Hilbert's basis theorem

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

Statement[edit]

If R a ring, let R[X] denote the ring of polynomials in the indeterminate X over R. Hilbert proved 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 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 (1890) 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.

Proof[edit]

Theorem. If R is a left (resp. right) Noetherian ring, then the polynomial 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[edit]

Suppose {\mathfrak {a}}\subseteq R[X] were a non-finitely generated left-ideal. Then by recursion (using the axiom of dependent choice) there is a sequence \{f_{0},f_{1},\dotsc \} of polynomials such that if {\mathfrak {b}}_{n} is the left ideal generated by f_{0},\dotsc ,f_{n-1} then f_{n} in {\mathfrak {a}}\setminus {\mathfrak {b}}_{n} is of minimal degree. It is clear that \{\deg(f_{0}),\deg(f_{1}),\dotsc \} is a non-decreasing sequence of naturals. 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},\dotsc . Since R is Noetherian the chain of ideals (a_{0})\subset (a_{0},a_{1})\subset (a_{0},a_{1},a_{2})\dotsc must terminate. Thus {\mathfrak {b}}=(a_{0},\dotsc ,a_{N-1}) for some integer 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[edit]

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},\dotsc ,f_{N-1}. Let d be the maximum of the set \{\deg(f_{0}),\dotsc ,\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)},\cdots ,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 is a left-linear combination
a=\sum _{j}u_{j}a_{j}
of the coefficients of the f_{j}. Consider
h_{0}\triangleq \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 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}\triangleq \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[edit]

Let R be a Noetherian commutative ring. Hilbert's basis theorem has some immediate corollaries.

  1. By induction we see that R[X_{0},\dotsc ,X_{n-1}] will also be Noetherian.
  2. Since any affine variety over 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. the intersection of finitely many hypersurfaces.
  3. If A is a finitely-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 is finitely presented.

Mizar System[edit]

The Mizar project has completely formalized and automatically checked a proof of Hilbert's basis theorem in the HILBASIS file.

References[edit]