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The polynomial ring in countably many variables is not Noetherian

Example

Let $k$ be a field and, for $m \in N$ , write $R_{m} : = k [x_{1}, \dots, x_{m}]$ for the iterated polynomial ring, so that $R_{0} = k$ and $R_{m + 1} = R_{m} [x_{m + 1}]$ (Polynomial rings in finitely many commuting indeterminates by iteration). Identify $R_{m}$ with the subring of constant polynomials in $R_{m + 1}$ , which is the identification the published iterative definition already makes and which is legitimate because the constant-polynomial map is an injective unital ring homomorphism (Polynomial convolution makes $R [x]$ a commutative ring containing $R$ as its constant subring). Under it

$R_{0} \subseteq R_{1} \subseteq R_{2} \subseteq \dots,$

and the union

$R_{\infty} : = ⋃_{m \in N} R_{m}$

carries well-defined operations, making it a commutative ring: the polynomial ring in the countably many indeterminates $x_{1}, x_{2}, \dots$ over $k$ .

Then $R_{\infty}$ is not Noetherian. Writing $I_{m} : = (x_{1}, \dots, x_{m})$ for the ideal of $R_{\infty}$ generated by the first $m$ indeterminates, with $I_{0} = (\emptyset) = 0$ , the chain

$I_{0} ⊊ I_{1} ⊊ I_{2} ⊊ \dots$

is strictly ascending and therefore never stabilises.

Facts & Assumptions

Given: A field $k$ , the rings $R_{m} = k [x_{1}, \dots, x_{m}]$ for $m \in N$ , and their union $R_{\infty}$ .

[L1]

Polynomial rings in finitely many commuting indeterminates are defined by $R [x_{1}, \dots, x_{0}] : = R$ and $R [x_{1}, \dots, x_{n + 1}] : = R [x_{1}, \dots, x_{n}] [x_{n + 1}]$ ; at each stage the coefficient ring embeds as the constant polynomials, so all preceding indeterminates remain present (Polynomial rings in finitely many commuting indeterminates by iteration).

[L2]

For every commutative ring $R$ the polynomial ring $R [x]$ is a commutative ring, and the constant-polynomial map $R \to R [x]$ is an injective unital ring homomorphism (Polynomial convolution makes $R [x]$ a commutative ring containing $R$ as its constant subring).

[L3]

A subset $S$ of a ring is a subring when $1 \in S$ and $S$ is closed under addition, additive inverses and multiplication; it is then a ring with the same zero and identity (Subring: a subset containing $1_{R}$ and closed under addition, additive inverses and multiplication).

[L4]

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).

[L5]

In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and the empty sum is included and equals $0$ (In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$ ).

[L6]

For commutative rings $R, S$ , a unital ring homomorphism $φ ⁣ : R \to S$ and $s \in S$ , there is a unique unital ring homomorphism $R [x] \to S$ extending $φ$ on constants and sending $x$ to $s$ (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[L7]

For a commutative ring, being Noetherian is equivalent to every ascending chain of ideals indexed by $N$ stabilising (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).

Verification

technique · direct

1.1L1L2L3given

Under the identification of $R_{m}$ with the constants of $R_{m + 1}$ , each $R_{m}$ is a subring of $R_{m + 1}$ , so the family is increasing and any two of its members are comparable.

2.1L2L3step 1.1

The union $R_{\infty}$ is a commutative ring. Given $f, g \in R_{\infty}$ there is $m$ with $f, g \in R_{m}$ , by comparability of the two stages containing them; define $f + g$ and $f g$ there. The value does not depend on the stage chosen, because a larger stage contains the smaller as a subring and the operations of a subring are the restrictions of the ambient ones. Each ring axiom involves finitely many elements, which again lie in a common $R_{m}$ , where the axiom holds. The identity is $1_{k}$ and the zero is $0_{k}$ .

3.1L4L5step 2.1

For $m \in N$ let $I_{m} : = (x_{1}, \dots, x_{m})$ be the ideal of $R_{\infty}$ generated by $x_{1}, \dots, x_{m}$ ; at $m = 0$ the generating set is empty and $I_{0} = 0$ . Since the generating sets increase with $m$ , so do the ideals: $I_{m} \subseteq I_{m + 1}$ .

4.1L1L4L6step 1.1step 3.1

The inclusions are strict, because $x_{m + 1} \notin I_{m}$ . Suppose $x_{m + 1} = \sum_{i = 1}^{m} f_{i} x_{i}$ with $f_{i} \in R_{\infty}$ ; all the $f_{i}$ lie in a common $R_{N}$ with $N \geq m + 1$ , so the equation holds in $R_{N} = k [x_{1}, \dots, x_{N}]$ . Iterating the one-variable universal property along the tower defining $R_{N}$ produces a $k$ -algebra homomorphism $θ ⁣ : R_{N} \to k [T]$ with $θ (x_{m + 1}) = T$ and $θ (x_{j}) = 0$ for every $j \leq N$ with $j \neq m + 1$ . Applying $θ$ to the supposed equation gives $T = \sum_{i = 1}^{m} θ (f_{i}) \cdot 0 = 0$ in $k [T]$ , which is false. So $x_{m + 1} \in I_{m + 1} ∖ I_{m}$ and $I_{m} ⊊ I_{m + 1}$ .

5.1L7step 3.1step 4.1∎

The chain $I_{0} \subseteq I_{1} \subseteq \dots$ is an ascending chain of ideals of $R_{\infty}$ indexed by $N$ in which every inclusion is strict, so no index $N$ has $I_{n} = I_{N}$ for all $n \geq N$ : already $I_{N + 1} \neq I_{N}$ . The ascending chain condition therefore fails, and $R_{\infty}$ is not Noetherian.

Remarks

Every finite stage is Noetherian and the union is not. Each $R_{m}$ is Noetherian by If $R$ is Noetherian then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ ; the Noetherian condition is not preserved by unions of increasing chains of rings, and this is the standard witness for that.
The identification of $R_{m}$ with the constants of $R_{m + 1}$ is the one the published definition makes. Polynomial rings in finitely many commuting indeterminates by iteration states that the coefficient ring embeds at each stage and that all preceding indeterminates remain present; the union above is taken along exactly those embeddings, and the injectivity that makes the identification harmless is Polynomial convolution makes $R [x]$ a commutative ring containing $R$ as its constant subring.
$T$ is needed in the substitution. Sending every indeterminate to $0$ would kill $x_{m + 1}$ as well and prove nothing; the target $k [T]$ is what keeps one indeterminate alive while all the others vanish.

Depends on

Used by

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Sources

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., (16.1) (standard reference, not scraped)
B. Totaro, Commutative Algebra (Michaelmas 2011), notes by Z. Norwood, §8 (standard reference, not scraped)