integral extensions

**Banach** · October 11th 2008, 02:51 AM

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

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

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

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

Has somebody got answers or ideas?
Banach

**NonCommAlg** · October 11th 2008, 08:54 AM

Originally Posted by Banach

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

(i) $B[X]$ is integral over $A[X]$

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

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

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

we have the ring monomorphism $f: A/\bold{a} \longrightarrow B/\bold{b}$ defined by $f(\alpha + \bold{a})=\alpha + \bold{b}.$ also note that we can define $\forall \alpha \in A, \ \forall \beta \in B: \ (\alpha + \bold{a})(\beta + \bold{b})=\alpha \beta + \bold{b}.$ see that

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

**Banach** · October 15th 2008, 06:38 AM

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

**Banach** · October 15th 2008, 12:32 PM

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

Thread: integral extensions

LinkBack

Thread Tools

Display

integral extensions

Similar Math Help Forum Discussions

Separable Extensions

Algebraic Extensions

Extensions

Field Extensions

Field Extensions

Search Tags