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In differential geometry, there exists an immersion of a curve into a sphere whose image is dense. I wanted to know if this paradox still exists in algebraic geometry. I think the answer should be no.

Let $A \rightarrow B$ be an injective homomorphism of rings of finite type. Suppose that $A$ and $B$ are Noetherian domains. Is it true that $$ \mathrm{dim}\,A \leq \mathrm{dim}\,B \quad ?$$

I made a bit of progress.

If $A$ and $B$ are finite type $k$-algebras then the statement is true. In this case, the injection $A \rightarrow B$ gives an injection of their fields of fractions $R(A) \rightarrow R(B)$. But then $\mathrm{dim}\,A = \mathrm{tr\,deg}_k R(A)$, $\mathrm{dim}\,B = \mathrm{tr\,deg}_k R(A)$ and $\mathrm{tr\,deg}_k R(B) \leq \mathrm{tr\,deg}_k R(B)$.

If the corresponding morphism $\mathrm{Spec}\, B \rightarrow \mathrm{Spec}\, A$ is closed then the statement follows from Lemma 02JX. This makes me think that there may be an issue if the map is not closed.

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  • $\begingroup$ What do you mean by "rings of finite type"? $\endgroup$
    – Will Sawin
    Commented Jun 7, 2021 at 22:24
  • $\begingroup$ I mean that $B$ is of finite type over $A$, so there is a surjection $A[x_1,...,x_r] \rightarrow B$. Sorry for the confusion. $\endgroup$
    – ofiz
    Commented Jun 7, 2021 at 22:34
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    $\begingroup$ What if $A$ is a DVR with uniformiser $\pi$ and $B = K = A[t]/(t\pi-1)$ its fraction field? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jun 7, 2021 at 22:35
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    $\begingroup$ As stated, no. You could have $A$ a discrete valuation ring (dimension $1$) and $B$ its field of fractions (dimension $0$). $\endgroup$
    – Will Sawin
    Commented Jun 7, 2021 at 22:36
  • $\begingroup$ That's a good point. Thank you both $\endgroup$
    – ofiz
    Commented Jun 8, 2021 at 12:30

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