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Every ideal of a localisation is generated by the images of any generating set of its contraction

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative and let $λ_{S} ⁣ : R \to S^{- 1} R$ , $r \mapsto r / 1$ , be the localisation map (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions). Let $J$ be an ideal of $S^{- 1} R$ and let

$a : = λ_{S}^{- 1} (J) = {r \in R : r / 1 \in J}$

be its contraction. Then $a$ is an ideal of $R$ , and for every subset $T \subseteq R$ with $a = (T)$ the ideal $J$ is generated in $S^{- 1} R$ by the image $λ_{S} (T) = {t / 1 : t \in T}$ . In particular, if $a = (a_{1}, \dots, a_{n})$ with $n \in N$ then $J = (a_{1} / 1, \dots, a_{n} / 1)$ .

Only the $S$ -saturated half of the ideal correspondence is used. Nothing here says that extension and contraction are mutually inverse on all ideals of $R$ ; that is false in general, and it is the contraction of an ideal of $S^{- 1} R$ , not an arbitrary ideal of $R$ , that this lemma starts from.

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , an ideal $J$ of $S^{- 1} R$ , and a subset $T \subseteq R$ generating the contraction $a = λ_{S}^{- 1} (J)$ .

[L1]

A subset $S \subseteq R$ of a commutative ring is multiplicative if $1 \in S$ and $s, t \in S$ implies $s t \in S$ ; the localisation $S^{- 1} R$ has elements $r / s$ with the displayed arithmetic, and the localisation map is the ring homomorphism $λ_{S} : R \to S^{- 1} R$ , $r \mapsto r / 1$ (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions).

[L2]

For a commutative ring $R$ and a multiplicative $S \subseteq R$ , extension $I \mapsto S^{- 1} I = {r / s : r \in I, s \in S}$ and contraction $J \mapsto λ_{S}^{- 1} (J)$ give inverse inclusion-preserving bijections between $S$ -saturated ideals of $R$ and ideals of $S^{- 1} R$ (Ideals of $S^{- 1} R$ correspond to $S$ -saturated ideals of $R$ , and prime ideals correspond to primes disjoint from $S$ ).

[L3]

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

[L4]

For a subset $T$ of a ring, $(T)$ is the intersection of all two-sided ideals of that ring containing $T$ (The ideal generated by a subset and principal ideals).

Proof

technique · direct

1.1

Fix the data of the Given line. The set $a = λ_{S}^{- 1} (J)$ is the preimage of an ideal under the ring homomorphism $λ_{S}$ .

L1given

2.1

The contraction of an ideal of $S^{- 1} R$ is one of the two directions of the ideal correspondence, so $a$ is an $S$ -saturated ideal of $R$ , and because extension and contraction are mutually inverse on that class of ideals, its extension recovers $J$ : $J = S^{- 1} a = {r / u : r \in a, u \in S}$ .

L2step 1.1

3.1

Every element of $J$ lies in the ideal generated by the image of $T$ . Indeed such an element is $r / u$ with $r \in a = (T)$ and $u \in S$ , and $r$ is then a finite sum $r = \sum_{i = 1}^{k} c_{i} t_{i}$ with $c_{i} \in R$ and $t_{i} \in T$ ; dividing by $u$ gives $r / u = \sum_{i = 1}^{k} (c_{i} / u) (t_{i} / 1)$ , a finite $S^{- 1} R$ -linear combination of elements of the image of $T$ . The empty sum $k = 0$ gives $r = 0$ and $r / u = 0$ , which lies in every ideal.

L3L4step 2.1algebra

3.2

Conversely each $t / 1$ with $t \in T$ lies in $J$ , because $t \in a = λ_{S}^{- 1} (J)$ says exactly that $λ_{S} (t) = t / 1$ belongs to $J$ ; and $J$ is an ideal, so it contains the ideal generated by that image.

L4step 2.1

4.1

The two inclusions of steps 3.1 and 3.2 give $J = (λ_{S} (T))$ . Taking $T = {a_{1}, \dots, a_{n}}$ finite yields $J = (a_{1} / 1, \dots, a_{n} / 1)$ .

L4step 3.1step 3.2∎

Remarks

What the cited correspondence does and does not give. It is a bijection between the $S$ -saturated ideals of $R$ and all ideals of $S^{- 1} R$ (Ideals of $S^{- 1} R$ correspond to $S$ -saturated ideals of $R$ , and prime ideals correspond to primes disjoint from $S$ ). An arbitrary ideal $b$ of $R$ need not be $S$ -saturated, and then $λ_{S}^{- 1} (S^{- 1} b)$ is strictly larger than $b$ . The proof above never applies the correspondence to such a $b$ : it starts from an ideal of $S^{- 1} R$ and contracts.
The generating set is not required to be finite. The argument uses only that each element of $a$ is a finite $R$ -linear combination of elements of $T$ ; the finite case is stated separately because it is the one the Noetherian application needs.
The degenerate localisation is covered. If $0 \in S$ then $S^{- 1} R$ is the zero ring, its only ideal is $0$ , and the contraction is all of $R$ ; the conclusion holds because every subset of the zero ring generates its only ideal.

Depends on

Used by

Every quotient and every localisation of a Noetherian ring is Noetherian Theorem

Dependency tree · two levels

13 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

B. Totaro, Commutative Algebra (Michaelmas 2011), notes by Z. Norwood, §8 (standard reference, not scraped)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., (16.7) (standard reference, not scraped)