# Hilbert's basis theorem

Hilbert's basis theorem In mathematics, specifically commutative algebra, Hilbert's basis theorem says that a polynomial ring over a Noetherian ring is Noetherian.

Contents 1 Statement 2 Proof 2.1 First Proof 2.2 Second Proof 3 Applications 4 Formal proofs 5 References 6 Further reading Statement If {displaystyle R} is a ring, let {displaystyle R[X]} denote the ring of polynomials in the indeterminate {displaystyle X} over {displaystyle R} . Hilbert proved that if {displaystyle R} is "not too large", in the sense that if {displaystyle R} is Noetherian, the same must be true for {displaystyle R[X]} . Formally, Hilbert's Basis Theorem. If {displaystyle R} is a Noetherian ring, then {displaystyle R[X]} is a Noetherian ring.

Corollary. If {displaystyle R} is a Noetherian ring, then {displaystyle 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. Hilbert 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.[1] 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.

Proof Theorem. If {displaystyle R} is a left (resp. right) Noetherian ring, then the polynomial ring {displaystyle 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 {displaystyle {mathfrak {a}}subseteq R[X]} is a non-finitely generated left ideal. Then by recursion (using the axiom of dependent choice) there is a sequence of polynomials {displaystyle {f_{0},f_{1},ldots }} such that if {displaystyle {mathfrak {b}}_{n}} is the left ideal generated by {displaystyle f_{0},ldots ,f_{n-1}} then {displaystyle f_{n}in {mathfrak {a}}setminus {mathfrak {b}}_{n}} is of minimal degree. It is clear that {displaystyle {deg(f_{0}),deg(f_{1}),ldots }} is a non-decreasing sequence of natural numbers. Let {displaystyle a_{n}} be the leading coefficient of {displaystyle f_{n}} and let {displaystyle {mathfrak {b}}} be the left ideal in {displaystyle R} generated by {displaystyle a_{0},a_{1},ldots } . Since {displaystyle R} is Noetherian the chain of ideals {displaystyle (a_{0})subset (a_{0},a_{1})subset (a_{0},a_{1},a_{2})subset cdots } must terminate. Thus {displaystyle {mathfrak {b}}=(a_{0},ldots ,a_{N-1})} for some integer {displaystyle N} . So in particular, {displaystyle a_{N}=sum _{i

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