MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Some hours ago, a question was posted, asking (citation by heart, not literally)

Let $R$ be a commutative ring (with unit). What can be said about the Krull dimension of an algebraic, non-integral extension $S = R[x]$ of $R$ ?

I don't know why the question disappeared in the meanwhile. Nevertheless I find it interesting and repost it therefore.

Let $R$ be noetherian (with finite dimension). Since $S = R[x]$ is a quotient of a polynomial ring $R[T]$, there are epimorphisms $R[T] \to S \to R$ showing $\dim R + 1 = \dim R[T] \ge \dim S \ge \dim R$. Thus $\dim S = \dim R$ or $\dim S = \dim R +1$.

Both cases can occur:

1) Let $R = \mathbb{Z}$ and $S=R[T]/(pT) = R[x]$ with $x = T + (pT)$. Since $x$ is annulated by the polynomial $pX$, the extension is (non-integral) algebraic and since $pT \in R[T]$ is homogenous and regular, $\dim S = \dim R[T] -1 = \dim R = 1$.

2) Let $R =\mathbb{Z}/p^2$ and $S = R[T]/(pT)$. As above the extension is (non-integral) algebraic and since $pT$ is nilpotent $\dim S = \dim R[T] = \dim R + 1 \;(= 1)$.

share|cite|improve this question
Another example of an algebraic non-integral extension is adjunction of an inverse, as in the previous question referred to by the now-deleted question. For example, if $R = \mathbb{Q}[[t]]$ and $x=1/t$, then $x$ is a root of the polynomial $f(X) = tX-1$. In this case, $S$ is the quotient field of $R$. This shows that there is in general no epimorphism $S \to R$ as in your question, and dim S is not necessarily $\geq$ dim R. – Graham Leuschke Jun 24 '11 at 13:01
Graham, thanks for pointing out. Now, to state it correctly: Let $S=R[T]/(f)$, $f$ having zeroth term $f_0$, $R$ still noetherian. Then there are surjections $R[T] \to S \to S/(x)$ and thus $\dim R + 1 \ge \dim S \ge \dim S/(x)$ where $S/(x) \cong R/(f_0)$. So, $\dim S \ge \dim R$ if e.g. $f_0$ is in the radical of $R$. – Ralph Jun 24 '11 at 15:01
(continue) Assume $f_0$ is not a unit in $R$. Then $\dim R/(f_0) \ge \dim R -1$. This yields $$ \dim R + 1 \ge \dim S \ge \dim R -1.$$ Therefore the only critical case in regard to bound the possibilities of $\dim S$ is when $f_0$ is a unit. – Ralph Jun 24 '11 at 19:10
@Graham: In a former, now deleted comment, you said, in order to use the correct wording, I should use "surjective homomorphism" instead of "epimorphism" in my answer above. But note that "surjective" implies "epi" (while in general not every epi is surjective). – Ralph Jun 24 '11 at 19:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.