Consider a homogeneous polynomial of several variables $f(x_1,x_2,\ldots,x_n)$ with the leading term (with respect to lex ordering) having the maximum degree of any $x_i$s to be $k$. Take the ideal I generated by the polynomials $\langle x_1^{k+1}-1, x_2^{k+1}-1,\ldots, x_n^{k+1}-1\rangle$. It can be seen that the polynomials that generate the ideal are a Groebner basis for the ideal.
Now, if we reduce $f$ modulo $I$. what is the remainder we could expect? From the division algorithm for the multivariate case, it is clear that the maximum degree of leading term in the remainder for each $x_i$s would be at most $k$. But, since $x_i^{k+1}$ cannot be converted to $x_i^k$ by any polynomial multiplication, I expect that the remainder must be $f$. But, when I ran an implementation on Sage, I got the remainder not equal to $f$ for the specific case $k=3$. The same process when I applied to reduce the polynomial with respect to the basis $\langle x_1^{k+2}-1,x_2^{k+2}-1,\ldots,x_n^{k+2}-1\rangle$, I got the remainder as $f$.
For example, take the polynomial $$f=(x_1-x_2)(x_1-x_3)(x_1-x_6)(x_2-x_3)(x_2-x_7)(x_3-x_5)(x_3-x_9)(x_5-x_8)(x_5-x_{10})(x_8-x_4)$$ where $x_3,x_4,x_5,x_1,x_2, x_6, x_7,x_8, x_9, x_{10}$ are variables, which is homegeneous. Now, reduction of this modulo the ideal $\langle x_1^4-1, x_2^4-1, x_3^4-1, x_4^4-1,x_5^4-1,x_6^4-1, x_7^4-1, x_8^4-1,x_9^4-1,x_{10}^4-1\rangle$ yields a non-zero remainder not equal to $f$. The sagecode used is:
sagemath: R.<x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10}>= PolynomialRing(QQ,order='lex')
sagemath: f=(x_1-x_2)*(x_1-x_3)*(x_1-x_6)*(x_2-x_3)*(x_2-x_7)*(x_3-x_5)*(x_3-x_9)*(x_5-x_8)*(x_5-x_{10})*(x_8-x_4)
sagemath: I=Ideal(x_1^4-1, x_2^4-1, x_3^4-1, x_4^4-1,x_5^4-1,x_6^4-1, x_7^4-1, x_8^4-1,x_9^4-1,x_{10}^4-1)
sagemath: h=f.reduce(I)
sagemath: h
Whereas when $f$ is reduced modulo $\langle x_i^5-1\rangle\ \ i\in\{1,2,\ldots,10\}$ yields remainder equal to $f$. Any explanation for this phenomenon? Thanks beforehand.