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.

Contenu 1 Déclaration 2 Preuve 2.1 First Proof 2.2 Second Proof 3 Applications 4 Formal proofs 5 Références 6 Further reading Statement If {style d'affichage R} is a ring, laisser {style d'affichage R[X]} denote the ring of polynomials in the indeterminate {style d'affichage X} plus de {style d'affichage R} . Hilbert proved that if {style d'affichage R} est "not too large", in the sense that if {style d'affichage R} is Noetherian, the same must be true for {style d'affichage R[X]} . Officiellement, Hilbert's Basis Theorem. Si {style d'affichage R} is a Noetherian ring, alors {style d'affichage R[X]} is a Noetherian ring.

Corollaire. Si {style d'affichage R} is a Noetherian ring, alors {style d'affichage R[X_{1},pointsc ,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. Si {style d'affichage R} is a left (resp. droit) Noetherian ring, then the polynomial ring {style d'affichage R[X]} is also a left (resp. droit) Noetherian ring.

Remarque. We will give two proofs, in both only the "la gauche" case is considered; the proof for the right case is similar. First Proof Suppose {style d'affichage {mathfrak {un}}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 {style d'affichage {F_{0},F_{1},ldots }} telle que si {style d'affichage {mathfrak {b}}_{n}} is the left ideal generated by {style d'affichage f_{0},ldots ,F_{n-1}} alors {style d'affichage f_{n}dans {mathfrak {un}}setmoins {mathfrak {b}}_{n}} is of minimal degree. Il est clair que {style d'affichage {degré(F_{0}),degré(F_{1}),ldots }} is a non-decreasing sequence of natural numbers. Laisser {style d'affichage a_{n}} be the leading coefficient of {style d'affichage f_{n}} et laissez {style d'affichage {mathfrak {b}}} be the left ideal in {style d'affichage R} generated by {style d'affichage a_{0},un_{1},ldots } . Depuis {style d'affichage R} is Noetherian the chain of ideals {style d'affichage (un_{0})subset (un_{0},un_{1})subset (un_{0},un_{1},un_{2})subset cdots } must terminate. Ainsi {style d'affichage {mathfrak {b}}=(un_{0},ldots ,un_{N-1})} for some integer {displaystyle N} . So in particular, {style d'affichage a_{N}=somme _{jeand claim also {style d'affichage {mathfrak {un}}sous-ensemble {mathfrak {un}}^{*}} . Suppose for the sake of contradiction this is not so. Then let {displaystyle hin {mathfrak {un}}setmoins {mathfrak {un}}^{*}} be of minimal degree, and denote its leading coefficient by {style d'affichage a} . Cas 1: {displaystyle deg(h)geek d} . Regardless of this condition, Nous avons {displaystyle ain {mathfrak {b}}} , so is a left linear combination {displaystyle a=sum _{j}tu_{j}un_{j}} of the coefficients of the {style d'affichage f_{j}} . Envisager {style d'affichage h_{0}triangleq sum _{j}tu_{j}X^{degré(h)-degré(F_{j})}F_{j},} which has the same leading term as {style d'affichage h} ; moreover {style d'affichage h_{0}dans {mathfrak {un}}^{*}} tandis que {displaystyle hnotin {mathfrak {un}}^{*}} . Par conséquent {displaystyle h-h_{0}dans {mathfrak {un}}setmoins {mathfrak {un}}^{*}} et {displaystyle deg(h-h_{0})

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