Throughout, \(k\) is a field, \(R = k[x_1,\ldots,x_n]\), and \(J\subseteq R\) is an ideal.
Buchberger’s algorithm
Fix a monomial order, such as lexicographic order or graded reverse lexicographic order.
For each pair of elements of the current basis, form their \(S\)-polynomial, reduce it by division against the current basis, and adjoin any nonzero remainder to the basis.
Stop when every \(S\)-polynomial of a pair of basis elements reduces to zero.
By Buchberger’s criterion, a finite generating set of \(J\) is a Gröbner basis if and only if every \(S\)-polynomial of a pair of its elements reduces to zero. The algorithm terminates because the ideals generated by the leading terms of the successive bases form a strictly ascending chain of monomial ideals, which stabilizes by Dickson’s lemma.
Applications of Gröbner bases
Let \(G\) be a Gröbner basis of \(J\).
Ideal membership. \(f\in J\) if and only if the remainder of \(f\) on division by \(G\) is zero.
Elimination. For lexicographic order with \(x_1>\cdots>x_n\), \(G\intersect k[x_{i+1},\ldots,x_n]\) is a Gröbner basis of the elimination ideal \(J\intersect k[x_{i+1},\ldots,x_n]\); this solves polynomial systems by back-substitution and computes implicitizations.
Dimension and degree. For any monomial order, the monomials not in \(\operatorname{LT}(J)\) form a \(k\)-basis of both \(R/J\) and \(R/\operatorname{LT}(J)\). For a graded order, \(R/J\) and \(R/\operatorname{LT}(J)\) therefore have the same Hilbert function when \(J\) is homogeneous, and the same affine Hilbert function in general, hence the same Krull dimension and the same degree. For \(k\) algebraically closed, \(\dim V(J) = \dim R/J\), and the degree of \(R/J\) equals \(\deg V(J)\) when \(J\) is radical.
Equality of ideals. For a fixed monomial order, each ideal has a unique reduced Gröbner basis, so two ideals are equal if and only if their reduced Gröbner bases are equal.
For \(J = (x^2)\subseteq k[x,y]\), \(V(J)\) is the line \(x = 0\), of degree \(1\), while \(R/J\) has degree \(2\); the radical \(\sqrt J = (x)\) gives degree \(1\).
Ideals and varieties
Let \(k\) be algebraically closed. By the Nullstellensatz, \(I(V(J)) = \sqrt J\). By the Rabinowitsch trick, \(f\in\sqrt J\) if and only if \(1\in J + \gens{1-yf}\subseteq R[y]\), which is an ideal membership test. In particular \(V(J) = \emptyset\) if and only if \(1\in J\), if and only if the reduced Gröbner basis of \(J\) is \(\ts 1\).