Английская Википедия:Hilbert's basis theorem
Шаблон:Short description Шаблон:Use shortened footnotes In mathematics, specifically commutative algebra, Hilbert's basis theorem says that a polynomial ring over a Noetherian ring is Noetherian.
Statement
If <math>R</math> is a ring, let <math>R[X]</math> denote the ring of polynomials in the indeterminate <math>X</math> over <math>R</math>. Hilbert proved that if <math>R</math> is "not too large", in the sense that if <math>R</math> is Noetherian, the same must be true for <math>R[X]</math>. Formally,
Hilbert's Basis Theorem. If <math>R</math> is a Noetherian ring, then <math>R[X]</math> is a Noetherian ring.[1]
Corollary. If <math>R</math> is a Noetherian ring, then <math>R[X_1,\dotsc,X_n]</math> 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.Шаблон:R
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 <math>R</math> is a left (resp. right) Noetherian ring, then the polynomial ring <math>R[X]</math> 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 <math>\mathfrak a \subseteq R[X]</math> is a non-finitely generated left ideal. Then by recursion (using the axiom of dependent choice) there is a sequence of polynomials <math>\{ f_0, f_1, \ldots \}</math> such that if <math>\mathfrak b_n</math> is the left ideal generated by <math>f_0, \ldots, f_{n-1}</math> then <math>f_n \in \mathfrak a \setminus \mathfrak b_n</math> is of minimal degree. By construction, <math>\{\deg(f_0), \deg(f_1), \ldots \}</math> is a non-decreasing sequence of natural numbers. Let <math>a_n</math> be the leading coefficient of <math>f_n</math> and let <math>\mathfrak{b}</math> be the left ideal in <math>R</math> generated by <math>a_0,a_1,\ldots</math>. Since <math>R</math> is Noetherian the chain of ideals
- <math>(a_0)\subset(a_0,a_1)\subset(a_0,a_1,a_2) \subset \cdots</math>
must terminate. Thus <math>\mathfrak b = (a_0,\ldots ,a_{N-1})</math> for some integer <math>N</math>. So in particular,
- <math>a_N=\sum_{i<N} u_{i}a_{i}, \qquad u_i \in R.</math>
Now consider
- <math>g = \sum_{i<N}u_{i}X^{\deg(f_{N})-\deg(f_{i})}f_{i},</math>
whose leading term is equal to that of <math>f_N</math>; moreover, <math>g\in\mathfrak b_N</math>. However, <math>f_N \notin \mathfrak b_N</math>, which means that <math>f_N - g \in \mathfrak a \setminus \mathfrak b_N</math> has degree less than <math>f_N</math>, contradicting the minimality.
Second proof
Let <math>\mathfrak a \subseteq R[X]</math> be a left ideal. Let <math>\mathfrak b</math> be the set of leading coefficients of members of <math>\mathfrak a</math>. This is obviously a left ideal over <math>R</math>, and so is finitely generated by the leading coefficients of finitely many members of <math>\mathfrak a</math>; say <math>f_0, \ldots, f_{N-1}</math>. Let <math>d</math> be the maximum of the set <math>\{\deg(f_0),\ldots, \deg(f_{N-1})\}</math>, and let <math>\mathfrak b_k</math> be the set of leading coefficients of members of <math>\mathfrak a</math>, whose degree is <math>\le k</math>. As before, the <math>\mathfrak b_k</math> are left ideals over <math>R</math>, and so are finitely generated by the leading coefficients of finitely many members of <math>\mathfrak a</math>, say
- <math>f^{(k)}_{0}, \ldots, f^{(k)}_{N^{(k)}-1}</math>
with degrees <math>\le k</math>. Now let <math>\mathfrak a^*\subseteq R[X]</math> be the left ideal generated by:
- <math>\left\{f_{i},f^{(k)}_{j} \, : \ i<N,\, j<N^{(k)},\, k<d \right\}\!\!\;.</math>
We have <math>\mathfrak a^*\subseteq\mathfrak a</math> and claim also <math>\mathfrak a\subseteq\mathfrak a^*</math>. Suppose for the sake of contradiction this is not so. Then let <math>h\in \mathfrak a \setminus \mathfrak a^*</math> be of minimal degree, and denote its leading coefficient by <math>a</math>.
- Case 1: <math>\deg(h)\ge d</math>. Regardless of this condition, we have <math>a\in \mathfrak b</math>, so <math>a</math> is a left linear combination
- <math>a=\sum_j u_j a_j</math>
- of the coefficients of the <math>f_j</math>. Consider
- <math>h_0 =\sum_{j}u_{j}X^{\deg(h)-\deg(f_{j})}f_{j},</math>
- which has the same leading term as <math>h</math>; moreover <math>h_0 \in \mathfrak a^*</math> while <math>h\notin\mathfrak a^*</math>. Therefore <math>h - h_0 \in \mathfrak a\setminus\mathfrak a^*</math> and <math>\deg(h - h_0) < \deg(h)</math>, which contradicts minimality.
- Case 2: <math>\deg(h) = k < d</math>. Then <math>a\in\mathfrak b_k</math> so <math>a</math> is a left linear combination
- <math>a=\sum_j u_j a^{(k)}_j</math>
- of the leading coefficients of the <math>f^{(k)}_j</math>. Considering
- <math>h_0=\sum_j u_j X^{\deg(h)-\deg(f^{(k)}_{j})}f^{(k)}_{j},</math>
- we yield a similar contradiction as in Case 1.
Thus our claim holds, and <math>\mathfrak a = \mathfrak a^*</math> which is finitely generated.
Note that the only reason we had to split into two cases was to ensure that the powers of <math>X</math> multiplying the factors were non-negative in the constructions.
Applications
Let <math>R</math> be a Noetherian commutative ring. Hilbert's basis theorem has some immediate corollaries.
- By induction we see that <math>R[X_0,\dotsc,X_{n-1}]</math> will also be Noetherian.
- Since any affine variety over <math>R^n</math> (i.e. a locus-set of a collection of polynomials) may be written as the locus of an ideal <math>\mathfrak a\subset R[X_0, \dotsc, X_{n-1}]</math> 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.
- If <math>A</math> is a finitely-generated <math>R</math>-algebra, then we know that <math>A \simeq R[X_0, \dotsc, X_{n-1}] / \mathfrak a</math>, where <math>\mathfrak a</math> is an ideal. The basis theorem implies that <math>\mathfrak a</math> must be finitely generated, say <math>\mathfrak a = (p_0,\dotsc, p_{N-1})</math>, i.e. <math>A</math> 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.
