# $G$ is a Grobner basis if and only if $\overline{f}^{G} = 0$

Claim: $$G = \{g_1, \cdots, g_s\}$$ is a Grobner basis if and only if for all $$f \in I \subseteq k[x_1, \cdots, x_n]$$ we have $$\overline{f}^{G} = 0$$. Here, the overline notation means remainder upon division by G.

I am having difficulty showing that if $$\overline{f}^{G} = 0$$, then $$G$$ is a Grobner basis.

To be a Grobner basis means:

$$\langle LT(I) \rangle = \langle LT(g_1), \cdots, LT(g_s) \rangle$$

Since each $$g_i$$ is contained in the ideal $$I$$, we know $$\supseteq$$ for the above equation. But this does not utilize the assumption. How do we show the $$\subseteq$$ using our assumption?

It suffices to pick a nonzero $$f\in I$$ and show that $$LT(f)\in\langle LT(g_1),\ldots LT(g_s)\rangle$$. By properties of monomial ideals, we just need to show that $$LT(f)$$ is divisible by one of the $$LT(g_i)$$. Since $$f$$ has a remainder of zero upon division by $$G$$, then it must be the case that $$LT(f)$$ is divisible by one of the $$LT(g_i)$$ just by using the definition of the division algorithm. Otherwise, if $$LT(f)$$ weren't divisible by any $$LT(g_i)$$, then $$LT(f)$$ would show up as a nonzero term in the remainder (but this cannot happen since the remainder was assumed to be zero).