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can you please help me understand something.

I defined minimal GB like this:

Let $G = \{g1, . . . , gs\}$ be a GB of an ideal $I ⊂ k[x1, . . . , xn]$. Then $G$ is a minimal GB if and only if for each $i = 1, . . . , s$, the polynomial $LC(gi) = 1$ and its leading monomial $LM(gi)$ does not divide $LM(gj)$ for any $j$ different than $i$.

and I showed that if:

$G = \{g1, . . . , gs\}$ and $H = \{h1, . . . , ht\}$ are two minimal GB for $I$ then $s = t$ and, after renumbering as necessary, $LT(gi) = LT(hi)$ for $i = 1, . . . , s$.

Now I have a definition for reduced GB:

Let $G = \{g1, . . . , gs\}$ be a GB of an ideal $I ⊂ k[x1, . . . , xn]$. Then $G$ is a reduced GB if and only if for each $i = 1, . . . , s$ $LC(gi) = 1$ and its leading monomial $LM(gi)$ does not divide any term of any $gj$ for any $j$ different then $i$.

Now i want to show uniqueness of reduced GB and the proof goes like this:

Suppose that $\{f1, . . . , fs\}$ and $\{g1, . . . , gs\}$ are both reduced and ordered so that $LT(fi) = LT(gi)$ for each $i$. Consider $fi − gi ∈ I$. If it is not zero, then its leading term must be a term that appeared in $fi$ or in $gi$ . In either case, this contradicts the bases being reduced, so in fact $fi = gi$ as claimed.

I don't understand why is there a contradiction with the bases being reduced.

Prajwal Kansakar
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Petra
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1 Answers1

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There are some $f_j$ and $g_k$ whose leading terms divide the leading term of $f_i-g_i$. They must be $j,k \neq i$ since $LT(f_i-g_i)$ is strictly less than $LT(f_i)$ and $LT(g_i)$. If $LT(f_i-g_i)$ comes from $f_i$, then $f_i$ has a term divisible by $LT(f_j)$; similarly if $LT(f_i-g_i)$ comes from $g_k$.

Zach Teitler
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