Hilbert's basis theorem

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

Statement

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

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 $\mathfrak a \subseteq R[X]$ 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

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: $\deg(h) = k < d$. 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

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

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

References

• Cox, Little, and O'Shea, Ideals, Varieties, and Algorithms, Springer-Verlag, 1997.