Hilbert's basis theorem

In mathematics, specifically commutative algebra, Hilbert's basis theorem says that a polynomial ring over a Noetherian ring is Noetherian.Hilbert's Basis Theorem. If R {displaystyle R} is a Noetherian ring, then R [ X ] {displaystyle R} is a Noetherian ring.Corollary. If R {displaystyle R} is a Noetherian ring, then R [ X 1 , … , X n ] {displaystyle R} is a Noetherian ring. In mathematics, specifically commutative algebra, Hilbert's basis theorem says that a polynomial ring over a Noetherian ring is Noetherian. If R {displaystyle R} is a ring, let R [ X ] {displaystyle R} denote the ring of polynomials in the indeterminate X {displaystyle X} over R {displaystyle R} . Hilbert proved that if R {displaystyle R} is 'not too large', in the sense that if R {displaystyle R} is Noetherian, the same must be true for R [ X ] {displaystyle R} . Formally, 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. Hilbert (1890) 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 Gröbner bases. Remark. We will give two proofs, in both only the 'left' case is considered; the proof for the right case is similar. Suppose a ⊆ R [ X ] {displaystyle {mathfrak {a}}subseteq R} is a non-finitely generated left-ideal. Then by recursion (using the axiom of dependent choice) there is a sequence { f 0 , f 1 , … } {displaystyle {f_{0},f_{1},ldots }} of polynomials such that if b n {displaystyle {mathfrak {b}}_{n}} is the left ideal generated by f 0 , … , f n − 1 {displaystyle f_{0},ldots ,f_{n-1}} then f n ∈ a ∖ b n {displaystyle f_{n}in {mathfrak {a}}setminus {mathfrak {b}}_{n}} is of minimal degree. It is clear that { deg ⁡ ( f 0 ) , deg ⁡ ( f 1 ) , … } {displaystyle {deg(f_{0}),deg(f_{1}),ldots }} is a non-decreasing sequence of naturals. Let a n {displaystyle a_{n}} be the leading coefficient of f n {displaystyle f_{n}} and let b {displaystyle {mathfrak {b}}} be the left ideal in R {displaystyle R} generated by a 0 , a 1 , … {displaystyle a_{0},a_{1},ldots } . Since R {displaystyle R} is Noetherian the chain of ideals must terminate. Thus b = ( a 0 , … , a N − 1 ) {displaystyle {mathfrak {b}}=(a_{0},ldots ,a_{N-1})} for some integer N {displaystyle N} . So in particular,