Gröbner Basis

From ProofWiki
Jump to navigation Jump to search


Let $F$ be a finite set of polynomials.

Let $\map {\operatorname {SP} } {f_1, f_2}$ be the $S$-polynomial of $f_1$ and $f_2$.

Let $g$ be a polynomial.

Let $\map {\operatorname {RF} } g$ be the reduced form of $g$.

Then $F$ is a Gröbner basis if and only if:

$\forall f_1, f_2 \in \map F {\map {\operatorname {RF} } {F, \map {\operatorname {SP} } {f_1, f_2} } = 0}$


Necessary Condition

Let $F$ be a Gröbner basis.

Let $f_1, f_2 \in F$.


$\map {\operatorname {SP} } {f_1, f_2} \in \map {\operatorname {Ideal} } F$

That is:

$\map {\operatorname {SP} } {f_1, f_2} \equiv_F 0$

By the relation between reduction and congruence this implies that:

$\map {\operatorname {SP} } {f_1, f_2} \leftrightarrow_F^* 0$


$\map {\operatorname {RF} } {F, \map {\operatorname {SP} } {f_1, f_2} } = \map {\operatorname {RF} } {F, 0} = 0$

because $F$ is a Gröbner basis (and because of the equivalence between the Church-Rosser property and the normal form Church-Rosser property).


Sufficient Condition


$\forall f_1, f_2 \in \map F {\map {\operatorname {RF} } {F, \map {\operatorname {SP} } {f_1, f_2} } = 0}$

By the generalized Newman lemma and:

$\gets_F \mathop \subseteq \succ$

it suffices to prove local connectibility.

That is, it suffices to prove that under the assumption:

$g_1 \to_F h \gets_F g_2$

we have:

$g_1 \stackrel {\prec h} {\longleftrightarrow^*_F} g_2$

Let $\map {\operatorname {LPP} } f$ be the leading power product of $f$.

Then by the assumption, there exist $f_1, f_2 \in F$ and $t_1, t_2 \in \map S h$ with:

$\map {\operatorname {LPP} } {f_1} \mathop | t_1$


$\map {\operatorname {LPP} } {f_2} \mathop | t_2$

such that:

$h \rightarrow_{f_1, t_1} g_1$


$h \rightarrow_{f_2, t_2} g_2$

Now we have three cases.

Case $t_1 \succ t_2$:

In this case:

\(\ds g_1\) \(=\) \(\ds \map H {h, t_1} + 0 \cdot t_1 + \map B {h, t_1, t_2} + \map C {h, t_2} \cdot t_2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \map L {h, t_2} - \map C {h, t_1} \cdot u_1 \cdot \map R {f_1}\)


\(\ds g_2\) \(=\) \(\ds \map H {h, t_1} + \map C {h, t_1} \cdot t_1 + \map B {h, t_1, t_2} + 0 \cdot t_2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \map L {h, t_2} - \map C {h, t_2} \cdot u_2 \cdot \map R {f_2}\)


$u_1 := \dfrac {t_1} {\map {\operatorname {LPP} } {f_1} }$


$u_2 := \dfrac {t_2} {\map {\operatorname {LPP} } {f_1} }$


\(\ds g_2 \to_{f_1} g_{1, 2}\) \(=\) \(\ds \map H {h, t_1} + 0 \cdot t_1 + \map B {h, t_1, t_2} + 0 \cdot t_2 + \map L {h, t_2}\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \map C {h, t_1} \cdot u_1 \cdot \map R {f_1} - \map C {h, t_2} \cdot u_2 \cdot \map R {f_2}\)


$g_1 = h - \map C {h, t_1} \cdot u_1 \cdot f_1$


$g_{1, 2} = g_2 - \map C {h, t_1} \cdot u_1 \cdot f_1$

and, by assumption:

$h \to_F g_2$

Hence, by sum semi-compatibility:

$g_1 \downarrow_{\stackrel * F} g_{1, 2}$

and hence:

$g_1 \stackrel {\prec h} {\longleftrightarrow^*_F} g_2$

(Note that, in general, $g_1 \to_{f_1} g_{1, 2}$ need not be the case. Why not?)

Case $t_1 \prec t_2$:


Case $t := t_1 = t_2$:

In this case:

$g_1 = \map H {h, t} + 0 \cdot t + \map L {h, t} - \map C {h, t} \cdot u_1 \cdot \map R {f_1}$


$g_2 = \map H {h, t} + 0 \cdot t + \map L {h, t} - \map C {h, t} \cdot u_2 \cdot \map R {f_2}$


\(\ds g_1 - g_2\) \(=\) \(\ds - \map C {h, t} \cdot \paren {u_1 \cdot \map R {f_1} - u_2 \cdot \map R {f_2} }\)
\(\ds \) \(=\) \(\ds - \map C {h, t} \cdot \paren {u_1 \cdot f_1 - u_2 \cdot f_2}\)
\(\ds \) \(=\) \(\ds - \map C {h, t} \cdot v \cdot \map {\operatorname {SP} } {f_1, f_2}\)


$v := t / \map {\operatorname {LCM} } {\map {\operatorname {LPP} } {f_1}, \map {\operatorname {LPP} } {f_2} }$

We have assumed that:

$\map {\operatorname {RF} } {F, \map {\operatorname {SP} } {f_1, f_2} } = 0$

That is:

$\map {\operatorname {SP} } {f_1, f_2} \to_{\stackrel * F} 0$

Hence, by product compatibility:

$g_1 - g_2 = - \map C {h, t} \cdot v \cdot \map {\operatorname {SP} } {f_1, f_2} \to_{\stackrel * F} 0$

This means that there exists a sequence $p \in P^*$ such that:

\(\ds p_1\) \(=\) \(\ds g_1 - g_2\)
\(\text {(1)}: \quad\) \(\ds \forall 1 \le i < \size p: \, \) \(\ds p_i\) \(\to_F\) \(\ds p_{i + 1}\)


$p_{\size p} = 0 $

Furthermore note that, because of $\to_F \subseteq \succ$:

$\forall 1 \le i < \size p: p_i\preceq g_1 -g_2 \prec h$

Thus, by sum semi-compatibility applied to $(1)$:

$g_1 = p_1 + g_2$
$\forall 1 \le i < \size p: p_i + g_2 \downarrow_{\stackrel * F} p_{i + 1} + g_2$
$g_2 = p_{\size p} + g + 2$

Also, we have:

$\forall 1 \le i < \size p: p_i + g_2 \prec h$


$\forall 1 \le i < \size p: \map H {p_i + g_2, t} = \map H {h, t} \wedge \map C {p_i + g_2, t} = 0$

Thus, summarizing:

$g_1 \stackrel {\prec h} {\longleftrightarrow^*_F} g_2$ also in this case.

Buchberger's Algorithm

Input: a finite set of polynomials $F$

Output: $\map {\operatorname {GB} } F$, a Gröbner bases of $\map {\operatorname {Ideal} } F$

  • $\map {\operatorname {Gröbner-Basis} } F := \map {\operatorname {GB} } {F, \set {f_1, f_2} | f_1, f_2 \in F}$
  • $\map {\operatorname {GB} } {F, \O} := F$
  • $\map {\operatorname {GB} } {F, \set {f_1, f_2} \cup B} := \begin{cases} \map {\operatorname {GB} } {F, B} & : h = 0 \\ \map {\operatorname {GB} } {F \cup h, B \cup \set {h, f | f \in F} } & : \text{otherwise} \end{cases}$


$h := \map {\operatorname {RF} } {F, \map {\operatorname {SP} } {f_1, f_2} }$

Source of Name

This entry was named for Wolfgang Gröbner.


  • 2006: Bruno BuchbergerAn Algorithm for Finding the Basis Elements of the Residue Class Ring of a Zero Dimensional Polynomial Ideal (J. Symb. Comput. Vol. 41: pp. 471 – 511)
(translated by Michael P. Abramson from "Ein Algorithmus zum Auffinden der Basiselemente des Restklassenringes nach einem nulldimensionalen Polynomideal")