in mathematics, specifically commutative algebra, hilbert's basis theorem says that a polynomial ring over a noetherian ring is noetherian. ==statement== if r a ring, let r[x] denote the ring of polynomials in the indeterminate x over r. hilbert proved that if r is "not too large", in the sense that if r is noetherian, the same must be true for r[x]. formally, hilbert's basis theorem. if r is a noetherian ring, then r[x] is a noetherian ring. corollary. if r is a noetherian ring, then r[x_1,\dotsc,x_n] is a noetherian ring. this can be translated into algebraic geometry as follows: every algebraic set over a field can be described as the set of common roots of finitely many polynomial equations.  proved the theorem (for the special case of polynomial rings over a field) in the course of his proof of finite generation of rings of invariants. hilbert produced an innovative proof by contradiction using mathematical induction; his method does not give an algorithm to produce the finitely many basis polynomials for a given ideal: it only shows that they must exist. one can determine basis polynomials using the method of grÃ¶bner bases. ==proof== :theorem. if r is a left (resp. right) noetherian ring, then the polynomial ring r[x] is also a left (resp. right) noetherian ring. remark. we will give two proofs, in both only the "left" case is considered, the proof for the right case is similar. ===first proof=== suppose \mathfrak a \subseteq r[x] were a non-finitely generated left-ideal. then by recursion (using the axiom of dependent choice) there is a sequence \{f_0, f_1, \dotsc\} of polynomials such that if \mathfrak b_n is the left ideal generated by f_0, \dotsc, f_{n-1} then f_n in \mathfrak a \setminus \mathfrak b_n is of minimal degree. it is clear that \{\deg(f_0), \deg(f_1), \dotsc, \} is a non-decreasing sequence of naturals. let a_n be the leading coefficient of f_n and let \mathfrak b be the left ideal in r generated by a_0,a_1,\dotsc. since r is left-noetherian, we have that \mathfrak b must be finitely generated; and since the a_n comprise an r-basis, it follows that for a finite amount of them, say a_0,\dotsc,a_{n-1}, will suffice. so for example, :a_n=\sum_{i<n}u_{i}a_{i}, \qquad u_i \in r. now consider :g \triangleq \sum_{i<n}u_{i}x^{\deg(f_{n})-\deg(f_{i})}f_{i}, whose leading term is equal to that of f_n; moreover, g\in\mathfrak b_n. however, f_n \notin \mathfrak b_n, which means that f_n - g \in \mathfrak a \setminus \mathfrak b_n has degree less than f_n, contradicting the minimality. ===second proof=== let \mathfrak a \subseteq r[x] be a left-ideal. let \mathfrak b be the set of leading coefficients of members of \mathfrak a. this is obviously a left-ideal over r, and so is finitely generated by the leading coefficients of finitely many me
Commutative_algebra Invariant_theory Theorems_in_abstract_algebra