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Tag 0307

Chapter 10: Commutative Algebra > Section 10.35: Finite and integral ring extensions

Lemma 10.35.11. Integral closure commutes with localization: If $A \to B$ is a ring map, and $S \subset A$ is a multiplicative subset, then the integral closure of $S^{-1}A$ in $S^{-1}B$ is $S^{-1}B'$, where $B' \subset B$ is the integral closure of $A$ in $B$.

Proof. Since localization is exact we see that $S^{-1}B' \subset S^{-1}B$. Suppose $x \in B'$ and $f \in S$. Then $x^d + \sum_{i = 1, \ldots, d} a_i x^{d - i} = 0$ in $B$ for some $a_i \in A$. Hence also $$ (x/f)^d + \sum\nolimits_{i = 1, \ldots, d} a_i/f^i (x/f)^{d - i} = 0 $$ in $S^{-1}B$. In this way we see that $S^{-1}B'$ is contained in the integral closure of $S^{-1}A$ in $S^{-1}B$. Conversely, suppose that $x/f \in S^{-1}B$ is integral over $S^{-1}A$. Then we have $$ (x/f)^d + \sum\nolimits_{i = 1, \ldots, d} (a_i/f_i) (x/f)^{d - i} = 0 $$ in $S^{-1}B$ for some $a_i \in A$ and $f_i \in S$. This means that $$ (f'f_1 \ldots f_d x)^d + \sum\nolimits_{i = 1, \ldots, d} f^i(f')^if_1^i \ldots f_i^{i - 1} \ldots f_d^i a_i (f'f_1 \ldots f_dx)^{d - i} = 0 $$ for a suitable $f' \in S$. Hence $f'f_1\ldots f_dx \in B'$ and thus $x/f \in S^{-1}B'$ as desired. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 7243–7249 (see updates for more information).

    \begin{lemma}
    \label{lemma-integral-closure-localize}
    Integral closure commutes with localization: If $A \to B$ is a ring
    map, and $S \subset A$ is a multiplicative subset, then the integral
    closure of $S^{-1}A$ in $S^{-1}B$ is $S^{-1}B'$, where $B' \subset B$
    is the integral closure of $A$ in $B$.
    \end{lemma}
    
    \begin{proof}
    Since localization is exact we see that $S^{-1}B' \subset S^{-1}B$.
    Suppose $x \in B'$ and $f \in S$. Then
    $x^d + \sum_{i = 1, \ldots, d} a_i x^{d - i} = 0$
    in $B$ for some $a_i \in A$. Hence also
    $$
    (x/f)^d + \sum\nolimits_{i = 1, \ldots, d} a_i/f^i (x/f)^{d - i} = 0
    $$
    in $S^{-1}B$. In this way we see that $S^{-1}B'$ is contained in
    the integral closure of $S^{-1}A$ in $S^{-1}B$. Conversely, suppose
    that $x/f \in S^{-1}B$ is integral over $S^{-1}A$. Then we have
    $$
    (x/f)^d + \sum\nolimits_{i = 1, \ldots, d} (a_i/f_i) (x/f)^{d - i} = 0
    $$
    in $S^{-1}B$ for some $a_i \in A$ and $f_i \in S$. This means that
    $$
    (f'f_1 \ldots f_d x)^d +
    \sum\nolimits_{i = 1, \ldots, d}
    f^i(f')^if_1^i \ldots f_i^{i - 1} \ldots f_d^i a_i
    (f'f_1 \ldots f_dx)^{d - i} = 0
    $$
    for a suitable $f' \in S$. Hence $f'f_1\ldots f_dx \in B'$ and thus
    $x/f \in S^{-1}B'$ as desired.
    \end{proof}

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