False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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False statement: in a Noetherian ring there is a single bound on the number of generators an ideal needs

Statement

False claim. For every Noetherian commutative ring $R$ there is an $N \in N$ such that every ideal of $R$ can be generated by at most $N$ elements.

The Noetherian condition bounds the number of generators of each ideal separately (A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member) and asserts nothing uniform across the ideals of the ring.

Facts & Assumptions

Given: A field $k$ and the polynomial ring $k [x, y] = k [x] [y]$ (Polynomial rings in finitely many commuting indeterminates by iteration), each of whose elements is uniquely a $k$ -linear combination of the monomials $x^{a} y^{b}$ , $a, b \in N$ , with finitely many nonzero coefficients. For an ideal $I$ its powers are $I^{0} : = k [x, y]$ and $I^{n + 1} : = I^{n} I$ , where $I J$ is the ideal of finite sums $\sum_{l} i_{l} j_{l}$ with $i_{l} \in I$ and $j_{l} \in J$ (The sum $I + J$ and product $I J$ of two-sided ideals). For $f \in k [x, y]$ and $n \in N$ , write $f^{(n)}$ for the sum of the terms of $f$ whose monomials $x^{a} y^{b}$ have $a + b = n$ .

[L1]

For $S \subseteq R$ , $(S)$ is the intersection of all two-sided ideals containing $S$ , so $S \subseteq (S)$ (The ideal generated by a subset and principal ideals).

[L2]

In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$ (In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$ ).

[L3]

$R [x]$ is the set of finitely supported functions $N \to R$ , with coefficientwise addition and the convolution product $(a b)_{i} = \sum_{j + k = i} a_{j} b_{k}$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

[L4]

A vector space over a field $F$ is a set $V$ with an addition making $(V, +, 0_{V})$ an abelian group and a scalar multiplication $F \times V \to V$ satisfying $λ (u + v) = λ u + λ v$ , $(λ + μ) v = λ v + μ v$ , $(λ μ) v = λ (μ v)$ and $1_{F} v = v$ (Vector space over a field).

[L5]

The span of a subset of a vector space is the set of finite linear combinations of its elements, and a subset spans $V$ when its span is $V$ (Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ).

[L6]

A subset $S \subseteq V$ is linearly independent when every injective finite list $v ⁣ : n \to S$ satisfies: $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ implies $λ_{i} = 0_{F}$ for every $i < n$ (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent).

[L7]

If a vector space $V$ over a field $F$ has a spanning subset $S$ with $S \approx n$ for some $n \in N$ , then every linearly independent subset $L \subseteq V$ is finite and the unique $m \in N$ with $L \approx m$ satisfies $m \leq n$ (If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $N$ ).

[L8]

If $R$ is a Noetherian commutative ring then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ (If $R$ is Noetherian then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ ).

[L9]

Every field is a Noetherian ring (Fields and $Z$ are Noetherian, and so are their polynomial rings in finitely many variables).

Refutation

technique · direct

1.1

The ring $k [x, y]$ is Noetherian, being a polynomial ring in finitely many variables over the Noetherian ring $k$ .

L8L9given

1.2

Fix $n \in N$ and let $V_{n}$ be the set of $k$ -linear combinations of the $n + 1$ monomials $x^{i} y^{n - i}$ for $0 \leq i \leq n$ . It is a $k$ -vector space: it contains $0$ , is closed under addition and under multiplication by elements of $k$ read as constant polynomials, and the vector-space axioms are the corresponding identities in $k [x, y]$ . The monomials $x^{i} y^{n - i}$ , $0 \leq i \leq n$ , are distinct and form a linearly independent subset of $V_{n}$ , since a polynomial is a finitely supported family of coefficients and so vanishes exactly when all of its coefficients do; and they span $V_{n}$ by construction.

L3L4L5L6given

2.1

Two facts about the ideal $m^{n}$ , where $m = (x, y)$ . First, $x^{i} y^{n - i} \in m^{n}$ for $0 \leq i \leq n$ : at $n = 0$ the ideal is all of $k [x, y]$ , and if a product of $n$ factors from ${x, y}$ lies in $m^{n}$ then multiplying it by $x$ or by $y$ lands in $m^{n} m = m^{n + 1}$ . Second, every element of $m^{n}$ has all its monomials of total degree at least $n$ : this holds vacuously at $n = 0$ ; every element of $m$ is $a x + b y$ and so has all monomials of degree at least $1$ ; and an element of $m^{n + 1}$ is a finite sum of products $u v$ with $u \in m^{n}$ and $v \in m$ , whose monomials have degree at least $n + 1$ because degrees add on monomials.

L1L2L3step 1.2

3.1

Suppose $m^{n} = (g_{1}, \dots, g_{m})$ with $m \in N$ . By step 2.1 every monomial of every $g_{j}$ has degree at least $n$ , so $g_{j} = g_{j}^{(n)} + p_{j}$ with $g_{j}^{(n)} \in V_{n}$ and every monomial of $p_{j}$ of degree at least $n + 1$ . Fix $i$ with $0 \leq i \leq n$ and write $x^{i} y^{n - i} = \sum_{j} h_{j} g_{j}$ with $h_{j} \in k [x, y]$ . Splitting each $h_{j}$ into its constant term $λ_{j} \in k$ and a remainder whose monomials have degree at least $1$ , every contribution to $h_{j} g_{j}$ other than $λ_{j} g_{j}^{(n)}$ has all monomials of degree at least $n + 1$ . Comparing the coefficients of the monomials of degree exactly $n$ on both sides gives $x^{i} y^{n - i} = \sum_{j} λ_{j} g_{j}^{(n)}$ .

L2L3step 1.2step 2.1

4.1

So the set ${g_{1}^{(n)}, \dots, g_{m}^{(n)}}$ , which has at most $m$ elements and lies in $V_{n}$ , spans a subset of $V_{n}$ containing all $n + 1$ monomials $x^{i} y^{n - i}$ , hence spans $V_{n}$ . Those monomials form a linearly independent subset of $V_{n}$ with exactly $n + 1$ elements, so $n + 1 \leq m$ . Every generating list of $m^{n}$ therefore has at least $n + 1$ members.

L5L6L7step 1.2step 3.1

5.1

Now suppose the claim held for $R = k [x, y]$ , which is Noetherian by step 1.1, with bound $N$ . Taking $n = N$ in step 4.1, the ideal $(x, y)^{N}$ needs at least $N + 1$ generators, so it cannot be generated by at most $N$ elements. The claim is false.

step 1.1step 4.1∎

Remarks

What the Noetherian condition does say. Each ideal has some finite generating list; the length of that list is allowed to depend on the ideal, and here it does, without bound.
The bound is sharp in this example and is not claimed to be sharp in general. The ideal $(x, y)^{n}$ is generated by the $n + 1$ monomials of degree $n$ , so exactly $n + 1$ generators suffice; the refutation needs only the lower bound.
A uniform bound does exist in some Noetherian rings. In a principal ideal domain every ideal is generated by one element (Principal ideal domain), so the false claim is not false for lack of any instance; it is false because it is asserted for every Noetherian ring.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

45 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, A Primer of Commutative Algebra, v4.03, §3 (Aside 3.18) (standard reference, not scraped)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., §16 (standard reference, not scraped)