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


2 Answers 2


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).

  • $\begingroup$ very tricky proof! I don't see any bug. $\endgroup$ 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.


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