Timeline for Reference request for results that involve the transcendence degree
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 10, 2022 at 20:16 | comment | added | François Brunault | (1) and (2) are special cases of localisation. Let $R'=S^{-1} R$ be a localisation of a ring $R$. Let $I$ be an ideal of $R$. If the ideal $I'=S^{-1} I \subset R'$ satisfies $hI' \subset \langle f_1,\ldots,f_m\rangle$ for some $h,f_1,\ldots,f_m \in R'$, then by clearing denominators everywhere we get the analogous statement for $I$. Case (1) is $S = \mathbb{Z} \backslash \{0\}$ and (2) is $S = A \backslash Q$. | |
| Jan 10, 2022 at 20:13 | comment | added | François Brunault | Actually I don't think their second algebraic result is true: if $hQ$ were generated by $m-r$ elements then so would be $Q$. But a prime ideal is not always generated by a family of size equal to the height. This is false even over $\mathbb{C}$: there are algebraic curves in 3-space which cannot be defined by 2 equations alone. What is true (and may suffice for their applications ?), is that $hQ$ is contained in such an ideal. | |
| Jan 9, 2022 at 22:45 | comment | added | Bytegear | Thank you for this helpful answer. I hope it's ok if I ask two small questions. (1) How is the reduction from $\mathbb{Z}[x_1,\ ...\ ,x_n]$ to $\mathbb{Q}[x_1,\ ...\ ,x_n]$ justified. (2) Why does $h(Q) = n-r$ yield that $gQ$ is generated by $n-r$ elements for some $g$? | |
| Jan 7, 2022 at 0:13 | history | answered | François Brunault | CC BY-SA 4.0 |