Ideal equals the whole ring

Question

Show that in the polynomial ring $S=K[x_1, ..., x_n]$, having an ideal $I = (g_1, ..., g_m)$ and ${g_1, ..., g_m}$ a Groebner bases of $I$, then $I = S$ if and only if one of the $g_i$ is a nonzero constant polynomial.

I tried using using another result: under the same conditions, a polynomial $f \in S$ belongs to $I$ if and only if $f$ reduces to $0$ with respect to ${g_1, ..., g_m}$ but am not 100% sure if everything is correct.
I was also curious if there is another approach.

Answer 1

This answer just expands on user26857's excellent hint.

Note that $I=S$ if and only if $1 \in I$.

One direction is trivial. If one of the $g_i$ is a constant, then clearly $I=S$.

Now, if $I=S$, then $1 \in I$. But this implies that $1 \in in(I)=\langle in(g_1),\cdots,in(g_m) \rangle$. All the generators are monomials (i.e. of the form $\vec x^\vec a$ for $\vec a \in \mathbb N_0^n$).

But the only way $1$ can be in an ideal generated by monomials is if $1$ is a scalar multiple of one of the generators. Thus $1 = k \cdot in(g_l)$ for some $l$. Now this implies that $g_l=k$ as well, since $in$ should pick out the highest term.