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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?

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1 Answer 1

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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).

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