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In his book Undergraduate Commutative Algebra Miles Reid describes example of Noetherian ring such that its normalisation (integral closure in field of fractions) is not finite ring extension.

He takes $k$ to be a field, and $K$ an algebraic field extension that is not finite. Also $k[[x]],\; K[[X]]$ are formal power series rings over those fields and $k[[x]] \subset A \subset K[[X]]$ - intermediate ring with elements $f = \sum a_ix^i$ such that field $k(a_0,...a_i, ...)$ is a finite extension of $k$.

Then $A$ is DVR. Suppose there exist element $z = z_0 = \sum b_ix^i \in K[[X]]$ that do not belong to $A$ but is integral over it and moreover elements $z_n = \sum_{i \geqslant n} b_ix^{i-n} = \frac{z- \sum_{i = 1}^{n-1} b_ix^{i-n}}{x^n}$ are also integral over $A$. When such elements exist he claims that $A[z]$ is not finite extension over $A$. Otherwise it would be Noetherian $A$-module so that the ascending chain $Az_0 \subset Az_0 + Az_1 \subset ... \subset \sum_{i=0}^n Az_i \subset ...$ would stabilise. So all $z_j$ will be $A$-linear combination of some $z_0,...,z_n$.

Then the claim is that all leading coefficients $b_j$ of all the $z_j$ would have to be $k$-linear combinations of $b_0,..., b_n$, contradicting the assumption that $k(b_0,...,b_i,...)$ is infinite extension of $k$.

That where my question is. Why the fact that all $z_j$ are $A$-linear combination of $z_0,...,z_n$ implies that all leading coefficients $b_j$ of all the $z_j$ are $k$-linear combinations of $b_0,..., b_n$?

Thank you!

user26857
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user468985
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  • Not clear to me why $b_j$ are $k$-linear combinations of $b_0,\dots,b_n$, but certainly $b_j\in k(A_{0},\dots,A_n,b_0,\dots,b_n)$ for all $j$, and this should be enough to reach a contradiction. (Here $A_i$ stands for the sequence of coefficients of $a_i(x)$.) – user26857 Dec 15 '17 at 15:53
  • In fact, I think $b_j$ are $k(A_{0},\dots,A_n)$-linear combinations of $b_0,\dots,b_n$, for all $j\ge n+1$. – user26857 Dec 15 '17 at 15:58
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    Yes, I think that how it should be. We could write $z_{n+1}$ as $\sum_{i = 0}^n a_iz_i$ and then inductively see than any $b_j$ is in the $k(A_0,...A_n, b_0,...,b_n)$ in your notation. And that is enough as last field is finite over $k$ (by definition of $A$ and the fact that all $z_j$ are integral over $A$) . So statement about $k$-linear combination isn't needed and probably is not correct. – user468985 Dec 15 '17 at 16:59

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