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.


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).] ♦


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?


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}


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.


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}


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). ♦




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.


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


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.


One can generalize the localization functor.


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.


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.



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.


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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