Lemma 10.36.14. Let $R \to S$ be a ring map. Let $f_1, \ldots , f_ n \in R$ generate the unit ideal.
If each $R_{f_ i} \to S_{f_ i}$ is integral, so is $R \to S$.
If each $R_{f_ i} \to S_{f_ i}$ is finite, so is $R \to S$.
Lemma 10.36.14. Let $R \to S$ be a ring map. Let $f_1, \ldots , f_ n \in R$ generate the unit ideal.
If each $R_{f_ i} \to S_{f_ i}$ is integral, so is $R \to S$.
If each $R_{f_ i} \to S_{f_ i}$ is finite, so is $R \to S$.
Proof. Proof of (1). Let $s \in S$. Consider the ideal $I \subset R[x]$ of polynomials $P$ such that $P(s) = 0$. Let $J \subset R$ denote the ideal (!) of leading coefficients of elements of $I$. By assumption and clearing denominators we see that $f_ i^{n_ i} \in J$ for all $i$ and certain $n_ i \geq 0$. Hence $J$ contains $1$ and we see $s$ is integral over $R$. Proof of (2) omitted. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: