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:

- $\operatorname{N}_k(f) \in R[X_1^k,\dots, X_n^k]$.
- 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.