Does inverse limit preserve Krull dimension of any inverse system of rings (at least in the case of zero-dimensional)?

I mean, is the Krull dimension of the inverse limit zero when that of any of the rings of the inverse system is zero?

I could not reach a final answer by dealing with the standard characterization of inverse limit of $R_i$'s which is the following subring of $∏R_i$:

$I=\{ (r_i)\in ∏R_i : r_i={\mu}^j_i(r_j)$ whenever $i≼j\}$, where $\{R_i,{\mu}^j_i\}$ is the inverse system of $R_i$'s with order $≼$ betweeen the indices.

In the above characterization the so-called $\mathbb Z$-maps $\alpha_i:\varprojlim R_i→R_i$ with the properties ${\mu}^j_i \alpha_j=\alpha_i$ $(i≼j)$ are considered to be the restrictions of the projections $\pi_i$ from $∏R_i$ to $R_i$, and for a starting pace, I know that the image of any prime ideal under a ring homomorphism is a prime ideal. So, if $P$ is a prime ideal of $I$ then $\alpha_i(P)$ would be a prime, hence a maximal, ideal of $R_i$ for all $i$. Therefore, if $P⊆Q$ are prime ideals of $I$ we have $\alpha_i(P)=\alpha_i(Q)$ for all $i$. Then...?

Thanks for any cooperation!


1 Answer 1


No. Let $p$ be a prime number. The ring of $p$-adic integers $\mathbb Z_p$ can be constructed as an inverse limit of the rings $\mathbb Z/ p^n\mathbb Z$, with the maps $\mathbb Z/p^{n+1}\mathbb Z \rightarrow \mathbb Z/p^n\mathbb Z$ being the canonical surjections. All the rings $\mathbb Z/p^n\mathbb Z$ have Krull dimension zero (being finite), but $\mathbb Z_p$ has Krull dimension at least one, since it is an integral domain containing a nonzero prime ideal $p \mathbb Z_p$.

See also http://mathworld.wolfram.com/p-adicInteger.html

  • 2
    $\begingroup$ Just for the record: $\dim\mathbb Z_p=1$. $\endgroup$
    – user26857
    Mar 30, 2016 at 19:24

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