Hilbert's basis theorem

InmathematicsHilbert's basis theoremasserts that everyidealof apolynomial ringover afieldhas a finitegenerating set(a finitebasisin Hilbert's terminology).

In modernalgebra,ringswhose ideals have this property are calledNoetherian rings.Every field, and the ring ofintegersare Noetherian rings. So, the theorem can be generalized and restated as:every polynomial ring over a Noetherian ring is also Noetherian.

The theorem was stated and proved byDavid Hilbertin 1890 in his seminal article oninvariant theory,where he solved several problems on invariants. In this article, he proved also two other fundamental theorems on polynomials, theNullstellensatz(zero-locus theorem) and thesyzygy theorem(theorem on relations). These three theorems were the starting point of the interpretation ofalgebraic geometryin terms ofcommutative algebra.In particular, the basis theorem implies that everyalgebraic setis the intersection of a finite number ofhypersurfaces.

Another aspect of this article had a great impact on mathematics of the 20th century; this is the systematic use ofnon-constructive methods.For example, the basis theorem asserts that every ideal has a finite generator set, but the original proof does not provide any way to compute it for a specific ideal. This approach was so astonishing for mathematicians of that time that the first version of the article was rejected byPaul Gordan,the greatest specialist of invariants of that time, with the comment "This is not mathematics. This is theology."Later, he recognized "I have convinced myself that even theology has its merits."

Statement

If$R$is aring,let$R[X]$denote the ring ofpolynomialsin the indeterminate$X$over$R$.Hilbertproved that if$R$is "not too large", in the sense that if$R$is Noetherian, the same must be true for$R[X]$.Formally,

Hilbert's Basis Theorem.If$R$is a Noetherian ring, then$R[X]$is a Noetherian ring.

Corollary.If$R$is a Noetherian ring, then$R[X_{1},\dotsc,X_{n}]$is a Noetherian ring.

This can be translated intoalgebraic geometryas follows: everyalgebraic setover afieldcan be described as the set of commonrootsof 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 ofrings of invariants.

Hilbert produced an innovativeproof by contradictionusingmathematical induction;his method does not give analgorithmto produce the finitely many basis polynomials for a givenideal:it only shows that they must exist. One can determine basis polynomials using the method ofGröbner bases.

Proof

Theorem.If$R$is a left (resp. right)Noetherian ring,then thepolynomial ring$R[X]$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${\mathfrak {a}}\subseteq R[X]$is a non-finitely generated left ideal. Then by recursion (using theaxiom of dependent choice) there is a sequence of polynomials$\{f_{0},f_{1},\ldots \}$such that if${\mathfrak {b}}_{n}$is the left ideal generated by$f_{0},\ldots,f_{n-1}$then$f_{n}\in {\mathfrak {a}}\setminus {\mathfrak {b}}_{n}$is of minimaldegree.By construction,$\{\deg(f_{0}),\deg(f_{1}),\ldots \}$is a non-decreasing sequence ofnatural numbers.Let$a_{n}$be the leading coefficient of$f_{n}$and let${\mathfrak {b}}$be the left ideal in$R$generated by$a_{0},a_{1},\ldots$.Since$R$is Noetherian the chain of ideals

$(a_{0})\subset (a_{0},a_{1})\subset (a_{0},a_{1},a_{2})\subset \cdots$

must terminate. Thus${\mathfrak {b}}=(a_{0},\ldots,a_{N-1})$for someinteger$N$.So in particular,

$a_{N}=\sum _{i<N}u_{i}a_{i},\qquad u_{i}\in R.$

Now consider

$g=\sum _{i<N}u_{i}X^{\deg(f_{N})-\deg(f_{i})}f_{i},$

whose leading term is equal to that of$f_{N}$;moreover,$g\in {\mathfrak {b}}_{N}$.However,$f_{N}\notin {\mathfrak {b}}_{N}$,which means that$f_{N}-g\in {\mathfrak {a}}\setminus {\mathfrak {b}}_{N}$has degree less than$f_{N}$,contradicting the minimality.

Second proof

Let${\mathfrak {a}}\subseteq R[X]$be a left ideal. Let${\mathfrak {b}}$be the set of leading coefficients of members of${\mathfrak {a}}$.This is obviously a left ideal over$R$,and so is finitely generated by the leading coefficients of finitely many members of${\mathfrak {a}}$;say$f_{0},\ldots,f_{N-1}$.Let$d$be the maximum of the set$\{\deg(f_{0}),\ldots,\deg(f_{N-1})\}$,and let${\mathfrak {b}}_{k}$be the set of leading coefficients of members of${\mathfrak {a}}$,whose degree is$\leq k$.As before, the${\mathfrak {b}}_{k}$are left ideals over$R$,and so are finitely generated by the leading coefficients of finitely many members of${\mathfrak {a}}$,say

$f_{0}^{(k)},\ldots,f_{N^{(k)}-1}^{(k)}$

with degrees$\leq k$.Now let${\mathfrak {a}}^{*}\subseteq R[X]$be the left ideal generated by:

$\left\{f_{i},f_{j}^{(k)}\,:\ i<N,\,j<N^{(k)},\,k<d\right\}\!\!\;.$

We have${\mathfrak {a}}^{*}\subseteq {\mathfrak {a}}$and claim also${\mathfrak {a}}\subseteq {\mathfrak {a}}^{*}$.Suppose for the sake of contradiction this is not so. Then let$h\in {\mathfrak {a}}\setminus {\mathfrak {a}}^{*}$be of minimal degree, and denote its leading coefficient by$a$.

Case 1:$\deg(h)\geq d$.Regardless of this condition, we have$a\in {\mathfrak {b}}$,so$a$is a left linear combination

$a=\sum _{j}u_{j}a_{j}$

of the coefficients of the$f_{j}$.Consider

$h_{0}=\sum _{j}u_{j}X^{\deg(h)-\deg(f_{j})}f_{j},$

which has the same leading term as$h$;moreover$h_{0}\in {\mathfrak {a}}^{*}$while$h\notin {\mathfrak {a}}^{*}$.Therefore$h-h_{0}\in {\mathfrak {a}}\setminus {\mathfrak {a}}^{*}$and$\deg(h-h_{0})<\deg(h)$,which contradicts minimality.

Case 2:$\deg(h)=k<d$.Then$a\in {\mathfrak {b}}_{k}$so$a$is a left linear combination

$a=\sum _{j}u_{j}a_{j}^{(k)}$

of the leading coefficients of the$f_{j}^{(k)}$.Considering

$h_{0}=\sum _{j}u_{j}X^{\deg(h)-\deg(f_{j}^{(k)})}f_{j}^{(k)},$

we yield a similar contradiction as in Case 1.

Thus our claim holds, and${\mathfrak {a}}={\mathfrak {a}}^{*}$which is finitely generated.

Note that the only reason we had to split into two cases was to ensure that the powers of$X$multiplying the factors were non-negative in the constructions.

Applications

Let$R$be a Noetheriancommutative ring.Hilbert's basis theorem has some immediatecorollaries.

1. By induction we see that$R[X_{0},\dotsc,X_{n-1}]$will also be Noetherian.
2. Since anyaffine varietyover$R^{n}$(i.e. a locus-set of a collection of polynomials) may be written as the locus of an ideal${\mathfrak {a}}\subset R[X_{0},\dotsc,X_{n-1}]$and further as the locus of its generators, it follows that every affine variety is the locus of finitely many polynomials — i.e. theintersectionof finitely manyhypersurfaces.
3. If$A$is afinitely-generated$R$-algebra,then we know that$A\simeq R[X_{0},\dotsc,X_{n-1}]/{\mathfrak {a}}$,where${\mathfrak {a}}$is an ideal. The basis theorem implies that${\mathfrak {a}}$must be finitely generated, say${\mathfrak {a}}=(p_{0},\dotsc,p_{N-1})$,i.e.$A$isfinitely presented.

Formal proofs

Formal proofs of Hilbert's basis theorem have been verified through theMizar project(seeHILBASIS file) andLean(seering_theory.polynomial).

Further reading

• Cox, Little, and O'Shea,Ideals, Varieties, and Algorithms,Springer-Verlag, 1997.
• Reid, Constance.(1996).Hilbert.New York:Springer.ISBN0-387-94674-8.The definitive English-language biography of Hilbert.
• Roman, Stephen(2008),Advanced Linear Algebra,Graduate Texts in Mathematics(Third ed.), Springer,ISBN978-0-387-72828-5