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 left-Noetherian, we have that \mathfrak b must be finitely generated; and since the a_n comprise an R-basis, it follows that for a finite amount of them, say a_0,\dotsc,a_{N-1}, will suffice. So for example,

a_N=\sum_{i<N}u_{i}a_{i}, \qquad u_i \in R.

Now consider

g \triangleq \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 {}\le 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^{(k)}_{0}, \cdots, f^{(k)}_{N^{(k)}-1},

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

\left \{f_{i},f^{(k)}_{j} \ : \ 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)\ge 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: h = k < d. Then a\in\mathfrak b_k so is a left-linear combination
a=\sum_j u_j a^{(k)}_j
of the leading coefficients of the f^{(k)}_j. Considering
h_0 \triangleq\sum_j u_j X^{\deg(h)-\deg(f^{(k)}_{j})}f^{(k)}_{j},
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]