Gröbner Basis

Theorem
Let $F$ be a finite set of polynomials, $SP(f_{1},f_{2})$ S-polynomial of $f_{1}$ and $f_{2}$, and if $g$ is a polynomial and $RF(g)$ is reduced form of $g$. Then

F is a Gröbner basis $\Longleftrightarrow \forall f_{1}, f_{2} \in F (RF(F,SP(f_{1},f_{2}))=0)$.

Proof
The direction "=)" is easy. Namely, if f1 f2 2 F, then SP(f1 f2) 2 Ideal(F ), i.e. SP(f1 f2) F 0. By the relation between reduction and con- gruence this implies that SP(f1 f2) !F 0. Hence, RF( F , SP(f1 f2) ) = 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).  For the direction "(=", by the generalized Newman lemma and the fact that !F, it su ces to prove local connectibility, i.e. it su ces to prove that under the assumption g1 F h !F g2 we have                             h                                g1 ! F g2: 20                                                              Buchberger By the assumption, there exist f1 f2 2 F and t1 t2 2 S(h) with LPP(f1 ) j t1 and LPP(f2 ) j t2, such that h !f1 t1 g1 and h !f2 t2 g2. Now we have three cases. Case t1 t2: In this case, g1 = H(h t1) + 0 t1 +B(h t1 t2) + C(h t2) t2 + L(h t2) ; ;C(h t1) u1 R(f1) and g2 = H(h t1) + C(h t1) t1 + B(h t1 t2) + 0 t2 +L(h t2) ; ;C(h t2) u2 R(f2) where u1 := t1=LPP(f1) u2 := t2=LPP(f2): Furthermore, g2 !f1 g1 2 := H(h t1) + 0 t1+ B(h t1 t2) + 0 t2+ L(h t2) ; ;C(h t1) u1 R(f1) ; ;C(h t2) u2 R(f2): Now, g1 = h ; C(h t1) u1 f1 and g1 2 = g2 ; C(h t1) u1 f1 and, by assumption, h !F g2. Hence, by sum semi-compatibility, g1 #F g1 2 and, h hence, g1 ! F g2. (Note that, in general, g1 !f1 g1 2 need not be the case. Why not?) Case t1 t2: Analogous. Case t := t1 = t2: In this case, g1 = H(h t) + 0 t +L(h t) ; ;C(h t) u1 R(f1) and g2 = H(h t) + 0 t +L(h t) ; ;C(h t) u2 R(f2): Hence, g1 ; g2 = ;C(h t) (u1 R(f1) ; u2 R(f2 )) = = ;C(h t) (u1 f1 ; u2 f2) = ;C(h t) v SP (f1 f2) where v := t= LCM( LPP(f1),LPP(f2) ). 21 Introduction to Gröbner Bases We have assumed that RF(F ,SP(f1 f2)) = 0, i.e. SP(f1 f2) !F 0. Hence, by product compatibility, g1 ; g2 = ;C(h t) v SP (f1 f2) !F 0. This means that there exists a sequence p 2 P* such that p 1 = g 1 ; g2                         81 i < jpj (pi !F pi+1 )                    (*) and pjpj = 0: Furthermore note that, because of !F , 81 i jpj ( pi g1 ; g2 h): Thus, by sum semi-compatibility applied to (*), g1 = p1 + g2                   81 i < jpj ( pi + g2 #F pi+1 + g2) g2 = pjpj + g2: Also, we have 81 i jpj ( pi + g2 h) because 81 i jpj ( H(pi + g2 t) = H(h t) ^ C(pi + g2 t) = 0): h Thus, summarizing, g1 ! F g2 also in this case.