Ideal membership and change of fields

Question

Let $R=k[x_1,...,x_n]$ be a polynomial ring over a field. Let $f$ be a homogeneous polynomial and $I=(f_1,...,f_m)$ a homogeneous ideal.

With Macaulay2, one can compute the Groebner basis of $I$ when $k=\mathbb{Q}$ or when $k=\mathbb{F}_p$. This helps one decide whether $f$ is in $I$ or not. This is of course just a rephrase of the ideal membership problem.

Question 1: Suppose we know with M2 that $f\in I$ when $k$ is the field of rationals or the finite field of prime cardinality. Is it true that $f\in I$ when $k$ is a different field?

Naive partial answer: Sometimes. Say we run M2 for $k=\mathbb{Q}$ and obtain $f\in I$. Then we can then compute the presentation: $$ f=g_1f_1+\cdots+g_mf_m. $$ Suppose all the coefficients of $g_1,...,g_m$ are integers. Then we can conclude that $f\in I$ no matter what $k$ is because the integers transfer nicely to other fields. On the other hand, suppose all the coefficients of $g_1,...,g_m$ are (integer) multiples of $1/2$. Then we can conclude that $f\in I$ for any $k$ as long as $char(k)\neq 2$.

Here is a variation of Ques 1:

Question 2: Suppose we know with M2 that $f\notin I$ when $k$ is the field of rationals or the finite field of prime cardinality. Is it true that $f\notin I$ when $k$ is a different field?

For this one, I do not even have a naive answer. In fact, I'm not even sure how to approach it. My naive way would be assuming that $$ f=g_1f_1+\cdots + g_mf_m. $$ Then obtain polynomial equations and try to find a contradiction. This should work well if $f,f_1,...,f_m$ are of the same degree, and it gets harder when there is a big difference between the degrees.

Question: Are there other approaches for these problems?

The examples I'm working on: Let $X,Y$ be square generic matrices of the same size. COnsider the ideal $I=I_1(XY-YX)$, i.e., $I$ is generated by all entries of $XY-YX$. Let $f$ be a $n$-minor of the matrix concatenation $(X|Y)$, for some $n$. Is it true that $f\in I$?

Of course, with brute force, one may determine this when $n=2$. When $n$ is larger, M2 can show you some answers. But the questions come up: how about when $k$ is a field where you can't run on M2?

Answer 1

The ideal membership question is completely algorithmic, and indeed you can solve is using the reduced Gröbner basis of your ideal (for any ordering of your choice). The reduced Gröbner basis of $(f_1,\ldots,f_m)$ is computed via the Buchberger algorithm, and it is clear from the way the algorithm works, that everything happens over the subfield of $k$ generated by the coefficients of $f_1,\ldots,f_m$.

I am assuming that the coefficients of $f_1,\ldots,f_m, f$ are integers, since otherwise your question does not make much sense to begin with. In that case, what said above ensures that the reduced Gröbner basis is defined over the prime subfield of $k$, so it is indeed enough to figure it out over $\mathbb{Q}$ and all $\mathbb{F}_p$ to conclude the result for any field.