# 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 lie 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 $$\zeta \in \mu_k$$ 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 $$k^n$$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. Commented Jul 23, 2021 at 10:55

Here is a strengthening of the result established by Will Sawin:

Claim 1. Let $$R$$ be a commutative ring with identity. Let $$n$$ and $$k$$ be positive integers and let $$\prec$$ be a monomial order on $$R[X_1, \dots, X_n]$$. Let $$\mathcal{R}_k := \{0, \dots, k - 1\}^n$$, $$A := R_{\prec}\langle X_1,\dots, X_n\rangle$$ and $$B := R_{\prec}\langle X_1^k,\dots, X_n^k\rangle$$. Then $$A$$ is freely generated by $$\left\{X^{\alpha} \, \vert \, \alpha \in \mathcal{R}_k\right\}$$ over $$B$$.

where we have used the notation $$X^{\alpha} := X_1^{\alpha_1} \cdot X_2^{\alpha_2} \cdots X_n^{\alpha_n}$$ for $$\alpha \in \mathbb{Z}^n$$.

Corollary (Will Sawin's result). Let $$A$$ and $$B$$ be as in Claim 1. Then $$A$$ is integral over $$B$$.

Proof. The result follows immediately from the classical characterization of integral elements [1, Theorem 9.1.i].

Our proof of Claim 1 re-uses extensively the ideas of Will Sawin and doesn't claim to add anything original. This proof relies on the following four lemmas.

Lemma 1. Let $$R$$ be a commutative ring with identity. Let $$n$$ and $$k$$ be positive integers. Let $$A: = R[X_1, \dots, X_n]$$. Let $$B := R[X_1^k,\dots, X_n^k]$$. Then $$A$$ is freely generated by $$\left\{X^{\alpha} \, \vert \, \alpha \in \mathcal{R}_k\right\}$$ over $$B$$.

Proof. Let $$f = \sum_{\beta} a_{\beta} X^{\beta} \in A$$ with every $$a_{\beta} \in R$$. Then $$f = \sum_{\alpha \in \mathcal{R}_k} b_{\alpha}(f) X^{\alpha}$$ where $$b_{\alpha}(f) \in B$$ is uniquely defined by $$b_{\alpha}(f) := \frac{1}{X^\alpha} \sum_{\beta \equiv \alpha \mod k \mathbb{Z}^n} a_{\beta} X^{\beta}$$. Conversely, if $$f = \sum_{\alpha \in \mathcal{R}_k} c_{\alpha} X^{\alpha}$$ for some $$c_{\alpha} \in B$$, then it is immediate to check that $$c_{\alpha} = b_{\alpha}(f)$$ for every $$\alpha \in \mathcal{R}_k$$. Therefore $$\sum_{\alpha \in \mathcal{R}_k} c_{\alpha} X^{\alpha} = 0$$, if and only if, $$c_{\alpha} = 0$$ for every $$\alpha$$.

The same proof yields:

Lemma 2. Let $$R$$ be a commutative ring with identity. Let $$n$$ and $$k$$ be positive integers and let $$C: = R[X_1^{\pm 1}, \dots, X_n^{\pm 1}], D := R[X_1^{\pm k},\dots, X_n^{\pm k}]$$. Then $$C$$ is freely generated by $$\left\{X^{\alpha} \, \vert \, \alpha \in \mathcal{R}_k\right\}$$ over $$D$$.

Here is the part where Galois comes into play.

Lemma 3. Let $$R$$ be a commutative ring with identity. Let $$n$$ and $$k$$ be positive integers. We denote by $$\Phi_k(X)$$ be the $$k$$th-cyclotomic polynomial over $$\mathbb{Q}$$ and we set $$R[\overline{\zeta_k}] := R[X] / (\Phi_k(X)).$$ Let $$\zeta_k := e^{2i \pi / k}$$ and let $$\zeta \mapsto \overline{\zeta}$$ denote the ring homomorphism $$\mathbb{Z}[\zeta_k] \rightarrow R$$ induced by the map $$\zeta_k \mapsto X + (\Phi_k(X))$$. We denote by $$\langle \zeta_k \rangle$$ the (mutiplicative) cyclic subgroup of the units of $$\mathbb{Z}[\zeta_k]$$ generated by $$\zeta_k$$. Let $$f \in R[X_1, \dots, X_n]$$ and define $$\operatorname{N}_k(f) := \prod_{(\zeta_1, \dots, \zeta_n) \in \langle \zeta_k \rangle^n} f(\overline{\zeta_1} X_1, \dots, \overline{\zeta_n} X_n) \in R[\overline{\zeta_k}][X_1, \dots,X_n].$$ Then we have:

1. $$\operatorname{N}_k(f) \in R[X_1^k,\dots, X_n^k]$$.
2. If $$f$$ is a $$\prec$$-monic polynomial for some monomial order $$\prec$$ on $$R[X_1, \dots, X_n]$$, then the leading coefficient of $$\operatorname{N}_k(f)$$ with respect to $$\prec$$ is $$-1$$ or $$1$$.

Proof. As observed by Will Sawin, it suffices to prove the assertions (1) and (2) when $$R$$ is the formal ring generated by the coefficients of $$f$$, i.e., we can assume, without loss of generality, that $$R = \mathbb{Z}[Y_0, \dots, Y_d]$$ and $$f = Y_0X^{\alpha_0} + Y_1 X^{\alpha_1} + \cdots + Y_d X^{\alpha_d} \in R[X_1, \dots, X_n]$$. Indeed, the result for an arbitrary ring $$S$$ can be then inferred by means the appropriate evaluation homomorphism from $$\mathbb{Z}[Y_0, \dots, Y_d]$$ to $$S$$. This reduction allows us to embed $$\mathbb{Z}[\zeta_k]$$ into $$R[\overline{\zeta_k}]$$, so that we can omit, from now on, the bar symbol. Clearly, we have $$\operatorname{N}_k(f) \in R[\zeta_k][X_1, \dots, X_n]$$ and $$\operatorname{N}_k(f)(\zeta_1 X_1, \dots, \zeta_n X_n) = \operatorname{N}_k(f)(X_1, \dots, X_n)$$ for every $$(\zeta_1, \dots, \zeta_n) \in \langle \zeta_k \rangle^n$$. It readily follows that $$\operatorname{N}_k(f) \in R[\zeta_k][X_1^k, \dots, X_n^k]$$. Let us show assertion (1), that is, $$\operatorname{N}_k(f) \in R[X_1^k, \dots, X_n^k]$$. Let $$\alpha \in \mathbb{N}^n$$, let $$c_{\alpha} \in R[\zeta_k] = \mathbb{Z}[\zeta_k][Y_0, \dots, Y_d]$$ be the coefficient of $$X^{\alpha}$$ in $$\operatorname{N}_k(f)$$ and let us write $$a_{\alpha} = \sum_h c_h h$$ where $$h$$ ranges in a finite set of monomials of $$R$$ and $$c_h \in \mathbb{Z}[\zeta_k]$$. Observe now that each $$\sigma \in \operatorname{Gal}(\mathbb{Q}(\zeta_k) / \mathbb{Q})$$ induces a ring automorphism of $$R[\zeta_k]$$ via $$\sum_{\zeta \in \langle \zeta_k \rangle} r_{\zeta} \zeta \mapsto \sum_{\zeta \in \langle \zeta_k \rangle} r_{\zeta} \sigma(\zeta)$$. Each such automorphism induces in turn a ring automorphism of $$R[\zeta_k][X_1, \cdots, X_n]$$ which leaves $$\operatorname{N}_k(f)$$ invariant. This implies in particular that $$\sigma(a_{\alpha}) = a_{\alpha}$$, and hence $$\sigma(c_h) = c_h$$ for every $$\sigma \operatorname{Gal}(\mathbb{Q}(\zeta_k) / \mathbb{Q})$$ and every monomial $$h$$ of $$c_{\alpha}$$. Hence $$c_h \in \mathbb{Q}$$ since $$\mathbb{Q}(\zeta_k) / \mathbb{Q}$$ is a Galois extension. As $$c_h$$ is clearly integral over $$\mathbb{Z}$$ and $$\mathbb{Z}$$ is integrally closed, we have $$c_h \in \mathbb{Z}$$. Thus, we have established that $$f \in R[X_1^k,\dots, X_n^k]$$. For assertion (2), we note that the leading coefficient of $$\operatorname{N}_k(f)$$ with respect to $$\prec$$ is a product of $$k$$-th roots of unity, hence a unit of $$\mathbb{Z}[\zeta_k]$$. As it lies in $$\mathbb{Z}$$ by (1), it belongs to $$\{-1, 1\}$$, the unit group of $$\mathbb{Z}$$.

We need one more result to derive Claim 1.

Lemma 4. Let $$R$$ be a commutative ring with identity. Let $$\prec$$ be a monomial order on $$A: = R[X_1, \dots, X_n]$$ with $$n > 0$$. Let $$S$$ a subring of $$R$$ and let $$f, g \in S[X_1, \dots, X_n]$$ with $$g$$ a $$\prec$$-monic polynomial, such that there is $$q \in A$$ satisfying $$f = qg$$. Then $$q \in S[X_1, \dots, X_n]$$.

Notation. For $$f = a_{\alpha_0} X^{\alpha_0} + a_{\alpha_0} X^{\alpha_0} + \cdots + a_{\alpha_d} X^{\alpha_d}$$ with $$\alpha_0 \prec \alpha_1 \prec \dots \prec \alpha_d$$ and $$\alpha_i \neq 0$$ for every $$i$$, we set $$\alpha_{\max}(f) = \alpha_d$$, $$\operatorname{LT}(f) := a_{\alpha_d} X^{\alpha_d}$$ (leading term), $$\operatorname{LC}(f) = a_{\alpha_d}$$ (leading coefficient).

Proof of Lemma 4. Since $$\prec$$ is a well order on the monomials of $$A$$ [2, Lemma 15.2], we can reason by induction on $$\alpha_{\max}(f)$$. If $$\alpha_{\max}(f) = (0, \dots, 0)$$, then $$\alpha_{\max}(g) = \alpha_{\max}(q) = (0, \dots, 0)$$ and we have $$\operatorname{LC}(q) = \operatorname{LC}(f) \in S$$. Let us assume now that $$\alpha_{\max}(f) \succ (0, \dots, 0)$$. Since $$g$$ is $$\prec$$-monic, we have $$\operatorname{LT}(f) = \operatorname{LC}(q) \operatorname{LT}(g)$$. In particular, we have $$\operatorname{LC}(q) \in S$$. Let $$h = \frac{X^{\alpha_{\max}(f)}}{\operatorname{LT}(g)}$$ and observe that $$f - h\operatorname{LC}(q)g = (q - h\operatorname{LC}(q))g$$. As $$\alpha_{\max}(f - h\operatorname{LC}(q)g) < \alpha_{\max}(f)$$, the induction hypothesis applies.

We are now in position to prove Claim 1.

Proof of Claim 1. Because of Lemma 1, it suffices to show that $$\frac{1}{f}$$, for $$f \in R[X_1, \dots, X_n]$$ a $$\prec$$-monic polynomial, can be written as a linear span of $$\left\{X^{\alpha} \, \vert \, \alpha \in \mathcal{R}_k\right\}$$ with coefficients in $$R_{\prec} \langle X_1^k, \dots, X_n^k \rangle$$. It follows from Lemmas 3 and 4 that $$\operatorname{N}_k(f) \in R[X_1^k, \dots, X_n^k]$$ and $$q := \frac{\operatorname{N}_k(f)}{f} \in R[X_1, \dots, X_n]$$. Since $$\operatorname{N}_k(f)$$ is $$\prec$$-monic (up to multiplication by $$-1$$), writing $$\frac{1}{f} = \frac{q}{\operatorname{N}_k(f)}$$ and expanding $$q$$ thanks to Lemma 1 yields the result.

[1] H. Matsumura, "Commutative Ring Theory", 1986.
[2] D. Eisenbud, "Commutative Algebra with a View Toward Algebraic Geometry", 1995.