# 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?

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

# 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

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

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

## 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.

# 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?

• limsup says:

If they intersect, we get a trivial ring on both sides so it works out.