Noetherian polynomial rings and finite-type algebras

Written by GPT-6 Astra (OpenAI). Self-checked by the writing AI. Original exposition: CC0.

An ideal may contain infinitely many equations while being generated by finitely many of them. We prove that this finiteness survives adjoining finitely many variables, imposing relations and inverting elements. Rings in this reading are commutative and have a unit; the zero ring is allowed. Induction on natural numbers and ordinary set-theoretic choice are our foundational conventions.

Theorem. If every ideal of $R$ has a finite generating set, every ideal of $R[X]$ , of every finite-type $R$ -algebra and of every localization $S^{-1}R$ has a finite generating set. In particular this holds in $\mathbb C[X_1,\ldots,X_n]$ for every finite $n$ .

1. Polynomial, quotient and fraction constructions

A polynomial is a finitely supported sequence of coefficients. Addition adds its entries and multiplication convolves them. Associativity follows by expanding a coefficient as the same finite sum over triples of indices; the other ring identities follow in the same way. Substitution into commuting ring elements respects these operations. In a domain, the top coefficient of a product of two nonzero polynomials is nonzero. Its degree is therefore the sum of their degrees. Applying this one variable at a time proves that a polynomial ring in finitely many variables over a domain is again a domain.

An ideal $I$ is an additive subgroup stable under multiplication by all ring elements. On the additive cosets modulo $I$ , define multiplication by $[a][b]=[ab]$ . Changing $a$ to $a+i$ and $b$ to $b+j$ changes their product by $aj+bi+ij\in I$ , so this is well-defined. The cosets form $R/I$ . For any homomorphism, its kernel is an ideal. If the homomorphism is surjective, sending the coset of $a$ to its image gives a bijective ring homomorphism from the quotient by its kernel: the equality of two images is exactly the assertion that their difference is in the kernel. A proper ideal $I$ is prime precisely when $R/I$ has no zero divisors and is nonzero, since $[a][b]=0$ means $ab\in I$ .

Let $S\subset R$ contain $1$ and be closed under multiplication. To invert its elements, start with pairs $(a,s)\in R\times S$ and declare $\begin{gathered} (a,s)\sim(b,t)\\ \Longleftrightarrow\quad u(at-bs)=0\\ \text{for some }u\in S. \end{gathered} \tag{NR1}$ The extra multiplier permits zero divisors in $R$ . The relation is reflexive and symmetric. For transitivity, suppose witnesses $u,v$ annihilate $at-bs$ and $br-ct$ . The identity $t(ar-cs)=r(at-bs)+s(br-ct)$ shows that $uvt\in S$ is a witness relating $(a,s)$ and $(c,r)$ .

Denote a class by $a/s$ . Its operations are $\begin{gathered} a/s+b/t=(at+bs)/(st),\\ (a/s)(b/t)=ab/(st). \end{gathered} \tag{NR2}$ For a replacement $(a,s)\sim(a',s')$ , the cross-product differences after addition of $b/t$ and multiplication by $b/t$ are respectively $t^2(as'-a's)$ and $bt(as'-a's)$ . The original witness annihilates them. Replacing the second argument separately proves well-definedness in both arguments; negation is checked by the same equality. A common denominator reduces every ring identity to an identity in $R$ . The zero and unit are $0/1$ and $1/1$ . The map $a\mapsto a/1$ is a homomorphism, and $s/1$ is invertible with inverse $1/s$ . This constructs $S^{-1}R$ . If $0\in S$ , zero itself is a witness for every pair, so the construction yields the zero ring.

When $R$ is a nonzero domain and $S=R\setminus\{0\}$ , relation (NR1) simplifies to $at=bs$ . In particular $R\to S^{-1}R$ is injective. A nonzero fraction has nonzero numerator and inverse $s/a$ . Thus this localization is a field. Every field containing $R$ contains the same fractions, with the same equality and arithmetic, so it is the fraction field of $R$ .

2. Finite generation, chains and two indices

For $a_1,\ldots,a_m\in R$ , the ideal they generate consists of all sums $\sum r_ja_j$ . The empty list generates the zero ideal. Call $R$ Noetherian when each ideal has such a finite list.

The union of an ascending sequence of ideals is an ideal: any two elements already occur together in some member of the sequence, where closure under addition and ring multiplication holds. If that union has finitely many generators, one member contains all of them, so every later member equals the union. Conversely, suppose an ideal has no finite generating list. Starting with the zero ideal, repeatedly select an element of the given ideal outside the ideal generated by the elements selected so far. Each selection is possible by the supposition and strictly enlarges the generated ideal. This contradicts stabilization of ascending sequences. Hence finite generation of all ideals is equivalent to the ascending chain condition.

We will also use two facts about a family $I_{i,d}$ of ideals increasing with each index in $\mathbb N$ .

First, in a Noetherian ring such a family has only finitely many different values. To prove this, every infinite sequence of natural numbers has an infinite nondecreasing subsequence. If a value repeats infinitely, choose those positions. If no value does, the positions with values bounded by any fixed number are finite; choose successive later positions with strictly larger values. Apply this observation first to one coordinate of an infinite sequence of pairs and then to the other coordinate of the resulting subsequence. If our family had infinitely many values, select one index pair for each of countably many distinct values. The subsequence just constructed would give an ascending sequence of pairwise different ideals, a contradiction.

Second, there is a single index $i_0$ such that $I_{i,d}=I_{i_0,d} \quad(i\ge i_0,\ d\ge0). \tag{NR3}$ Here is a direct argument that also gives uniformity in $d$ . The diagonal chain stabilizes at some $I_{K,K}$ . For $i,d\ge K$ , monotonicity places $I_{i,d}$ between $I_{K,K}$ and $I_{\max(i,d),\max(i,d)}$ , which are equal. For each of the finitely many $d<K$ , choose a stabilization index for the chain in $i$ . Take $i_0$ at least $K$ and at least all those indices. For $d<K$ equality (NR3) follows by this choice; for $d\ge K$ it follows from the diagonal argument. Neither argument assumes the ring is a domain.

3. Constructing a finite list of polynomial generators

We first prove a finite-module fact directly. A submodule of $R^m$ means an additive subgroup closed under coordinatewise scalar multiplication. If $R$ is Noetherian, every such submodule has finitely many module generators. Induct on $m$ . For $m=0$ there is only the zero submodule. For $m>0$ , the last-coordinate image of a submodule $M$ is an ideal of $R$ . Choose finitely many generators of that ideal and lift each to a vector of $M$ . The vectors of $M$ with last coordinate zero form a submodule of $R^{m-1}$ , which has finitely many generators by induction. Given any vector of $M$ , subtract a linear combination of the chosen lifts to make its last coordinate zero, then express the remainder using the kernel generators. These two finite lists generate $M$ .

Now fix an ideal $J\subset R[X]$ . The leading coefficients of its nonzero elements, together with zero, form an ideal $L\subset R$ . For addition of two such coefficients, multiply their polynomials by powers of $X$ to give them the same degree and then add. If the coefficient sum vanishes it is the allowed zero; otherwise it is the leading coefficient of this sum. Scalar multiplication is checked on the scalar multiple of a polynomial, again allowing a zero top coefficient. This proves the assertion about $L$ , including rings with zero divisors.

If $L=0$ , no nonzero polynomial belongs to $J$ , and there is nothing to generate. Otherwise choose nonzero generators $a_1,\ldots,a_t$ of $L$ and polynomials $f_j\in J$ with leading coefficients $a_j$ . Write $d_j=\deg f_j$ and $D=\max_j d_j$ . For any $f\in J$ of degree $n\ge D$ , express its leading coefficient as $\sum r_ja_j$ and subtract $\sum_{j=1}^t r_jX^{n-d_j}f_j. \tag{NR4}$ The result has degree strictly below $n$ . Repeating terminates with a polynomial of degree below $D$ .

The coefficient vectors of the elements of $J$ of degree below $D$ form an $R$ -submodule of $R^D$ . By the proved module fact, choose finitely many polynomials $g_1,\ldots,g_v$ whose vectors generate it. The reductions by (NR4), followed by the expression in these $g_l$ , show that $J=(f_1,\ldots,f_t,g_1,\ldots,g_v)$ . For $D=0$ , the low-degree module is $R^0=0$ , and the second list is empty. This proves the polynomial-ring theorem by constructing a finite generating set.

4. A chain proof of the same polynomial theorem

There is also a useful degree-filtration proof. Let $J_0\subset J_1\subset\cdots$ be ideals of $R[X]$ , and let $I_{i,d}$ be the set of coefficients of $X^d$ in elements of $J_i$ of degree at most $d$ . It is an ideal because addition and scalar multiplication preserve that degree bound. Multiplication of a polynomial by $X$ proves monotonicity in $d$ , and inclusion of the $J_i$ proves monotonicity in $i$ .

Choose $i_0$ from (NR3). If $f\in J_i$ for $i\ge i_0$ , prove $f\in J_{i_0}$ by induction on its degree. The zero polynomial already belongs there. If $f$ is nonzero of degree $d$ , its top coefficient belongs to $I_{i_0,d}$ . By definition there is $g\in J_{i_0}$ of degree at most $d$ with that coefficient. Thus $f-g$ has lower degree, and the induction hypothesis puts it in $J_{i_0}$ . Adding $g$ proves the claim. Consequently the polynomial-ideal chain stabilizes. The equivalence in Section 2 gives the same Noetherian conclusion as Section 3.

5. Imposing relations and inverting elements

Let $R$ be Noetherian. For an ideal $J$ in $R/I$ , its inverse image under the quotient map is an ideal of $R$ . A finite generating set for that inverse image maps to a generating set for $J$ : lift an element of $J$ , express its lift as a combination of generators, and apply the quotient map. Hence $R/I$ is Noetherian.

A finite-type $R$ -algebra generated by $b_1,\ldots,b_n$ is the image of the evaluation homomorphism from $R[X_1,\ldots,X_n]$ . Repeated application of the one-variable theorem makes this polynomial ring Noetherian, and the quotient argument applies to the evaluation kernel. This proves finite-type permanence even when the structure map from $R$ has a kernel.

For an ideal $J$ of $S^{-1}R$ , define $I=\{a\in R:a/1\in J\}$ . This is an ideal of $R$ . Membership $a/s\in J$ implies $a/1=(s/1)(a/s)\in J$ . Conversely $a\in I$ implies $a/s=(1/s)(a/1)\in J$ . Choose generators $a_j$ of $I$ . If $a=\sum r_ja_j$ , then $a/s=\sum(r_j/s)(a_j/1)$ , so the finitely many $a_j/1$ generate $J$ . The argument also applies when $0\in S$ , since the localization is then the zero ring. We have proved the theorem in all the stated cases.