)$ is a prime ideal but not maximal.

it happens in a ring of Krull-dimension $\geq 2$. $\dim \mathbb{Z}[X] = 2$.

QU Binggang
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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.

user5826
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  • The maximal ideals of $\Bbb{Z}[x]$ are actually of the form $(p,f),$ where $f\in\Bbb{Z}[x]$ is irreducible modulo $p.$ – Stahl Sep 21 '20 at 07:45