Timeline for Does same group of units imply surjective contraction map on spectra
Current License: CC BY-SA 4.0
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| Jan 29, 2022 at 5:27 | comment | added | Hair80 | Thank you very much for your response, actually the example I am looking at consists in $B$ an order and $A$ a Dedekind domain not integral on $B$. Would you have any idea on what happens in this case? | |
| Jan 28, 2022 at 17:10 | comment | added | KConrad | The implication $(\Leftarrow)$ is true for all ring extensions. When $A$ is integral over $B$, without Dedekind or Krull dimension conditions, the equivalence is true since ${\mathfrak p}A \not= A$ for all prime ideals $\mathfrak p$ in $B$ (Lang, Algebraic Number Theory, Proposition 9, Chapter I), so also ${\mathfrak p} \cap A^\times = \emptyset$, as a nonempty intersection would make ${\mathfrak p}A = A$. Thus you may as well assume $A$ is not integral over $B$. | |
| Jan 28, 2022 at 16:20 | comment | added | Hair80 | Yes thank you, I meant actually $\mathfrak{p}$ does not contain units of $A$. I have edited the question. I hope now is more clear | |
| Jan 28, 2022 at 16:14 | history | edited | Hair80 | CC BY-SA 4.0 |
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| Jan 28, 2022 at 15:57 | comment | added | KConrad | You have a typographical error or missing hypothesis, since there is an easy counterexample: let $B = \mathbf Z$ and $A = \mathbf Z_{(p)}$ (localization of $\mathbf Z$ at the ideal $(p)$) or $\mathbf Z_p$ ($p$-adic integers). The only prime ideals in $A$ are $(0)$ and $pA$, which contract to $(0)$ and $p\mathbf Z$, neither of which contain a unit in $A$. | |
| Jan 28, 2022 at 14:55 | history | edited | Hair80 |
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| Jan 28, 2022 at 8:26 | history | edited | Hair80 | CC BY-SA 4.0 |
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| Jan 28, 2022 at 6:50 | history | asked | Hair80 | CC BY-SA 4.0 |