# 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.

Inhalt 1 Aussage 2 Nachweisen 2.1 First Proof 2.2 Second Proof 3 Anwendungen 4 Formal proofs 5 Verweise 6 Further reading Statement If {Anzeigestil R} is a ring, Lassen {Anzeigestil R[X]} denote the ring of polynomials in the indeterminate {Anzeigestil X} Über {Anzeigestil R} . Hilbert proved that if {Anzeigestil R} ist "not too large", in the sense that if {Anzeigestil R} is Noetherian, the same must be true for {Anzeigestil R[X]} . Formal, Hilbert's Basis Theorem. Wenn {Anzeigestil R} is a Noetherian ring, dann {Anzeigestil R[X]} is a Noetherian ring.

Logische Folge. Wenn {Anzeigestil R} is a Noetherian ring, dann {Anzeigestil R[X_{1},Punktec ,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. Wenn {Anzeigestil R} is a left (bzw. Rechts) Noetherian ring, then the polynomial ring {Anzeigestil R[X]} is also a left (bzw. Rechts) Noetherian ring.

Anmerkung. We will give two proofs, in both only the "links" case is considered; the proof for the right case is similar. First Proof Suppose {Anzeigestil {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 {Anzeigestil {f_{0},f_{1},Punkte }} so dass wenn {Anzeigestil {mathfrak {b}}_{n}} is the left ideal generated by {Anzeigestil f_{0},Punkte ,f_{n-1}} dann {Anzeigestil f_{n}in {mathfrak {a}}setminus {mathfrak {b}}_{n}} is of minimal degree. Es ist klar, dass {Anzeigestil {Grad(f_{0}),Grad(f_{1}),Punkte }} is a non-decreasing sequence of natural numbers. Lassen {Anzeigestil a_{n}} be the leading coefficient of {Anzeigestil f_{n}} und lass {Anzeigestil {mathfrak {b}}} be the left ideal in {Anzeigestil R} generated by {Anzeigestil a_{0},a_{1},Punkte } . Seit {Anzeigestil R} is Noetherian the chain of ideals {Anzeigestil (a_{0})Teilmenge (a_{0},a_{1})Teilmenge (a_{0},a_{1},a_{2})subset cdots } must terminate. Daher {Anzeigestil {mathfrak {b}}=(a_{0},Punkte ,a_{N-1})} for some integer {Anzeigestil N} . So in particular, {Anzeigestil a_{N}= Summe _{ichand claim also {Anzeigestil {mathfrak {a}}subsetq {mathfrak {a}}^{*}} . Suppose for the sake of contradiction this is not so. Then let {displaystyle hin {mathfrak {a}}setminus {mathfrak {a}}^{*}} be of minimal degree, and denote its leading coefficient by {Anzeigestil a} . Fall 1: {displaystyle deg(h)gek d} . Regardless of this condition, wir haben {displaystyle ain {mathfrak {b}}} , so is a left linear combination {displaystyle a=sum _{j}u_{j}a_{j}} of the coefficients of the {Anzeigestil f_{j}} . In Betracht ziehen {Anzeigestil h_{0}triangleq sum _{j}u_{j}X^{Grad(h)-Grad(f_{j})}f_{j},} which has the same leading term as {Anzeigestil h} ; Außerdem {Anzeigestil h_{0}in {mathfrak {a}}^{*}} während {displaystyle hnotin {mathfrak {a}}^{*}} . Deswegen {displaystyle h-h_{0}in {mathfrak {a}}setminus {mathfrak {a}}^{*}} und {displaystyle deg(h-h_{0})

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