Quotient Rings and Integral Extensions

Question

Suppose $S$ is an integral extension of $R$ and $I$ an ideal in $S$. Why is $S/I$ an integral ring extension of $R/(R \cap I)$?

To this question, Dummit and Foote says: Reducing the monic polynomial over $R$ satisfied by $s \in S$ modulo $I$ gives the monic polynomial satisfied by $\bar{s} \in S/I$ over $R/(R \cap I)$.

Can someone explain this remark in more detail?

If you try to use the definition of "integral ring extension" for $R/R\cap I\subseteq S/I$, there is one obvious idea... Have you tried it? — Mariano Suárez-Álvarez, Oct 18 '11 at 14:33

score 8 · Accepted Answer · edited Jul 17 '23 at 07:23

8

There a ring homomorphism $\phi:R\to R/R\cap I$ such that $\phi(r)=r+R\cap I$ for all $r\in R$. Using it, we can construct another homomorphism, $\Phi:R[X]\to(R/R\cap I)[X]$ which on a polynomial $p$ with coefficients in $R$ applies the function $\phi$ to each coefficient.

Now, if $\sigma\in S/I$ is any element, there exists an element $s\in S$ such that $\sigma=s+I$. Since $S$ is integral over $R$, there exists a monic polynomial $f\in R[X]$ such that $f(s)=0$. Now you can check that if $q=\Phi(f)\in(R/R\cap I)[X]$, then $q(\sigma)=0$.

edited Jul 17 '23 at 07:23

user26857

53,190

answered Oct 18 '11 at 14:49

Mariano Suárez-Álvarez

139,300

At first I was thinking $R/(R \cap I) \cong (R+I)/I$ but I don't think $R+I$ makes sense since $R$ is not an ideal of $S$. Is this correct? – user5826 Sep 29 '20 at 15:40

Quotient Rings and Integral Extensions

1 Answers1