Groebner basis over rings

Question

Let $I$ be an ideal in $A[x_1, \ldots, x_n]$, where $A$ is a Noetherian commutative ring, such that w.r.t. some monomial order it has a Groebner basis $G = \{g_1, \ldots, g_t\}$ with all the leading coefficients of each $g_i$ equal to $1$. For $S \subseteq \{x_1, \ldots, x_n\}$ it is given an $f \in A[S] \cap I$. Can we then assume that we always can choose $f'\in A[S] \cap I$ such that the leading coefficient of $f'$ is $1$?

Answer 1

You probably want to assume that $A$ is a domain. Otherwise the answer is no.

Assume there are elements $a,b \in A \setminus \{0\}$ with $ab =0$. Consider the ring $A[x_1,x_2]$ with the lexicographic order in which $x_1 > x_2$. Furthermore, consider the ideal $I = (x_1 - a x_2)$. The generator $x_1 - a x_2$ is a Grobner basis of this ideal.

One can easily compute the quotient ring: $$A[x_1,x_2] /I \simeq A[x_2], \quad x_1 \mapsto a x_2, x_2 \mapsto x_2.$$

Hence
$$I \cap A[x_1] = \operatorname{ker}( A[x_1] \to A[x_1,x_2] \to A[x_2] )$$ with the map $$A[x_1] \to A[x_2], \quad x_1 \mapsto a x_2 .$$

Thus $I \cap A[x_1]$ cannot contain monic polynomials. But $bx_1 = b (x_1 - ax_2) \in I \cap A[x_1]$.