Commutative Algebra 24

Quotient vs Localization

Taking the quotient A \mapsto A/\mathfrak a and localization A \mapsto S^{-1}A are two sides of the same coin when we look at \mathrm{Spec A}.

  • Quotient removes the “small” prime ideals in \mathrm{Spec A} – it only keeps the prime ideals containing \mathfrak a.
  • Localization removes the “large” prime ideals in \mathrm{Spec A} – it only keeps prime ideals disjoint from S.

It turns out these two operations commute.

Proposition 1.

Let \mathfrak a\subseteq A be an ideal and S\subseteq A a multiplicative subset.

  • Let T = \{\overline s \in A/\mathfrak a : s \in S\}, a multiplicative subset of A/\mathfrak a.
  • Let \mathfrak b = \mathfrak a (S^{-1}A), an ideal of S^{-1}A.

Then we have an isomorphism

(S^{-1}A) / \mathfrak b \stackrel\cong\longrightarrow T^{-1}(A/\mathfrak a), \quad \overline{a/s} \leftrightarrow \overline a/\overline s.

Proof

We will use the universal properties here.

The canonical homomorphism \phi : A\to S^{-1}A gives us \phi (\mathfrak a) \subseteq \mathfrak a (S^{-1}A) = \mathfrak b. Thus we get a ring homomorphism A/\mathfrak a \to (S^{-1}A)/\mathfrak b which takes \overline a \mapsto \overline {a/1}. This map takes every element of T to a unit so it factors through A/\mathfrak a \to T^{-1}(A/\mathfrak a) \to (S^{-1}A)/\mathfrak b. The latter map takes \overline a / \overline 1 \mapsto \overline {a/1} and thus \overline a / \overline s \mapsto \overline {a/s}.

The reverse map is left as an exercise. [Hint: start with A \to A/\mathfrak a \to T^{-1}(A/\mathfrak a).] ♦

Definition.

Let \mathfrak p \in \mathrm{Spec} A be a prime ideal. The residue field k(\mathfrak p) is defined equivalently as:

  • the field of fractions of A/\mathfrak p, or
  • the quotient of local ring A_{\mathfrak p} by its unique maximal ideal.

Thanks to proposition 1, the definition is sensible (let \mathfrak a = \mathfrak p and S = A - \mathfrak p).

Question to Ponder

If A = k[V] is the coordinate ring of a variety over algebraically closed k, what do elements of k(\mathfrak p) correspond to?

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Localization of Modules

Let M be an A-module and S a multiplicative subset of A.

Proposition 2.

The localized module S^{-1}M is defined as m/s with m\in M and s\in S under the equivalence

m/s = m'/s' \iff (\exists t\in S, \ t(s'm - sm') = 0).

Now S^{-1}M becomes an S^{-1}A-module via:

\frac m s, \frac {m'}{s'} \in S^{-1}M, \ \frac a t\in S^{-1}A \implies \begin{cases} \frac m s + \frac {m'}{s'} = \frac{s'm + sm'}{ss'}, \\ \frac a t \cdot \frac m s = \frac{am}{ts}.\end{cases}

Proof

Easy exercise (but tedious!). ♦

Proposition 3.

The S^{-1}A-module S^{-1}M satisfies the following universal property. We have the pair

(S^{-1}M, \phi : M \to S^{-1}M)

where S^{-1}M is an S^{-1}A-module, \phi(m) := \frac m 1 is A-linear, such that for any pair

(N, \psi : M\to N)

where N is an S^{-1}A-module and \psi is A-linear, there is a unique S^{-1}A-linear map f : S^{-1}M \to N such that f\circ \phi = \psi.

Note

Universal property thus says we get a bijection for any S^{-1}A-module N:

\begin{aligned} \mathrm{Hom}_{S^{-1}A}(S^{-1}M, N) \stackrel \cong \to &\mathrm{Hom}_A(M, N), \\ f \mapsto &f\circ \phi. \end{aligned}

Proof

Existence: given (N, \psi:M \to N), we define f:S^{-1}M \to N by \frac m s \mapsto \frac 1 s \cdot \psi(m). It is well-defined: if \frac m s = \frac {m'}{s'} then t(s'm) = t(sm') for some t\in S, and so ts'\cdot \psi(m) = ts\cdot \psi(m'), which gives \frac 1 s \cdot\psi(m) = \frac 1 {s'}\cdot \psi(m') since t acts bijectively on N. It is easy to check f is S^{-1}A-linear and f\circ \phi(m) = f(\frac m 1) = \psi(m).

Uniqueness: suppose f : S^{-1}M \to N is S^{-1}A-linear and f(\frac m 1) = \psi(m) for every m \in M. Then

\frac m s \in S^{-1}M \implies \psi(m) = f(\frac m 1) = f(\frac s 1\cdot\frac m s) = \frac s 1 f(\frac m s)

so f(\frac m s) =\frac 1 s \cdot \psi(m). ♦

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Functoriality

Definition.

Suppose f:M\to N is an A-linear map. We have an S^{-1}A-linear map

S^{-1}f : S^{-1}M \longrightarrow S^{-1}N, \quad \frac m s \mapsto \frac {f(m)}s.

Proof

Let us check the map is well-defined. If \frac m s = \frac {m'}{s'} then t(s'm - sm') = 0 for some t\in S. Applying f gives t(s'\cdot f(m) - s\cdot f(m')) = 0 and hence \frac{f(m)}s = \frac{f(m')}{s'}. Linearity is an easy exercise. ♦

Clearly S^{-1}(1_M) = 1_{S^{-1}M} and S^{-1}(g\circ f) = (S^{-1}g) \circ (S^{-1}f) for any f: N\to M and g:M\to P. Thus localization gives a functor

localization_functor

Furthermore, the functor is A-linear, i.e. for A-modules M and N,

a\in A,\ f, g : M\to N \implies \begin{cases} &S^{-1}f + S^{-1}g = S^{-1}(f+g), \\ &S^{-1}(af) = \frac a 1 S^{-1} f \end{cases}

In particular, it is additive.

Generalization

One can generalize the localization functor.

Definition.

Let B be an A-algebra and M an A-module. The B-module induced from M comprises of a pair

(M^B, \phi : M\to M^B)

where M^B is a B-module, \phi is A-linear, such that for any pair

(N, \psi : M\to N)

where N is a B-module and \psi is A-linear, there is a unique B-linear f : M^B \to N such that f\circ \phi =\psi.

Note

Again universal property says the following is a bijection for any B-module N:

\begin{aligned} \mathrm{Hom}_B(M^B, N) \stackrel \cong \to &\mathrm{Hom}_A(M, N), \\ f \mapsto &f\circ \phi. \end{aligned}

Exercise A

1. Prove uniqueneess of f in the above example.

2. Prove that for an ideal \mathfrak a \subseteq A and B = A/\mathfrak a, the B-module induced from M is M^B = M/\mathfrak a M. This explains what we meant when we said the construction is canonical.

Question to Ponder

Does M\mapsto M^B give an additive functor from the category of A-modules to the category of B-modules? Can all the necessary constructions and properties be deduced just from the universal property of induced modules?

In later articles, we will give an explicit construction of the induced module.

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Exactness

This result is of paramount importance.

Theorem 1.

The localization functor is exact, i.e. if N\stackrel f\to M \stackrel g \to P is exact, so is

S^{-1}N \stackrel {S^{-1}f}\longrightarrow S^{-1}M \stackrel{S^{-1}g} \longrightarrow S^{-1}P.

Proof

For any \frac n s \in S^{-1}N we have (S^{-1}g) (S^{-1}f)(\frac n s) = (S^{-1}g) (\frac {f(n)}s) = \frac{gf(n)}s = 0 and so \mathrm{im} (S^{-1}f) \subseteq \mathrm{ker} (S^{-1}g).

Conversely suppose (S^{-1}g)(\frac m s) = \frac{g(m)}s = 0. Then there exists t\in S, t\cdot g(m) = 0 so g(tm) = 0 and tm \in \mathrm{ker} g = \mathrm{im} f. We thus have tm = f(n) for some n\in N which gives us \frac m s = \frac{f(n)}{st} = (S^{-1}f)(\frac n {st}). ♦

Now all the properties we proved about exact additive functors can be ported over here.

1. If f : N\to M is injective, so is S^{-1} f : S^{-1}N \to S^{-1}M. Hence for A-submodule N\subseteq M we may regard S^{-1}N as an S^{-1}A-submodule of S^{-1}M.

2. If g : M\to P is surjective, so is S^{-1} g : S^{-1}M \to S^{-1}P.

3. For A-submodule N\subseteq M, we have S^{-1}(M/N) \cong (S^{-1}M)/(S^{-1}N).

4. For an A-linear map f:N\to M we have

F(\mathrm{ker} f) = \mathrm{ker} F(f), \quad F(\mathrm{coker} f) = \mathrm{coker} F(f), \quad F(\mathrm{im} f) = \mathrm{im} F(f).

5. For A-submodules N_1, N_2\subseteq M we have, as submodules of S^{-1}M,

S^{-1}(N_1 + N_2) = (S^{-1}N_1) + (S^{-1}N_2), \quad S^{-1}(N_1 \cap N_2) = (S^{-1}N_1) \cap (S^{-1}N_2).

Exercise B

Prove that for an ideal \mathfrak a\subseteq A,

  • the localized module S^{-1}\mathfrak a is the ideal \mathfrak a(S^{-1}A) we saw earlier,
  • for any A-module M we have (S^{-1}\mathfrak a)(S^{-1} M) = S^{-1}(\mathfrak a M) as S^{-1}A-submodules of S^{-1}M.

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2 Responses to Commutative Algebra 24

  1. Vanya says:

    Do we not require that the ideal \mathfrak{a} does not intersect S in Proposition 1?

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