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How do I see the following fact?

If $R$ has dimension $1$, and $a$ is a non-zerodivisor and non-unit, then $R/(a)$ has dimension $0$.

That is saying if $P_1\supset P_2\supset (a)$ are two prime ideals, then $P_1 = P_2$, i.e all prime ideals over $(a)$ are maximal.

user26857
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mez
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  • If $a$ is a unit, then $R/(a)=0$ has dimension $-\infty$. So you should really add "$a$ is a non-unit" as a condition. – Martin Brandenburg Jul 14 '15 at 09:13
  • It might help you to think of some concrete examples (e.g. $\mathbb{Z}$ or $k[x]$) and see what happens there. Also, don't think of maximal ideals as unique, which 'the' seems to imply. – HSN Jul 14 '15 at 09:16

2 Answers2

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$a$ is a non-zerodivisor, in particular $a$ is not contained in a minimal prime of $R$. In particular any prime over $(a)$ is of height at least $1$, hence a maximal ideal, since $R$ has dimension $1$.

Proof that a minimal prime consists of zero divisors:

Let $\mathfrak p_1, \dotsc \mathfrak p_s$ be the minimal primes of a noetherian ring $R$. $s$ is finite due to the noetherian property. We know the nilradical of $R$ is the intersection of the minimal primes, hence contains the product of the minimal primes. So $\mathfrak p_1 \dotsb \mathfrak p_s$ is nilpotent, which means there are minimal integers $n_1, \dotsc, n_s$, such that $$\mathfrak p_1^{n_1} \dotsb \mathfrak p_s^{n_s}=0.$$

Given any $x \in \mathfrak p_1$, choose $0 \neq y \in \mathfrak p_1^{n_1-1} \dotsb \mathfrak p_s^{n_s}$ and we have the desired $xy=0$.

Of course the argument requires $n_i \geq 1$ for all $i$. But this is an easy exercise. Literally one has to show that $\mathfrak p_2 \dotsb \mathfrak p_s$ (leaving out one of the minimal primes) is not nilpotent.

After some thought, one does not need the noetherian property: Let $\mathfrak p$ be a minimal prime. $\mathfrak p R_{\mathfrak p}$ is the only prime ideal of $R_{\mathfrak p}$, hence it coincides with the nilradical (This does not mean that $\mathfrak p R_{\mathfrak p}$ is nilpotent in the nonnoetherian case, but any element is nilpotent). Let $x \in \mathfrak p$. Then by the previous observation $x^n=0$ in $R_{\mathfrak p}$. The definition of localization yields some $t \notin \mathfrak p$, such that $x(x^{n-1}t)=x^nt=0$. If $n$ is chosen minimal we have $0 \neq x^{n-1}t$, hence $x$ is a zerodivisor.

MooS
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The property holds for arbitrary commutative rings.

Let $R$ be a commutative ring, and $a\in R$ a non-zero divisor. Then $\dim R/(a)\le\dim R-1$.

Set $n=\dim R<\infty$. Suppose the contrary, and let $P_0\subsetneq P_1\subsetneq\cdots\subsetneq P_n$ be a chain of prime ideals in $R/(a)$. Since $P_i$ writes as $p_i/(a)$ with $p_i$ a prime ideal of $R$ containing $(a)$, we get a chain of prime ideals $p_0\subsetneq p_1\subsetneq\cdots\subsetneq p_n$ in $R$. Now notice that $a\in p_0$ and $a$ non-zero divisor implies that $p_0$ is not minimal. (For a proof that works for arbitrary commutative rings see here.) This shows that $\dim R\ge n+1$, a contradiction.

user26857
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