Hilbert's basis theorem

Statement

If R is 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 proved the theorem (for the special case of multivariate polynomials over a field) in the course of his proof of finite generation of rings of invariants. The theorem is interpreted in algebraic geometry as follows: every algebraic set is the set of the common zeros of finitely many polynomials.

Hilbert's proof is highly non-constructive: it proceeds by induction on the number of variables, and, at each induction step uses the non-constructive proof for one variable less. Introduced more than eighty years later, Gröbner bases allow a direct proof that is as constructive as possible: Gröbner bases produce an algorithm for testing whether a polynomial belongs to the ideal generated by other polynomials. So, given an infinite sequence of polynomials, one can construct algorithmically the list of those polynomials that do not belong to the ideal generated by the preceding ones. Gröbner basis theory implies that this list is necessarily finite, and is thus a finite basis of the ideal. However, for deciding whether the list is complete, one must consider every element of the infinite sequence, which cannot be done in the finite time allowed to an algorithm.

Proof

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

Suppose \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 { 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 minimal degree. By construction, {\deg(f_0), \deg(f_1), \ldots } is a non-decreasing sequence of natural 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 some integer N. So in particular,

:a_N=\sum_{i

Now consider

:g = \sum_{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 \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}, \ldots, 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

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 a is a left linear combination