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 ${\displaystyle R}$ is a ring, let ${\displaystyle R[X]}$ denote the ring of polynomials in the indeterminate ${\displaystyle X}$ over ${\displaystyle R}$. Hilbert proved that if ${\displaystyle R}$ is "not too large", in the sense that if ${\displaystyle R}$ is Noetherian, the same must be true for ${\displaystyle R[X]}$. Formally,

Hilbert's Basis Theorem. If ${\displaystyle R}$ is a Noetherian ring, then ${\displaystyle R[X]}$ is a Noetherian ring.^[1]

Corollary. If ${\displaystyle R}$ is a Noetherian ring, then ${\displaystyle R[X_{1},\dotsc ,X_{n}]}$ 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 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.^[2]

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

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

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

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

Now consider

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

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

Second proof

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

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

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

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

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

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

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

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

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

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

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

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

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

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

we yield a similar contradiction as in Case 1.

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

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

Applications

Let ${\displaystyle R}$ be a Noetherian commutative ring. Hilbert's basis theorem has some immediate corollaries.

1. By induction we see that ${\displaystyle R[X_{0},\dotsc ,X_{n-1}]}$ will also be Noetherian.
2. Since any affine variety over ${\displaystyle R^{n}}$ (i.e. a locus-set of a collection of polynomials) may be written as the locus of an ideal ${\displaystyle {\mathfrak {a}}\subset R[X_{0},\dotsc ,X_{n-1}]}$ 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 ${\displaystyle A}$ is a finitely-generated ${\displaystyle R}$-algebra, then we know that ${\displaystyle A\simeq R[X_{0},\dotsc ,X_{n-1}]/{\mathfrak {a}}}$, where ${\displaystyle {\mathfrak {a}}}$ is an ideal. The basis theorem implies that ${\displaystyle {\mathfrak {a}}}$ must be finitely generated, say ${\displaystyle {\mathfrak {a}}=(p_{0},\dotsc ,p_{N-1})}$, i.e. ${\displaystyle A}$ is finitely presented.

Formal proofs

Formal proofs of Hilbert's basis theorem have been verified through the Mizar project (see HILBASIS file) and  Lean (see ring_theory.polynomial).

References

Further reading

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

• Roman, Stephen (2008), Advanced Linear Algebra, Graduate Texts in Mathematics (Third ed.), Springer, ISBN 978-0-387-72828-5

This page was last edited on 6 November 2023, at 13:57