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Let $R$ be a ring and consider a monomial order $<$ on $R[X_1,\ldots,X_n]$. A nonzero polynomial $f \in R[X_1,\ldots,X_n]$ is said to be monic if its leading coefficient with respect to $<$ is $1$. We will denote $R\langle X_1,\ldots,X_n \rangle$ the localization of $R[X_1,\ldots,X_n]$ at monic polynomials.

Let $k \in \mathbb{N}$ with $k \geq 2$. It is clear that the ring $R[X_1,\ldots,X_n]$ is integral over $R[X_1^k,\ldots,X_n^k]$. Is it true that $R\langle X_1,\ldots,X_n \rangle$ is integral over $R\langle X_1^k,\ldots,X_n^k \rangle$?

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Every element of $R \langle X_1,\dots, X_n\rangle$ has the form $\frac{a(X_1,\dots, X_N)}{b(X_1,\dots, X_n)}$ where $b$ is a monic polynomial.

This element satisfies the polynomial equation

$$ \prod_{ \zeta_1,\dots, \zeta_n \in \mu_k} \left( u - \frac{a ( \zeta_1 X_1,\dots, \zeta_n X_n)}{b ( \zeta_1 X_1,\dots, \zeta_n X_n)}\right)$$ where the product is over $n$-tuples of $k$'th roots of unity (i.e. a product of $k^n$ terms) so it suffices to check that the coefficients like in $R\langle X_1^k,\dots, X_n^k\rangle$.

It's best to check this by assuming that $R$ is the formal ring generated by the coefficients of $a$ and $b$, so that we can freely add the $k$th roots of unity to it and check that they cancel once we multiply everything out.

The coefficients clearly lie in $R (\mu_k, X_1,\dots, X_n)$ and they are invariant under the substitutions $X_i \to \zeta X_i$ for $\zeta \in \mu_n$ and $\zeta \to \zeta^{e}$ for $e \in (\mathbb Z/k)^\times$, thus they lie in the fixed field of those automorphims, which is $R(X_1^k, \dots, X_n^k)$.

To check they lie in $R \langle X_1^k, \dots, X_n^k\rangle$, it suffices to check that the denominator is monic. But the denominator is

$$ \prod_{ \zeta_1,\dots, \zeta_n \in \mu_k} b ( \zeta_1 X_1,\dots, \zeta_n X_n)$$ whose leading term is the $n^k$th power of the leading term of $b$, up to possible sign, and thus is monic (up to possible sign).

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  • $\begingroup$ very tricky proof! I don't see any bug. $\endgroup$ Jul 23 at 10:55
  • $\begingroup$ Line 6: lie instead of like. $\endgroup$
    – user26857
    Jul 27 at 21:30

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