# Localization at multivariate monic polynomials

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 $$R\langle X_1^k,\ldots,X_n^k \rangle$$?

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).

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