Thread: integral extensions

1. integral extensions

Hi! Let A be a subring of a ring $\displaystyle B$ which is integral over $\displaystyle A$. Then the following assertions hold:

(i) $\displaystyle B[X]$ is integral over $\displaystyle A[X]$
(ii) Let $\displaystyle b$ be an ideal of $\displaystyle B$ and define $\displaystyle a:=A \cap b$, then $\displaystyle B/b$ is integral over $\displaystyle A/a$.

ad (i)
Let $\displaystyle f \in B[X]$ (polynomial with degree n). I want to show that $\displaystyle (B[X])[f]$ is a finitely generated A[X]-module. I guess that $\displaystyle 1,f,f^2,...,f^(n-1)$ generates $\displaystyle (B[X])[f]$. Is this the right way? If yes, is it the right execution?

ad (ii)
Here, i have the following idea: Let $\displaystyle \overline{x} \in B/b$ and be $\displaystyle x$ any representative of this equivalence class. As$\displaystyle B$ is integer over $\displaystyle A$, there exists $\displaystyle f \in A[X]$, so that $\displaystyle f(x)=0$. Let $\displaystyle \pi$ be the canonical projection $\displaystyle B \rightarrow B/b$, then it follows that $\displaystyle \pi(f(x))=\overline{0}$, therefore $\displaystyle \pi(f(x))=\sum_{i=1}^n \overline{a_i} \cdot \overline{x}^i=\overline{0}$.
But is this polynomial now in $\displaystyle A/a[X]$?

Has somebody got answers or ideas?
Banach

2. Originally Posted by Banach
Hi! Let A be a subring of a ring $\displaystyle B$ which is integral over $\displaystyle A$. Then the following assertions hold:

(i) $\displaystyle B[X]$ is integral over $\displaystyle A[X]$
you only need to show that for any $\displaystyle f \in B[X],$ the ring $\displaystyle A[X][f]$ is contained in a ring $\displaystyle C \subseteq B[X]$ which is finitely generated $\displaystyle A[X]$-module. now let $\displaystyle f(X)=\sum_{j=0}^nb_jX^j.$

since $\displaystyle B$ is integral over $\displaystyle A,$ the ring $\displaystyle \overline{A}=A[b_0,b_1, \cdots, b_n]$ is finitely generated $\displaystyle A$-module. let $\displaystyle \overline{A}=\sum_{j=0}^mAc_j.$ clearly $\displaystyle \forall k : f^k \in \overline{A}[X].$ thus: $\displaystyle A[X][f] \subseteq \sum_{j=0}^mA[X]c_j. \ \ \ \Box$

(ii) Let $\displaystyle b$ be an ideal of $\displaystyle B$ and define $\displaystyle a:=A \cap b$, then $\displaystyle B/b$ is integral over $\displaystyle A/a$.
we have the ring monomorphism $\displaystyle f: A/\bold{a} \longrightarrow B/\bold{b}$ defined by $\displaystyle f(\alpha + \bold{a})=\alpha + \bold{b}.$ also note that we can define $\displaystyle \forall \alpha \in A, \ \forall \beta \in B: \ (\alpha + \bold{a})(\beta + \bold{b})=\alpha \beta + \bold{b}.$ see that

this is well-defined. now if $\displaystyle \beta \in B,$ then $\displaystyle \sum_{j=0}^n{\alpha_j}\beta^j=0,$ for some $\displaystyle \alpha_j \in A.$ thus: $\displaystyle \sum_{j=0}^n(\alpha_j + \bold{a})(\beta + \bold{b})^j=\sum_{j=0}^n(\alpha_j+\bold{a})(\beta^ j + \bold{b})=\sum_{j=0}^n\alpha_j \beta^j + \bold{b}=\bold{b}=0_{B/\bold{b}}. \ \ \ \Box$

3. Thank you for your helping answer and sorry for replying so late. Just one more question: What exactly do you mean by writing $\displaystyle \sum_{j=0}^mAc_j.$?

4. I thought about it. Now everything is clear. Thank you very much.