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Let $p(X)\in\mathbb{Z}[X]$ be a monic polynomial and let $A$ be a commutative ring in which every element is a zero of $p(X)$.

Prove that all prime ideals in $A$ are maximal.

By definition, give any prime ideal $I\leq A$, we are left to prove $A/I$ is field, i.e., for any $a\not\in I$, we want to find a nonzero element $z\in A$, such that $az-1\in I$.

$p(0)=0$ implies $p(X)=X^k(X^l+a_{l-1}X^{l-1}+\cdots+a_1X+a_0)$, such that $k\geq 1, a_0\neq 0$.

If $l=1$, then $p(a)=0, p(1)=0$ implies $a^k(a+a_0)=0,1+a_0=0$, so we can take $z=1$.

If $l\geq 2$, then $p(1+az)=0,\forall z\in A$, but how to proceed?

I have tried to use $\frac{p(x)-p(y)}{x-y}$ is also a polynomial with integers also in $\mathbb{Z}$, but got stuck, any hint?

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Hint: Finite domains are fields. – jspecter Jan 5 '13 at 13:59
@jspecter: your hint does not apply because $A$ is not supposed to be a domain. – Georges Elencwajg Jan 5 '13 at 14:15
But $A/I$ is supposed to be a domain. And in fact $A/I$ is what we want to be a field. – Hagen von Eitzen Jan 5 '13 at 14:25
@Hagen: yes, jspecter or you should post an answer along those lines. – Georges Elencwajg Jan 5 '13 at 14:58
@GeorgesElencwajg Did so and I'm surprised how little about $p$ was needed. – Hagen von Eitzen Jan 5 '13 at 15:31
up vote 1 down vote accepted

By hypothesis the unique morphism $f:\mathbb Z\to A$ makes $A$ integral over $\mathbb Z$.
Let $\ker(f)=n\mathbb Z $.
If $n\neq 0$, we have an inclusion of rings $\bar f:\mathbb Z/n\mathbb Z \hookrightarrow A$ with $A$ integral over $\mathbb Z/n\mathbb Z$, so that [Qing Liu, Prop. 2.5.10 (b)] $\dim(A)=\dim(\mathbb Z/n\mathbb Z)=0$, which means that all prime ideals in $A$ are maximal.

And the worrying case $n=0$ ? It fortunately cannot happen because it is already impossible that all elements of the subring $ f(\mathbb Z)\cong \mathbb Z$ of $A$ be roots of some non-zero polynomial in $\mathbb Z[X]$

Note carefully
It is perfectly posssible for $A$ to satisfy the hypothesis of the question and be (hugely) infinite.
For example every element of $(\mathbb Z/2\mathbb Z)^{\mathbb R}$ is a zero of $p(X)=X^2-X$

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thank you very much! I also doubt that $A$ might have infinite elements before, thanks for the simply counterexample. Seems always when I try to solve a problem by computation and got stuck, then I might ignore some known facts or theories, this time the dimension and the concept "integral". – ougao Jan 5 '13 at 15:18
@ Myself, hopelly I could switch to not always attack a problem by simply computation. – ougao Jan 5 '13 at 15:19

Let $I$ be a prime ideal and let $F:=A/I$. Thus $F$ is a domain and want to show that $F$ is a field. From $p\in\mathbb Z[X]$ we obtain $\overline p\in F[X]$ with $\overline p(a)=0$ for all $a\in F$. Since $p$ is monic, so is $\overline p$, especially, $\overline p\ne 0$. Let $q\in F[X]\setminus\{0\}$ be of minimal degree among all nontrivial polynomials with $q(a)=0$ for all $a\in F$. (We only need $\overline p$ for the existence of such $q$). From $q(0)=0$ we see that $q=X r $ for some $r\in F[X]\setminus\{0\}$. Since $F$ is a domain, we have $r(a)=0$ for all $a\ne 0$. By minimality of $q$, we conclude $r(0)\ne 0$, i.e. $r=Xs-c$ with $s\in F[X]$ and $c\in F\setminus\{0\}$. Thus $a\cdot s(a)=c$ for all $a\ne 0$. By substituting $ac$ for $a$, we find $ac\cdot s(ac)=c$ and hence $a\cdot s(ac)=1$ for all $a\ne 0$, i.e $F$ is indeed a field.

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this is exactly the type of answer I tried to find. – ougao Jan 5 '13 at 15:43
Dear Hagen, couldn't you more simply argue that the domain $F$ is finite because its elements are roots of the non-zero polynomial $\bar p\in F[X]$ ? Then the finite domain $F$ must necessarily be a field, as remarked by jspecter? – Georges Elencwajg Jan 5 '13 at 15:54
Erm, yes you are right of course, why didn't I? As an attempt of apology for not dseeing it: A proof of "every finite domain is a field" typically uses the pigeon-hole principle and is non-constructive, while I seem to have constructed an inverse for each $a$. At least once one has determined $q$. :) – Hagen von Eitzen Jan 5 '13 at 19:39

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