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Tag 032Q

Chapter 10: Commutative Algebra > Section 10.155: Japanese rings

Lemma 10.155.17. Let $R$ be a ring. If $R$ is Noetherian, a domain, and N-2, then so is $R[[x]]$.

Proof. Observe that $R[[x]]$ is Noetherian by Lemma 10.30.2. Let $R' \supset R$ be the integral closure of $R$ in its fraction field. Because $R$ is N-2 this is finite over $R$. Hence $R'[[x]]$ is finite over $R[[x]]$. By Lemma 10.36.9 we see that $R'[[x]]$ is a normal domain. Apply Lemma 10.155.16 to the element $x \in R'[[x]]$ to see that $R'[[x]]$ is N-2. Then Lemma 10.155.7 shows that $R[[x]]$ is N-2. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 42943–42947 (see updates for more information).

    \begin{lemma}
    \label{lemma-power-series-over-N-2}
    Let $R$ be a ring.
    If $R$ is Noetherian, a domain, and N-2, then so is $R[[x]]$.
    \end{lemma}
    
    \begin{proof}
    Observe that $R[[x]]$ is Noetherian by
    Lemma \ref{lemma-Noetherian-power-series}.
    Let $R' \supset R$ be the integral closure of $R$ in its fraction
    field. Because $R$ is N-2 this is finite over $R$. Hence $R'[[x]]$
    is finite over $R[[x]]$. By
    Lemma \ref{lemma-power-series-over-Noetherian-normal-domain}
    we see that $R'[[x]]$ is a normal domain.
    Apply Lemma \ref{lemma-tate-japanese} to the
    element $x \in R'[[x]]$ to see that $R'[[x]]$ is N-2. Then
    Lemma \ref{lemma-finite-extension-N-2} shows that $R[[x]]$ is N-2.
    \end{proof}

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