Let $K$ be a field, consider the polynomial ring in $n$ variables $R = K[X_1, \ldots, X_n]$ and let $F_1, \ldots, F_{n-1} \in R$ be arbitrary given polynomials. Let $S$ be the quotient ring $R/(F_1, \ldots, F_{n-1})$ and assume $S \neq 0$. For a fixed integer $m > 1$, is it possible that the residue classes of $X_1, \ldots, X_n$ in $S$ are all $m$-th powers?
I conjecture that the answer should be no (i.e. that for any choice of $F_1, \ldots, F_{n-1}$ and for any $m > 1$ one will have that at least one of the residue classes of the $X_1, \ldots, X_n$ is not an $m$-th power) and that this should use little more than the fact that $S$ has Krull dimension at least 1 (by the height theorem), but I cannot make it precise. Any ideas?
In the case where the characteristic of $K$ divides $m$ I might already have a proof of this conjecture; I am mainly interested in having a proof or counterexample in characteristic $0$.
In characteristic zero, the primitive element theorem makes it easy to construct counterexamples for any $n$ and $m$. Consider the extension $K(X_1,\ldots,X_n)\subseteq K(X_1^{\frac{1}{m}},\ldots,X_n^{\frac{1}{m}})$ which has degree $m^n$. Let $\alpha$ be a primitive element for this extension. Then, $X_1,\ldots,X_n$ are all $m$-th perfect powers in $R[\alpha]$ ; we can write each $X_k$ in the form $\frac{N_k(\alpha)}{D_k}$ where $N_k\in R[X]$ and $D_k\in R\setminus \lbrace 0\rbrace$. For a suitable denominator $D$ (one that is a multiple of each $D_k$, say), each $X_k$ will be of the form $P_k(\frac{\alpha}{D})$ where $P_k\in R$. Replacing $X_k$ with $\frac{X_k}{\alpha}$ if necessary, one can ensure that $P_k$ has zero as a constant term. Let $\beta=\frac{\alpha}{D}$ and let $M(X_1,\ldots,X_n,\beta)$ be its minimal polynomial. If $B$ is any polynomial in $X_1,X_2,\ldots,X_n$ such that $F_1=M(X_1,\ldots,X_n,B(X_1,\ldots,X_n))$ is nonzero and irreducible, in the quotient ring $\frac{R}{(F_1)}$, we have that the residue classes of $X_1,\ldots,X_n$ are all $m$-th perfect powers.
Here is an example that shows how it works for $n=m=2$ : we can take $\alpha=\sqrt{X_1}+\sqrt{X_2}$, then $\alpha^2=X_1+X_2+2\sqrt{X_1X_2}$ so $(\alpha^2-(X_1+X_2))^2-4X_1X_2=0$, and
$$ \sqrt{X_1}=\frac{\alpha^3-(X_2+3X_1)\alpha}{2(X_2-X_1)}, \sqrt{X_2}=\frac{\alpha^3-(X_1+3X_2)\alpha}{2(X_1-X_2)} $$
Next, let $\beta=\frac{\alpha}{X_2-X_1}$ ; then $(\beta^2-(X_1+X_2)(X_2-X_1)^2)^2=4X_1X_2(X_2-X_1)^4$ and $\sqrt{X_1}=\frac{(X_2-X_1)^2\beta^3-(X_2+3X_1)\beta}{2}$, $\sqrt{X_2}=\frac{(X_1-X_2)^2\beta^3-(X_1+3X_2)\beta}{2}$
If we take $B(X_1,X_2)=1$, we see that in the ring $\frac{R}{(F_1)}$ with $F_1=(1-(X_1+X_2)(X_2-X_1)^2)^2-4X_1X_2(X_2-X_1)^4$, $X_1$ and $X_2$ are both perfect squares.