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. Template:Harvs 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 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

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: $h=k<d$. 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

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.
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