In mathematics, specifically commutative algebra, Hilbert's basis theorem says that a polynomial ring over a Noetherian ring is Noetherian. If is a ring, let denote the ring of polynomials in the indeterminate over . Hilbert proved that if is "not too large", in the sense that if is Noetherian, the same must be true for . Formally, Hilbert's Basis Theorem. If is a Noetherian ring, then is a Noetherian ring. Corollary. If is a Noetherian ring, then 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. 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. Theorem. If is a left (resp. right) Noetherian ring, then the polynomial ring 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. Suppose is a non-finitely generated left ideal. Then by recursion (using the axiom of dependent choice) there is a sequence of polynomials such that if is the left ideal generated by then is of minimal degree. It is clear that is a non-decreasing sequence of natural numbers. Let be the leading coefficient of and let be the left ideal in generated by . Since is Noetherian the chain of ideals must terminate. Thus for some integer . So in particular, Now consider whose leading term is equal to that of ; moreover, . However, , which means that has degree less than , contradicting the minimality. Let be a left ideal. Let be the set of leading coefficients of members of .

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