$d \in \mathbb{Z}$ is a square-free integer ($d \ne 1$, and $d$ has no factors of the form $c^2$ except $c = \pm 1$), and let $R=\mathbb{Z}[\sqrt{d}]= \{ a+b\sqrt{d} \mid a,b \in \mathbb{Z} \}$. Prove that every nonzero prime ideal $P \subset R$ is a maximal ideal.

I have a possible outline which I think is good enough to follow.

I think that we need to first prove that every ideal $I \subset R$ is finitely generated.

So if $I$ is non-zero, then $I \cap \mathbb{Z}$ is a non-zero ideal in $\mathbb{Z}$.

Then I need to find $I \cap \mathbb{Z} = \{ xa \mid a \in \mathbb{Z} \}$ for some $x \in \mathbb{Z}$. That way if I let $J$ be the set of all integers $b$ such that $a+b\sqrt{d} \in I$ for some $a\in \mathbb{Z}$, then if there exists a integer $y$ such that $J=\{ yt \mid t\in \mathbb{Z} \}$, then there must exist $s \in \mathbb{Z}$ such that $s+y\sqrt{d} \in I$.

Then all I need to show is that $I = ( x,s+y\sqrt{d} )$.

Now I need to derive that the factor ring $R / P$ is a finite ring without zero divisors, also finite, then since every finite integral domain is a field, every prime ideal $P \subset R$ is a maximal ideal, then I'll be done.

  • $\begingroup$ It's not important for $d$ to be squarefree. The result is true for any $d$ that is not a square (including ${\mathbf Z}[\sqrt{12}]$, for instance). $\endgroup$ – KCd Dec 15 '13 at 3:59

Sounds good. In order to show that $I \cap \Bbb Z$ is a non-zero ideal, it is enough to notice that for $a+b\sqrt{d}\in I$ you have $(a+b\sqrt{d})(a-b\sqrt{d}) = a^2 -db^2 =:n\in \Bbb Z \cap I$. Now by writing $R = \Bbb Z[X]/(X^2-d)$ you can show that $R/P$ is a quotient of the finite ring $(\Bbb Z / n\Bbb Z)[X]/(X^2-d)$ where $X^2-d$ denotes the reduced polynomial in $(\Bbb Z/n \Bbb Z)[X]$. So indeed, $R/P$ is finite without zero divisors.

  • 1
    $\begingroup$ You omitted a crucial point: $n\ne 0$ because $\ldots$ $\endgroup$ – Bill Dubuque Dec 18 '13 at 20:12
  • $\begingroup$ Well, yes... but even if $d$ is a square we can choose a,b appropriately (e.g. by $a \mapsto a+1$) to get $n \neq 0$. $\endgroup$ – benh Dec 18 '13 at 23:13

If the ideal is prime, almost by definition the quotient has no zero divisors.

On the other hand, since $R$ is a finite generated abelian group, the quotient $R/P$ is also a finitely generated abelian group, and to show it is finite it is enough to show that $R/P$ has finite exponent.

If $P$ is non-zero, there is a non-zero element $x=a+b\sqrt d$ in $P$, and then $e=a^2-db^2=(a-b\sqrt d)x\in P$; you can check easily that $e\neq0$. It follows that the product of every element of $R/P$ by $e$ is zero, and therefore the exponent of the abelian group $R/P$ divides $e$.

  • $\begingroup$ I'm still unsure as to why the factor ring has no zero divisors. I know it may seem obvious but I can't seem to make the connection. $\endgroup$ – BlakeM Dec 18 '13 at 19:15
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    $\begingroup$ Suppose $a$ and $b$ are elements of $R$ such that the product in $R/P$ of their classes is zero. This means that $ab$ is an element of $P$ and, since the ideeal is prime, that one of $a$ or $b$ is in $P$. $\endgroup$ – Mariano Suárez-Álvarez Dec 18 '13 at 19:17
  • $\begingroup$ Great, this makes sense to me. $\endgroup$ – BlakeM Dec 18 '13 at 19:25

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