# Localization and Spectrum

Recall that the ideals of $S^{-1}A$ correspond to a subset of the ideals of A. If we restrict ourselves to prime ideals, we get the following nice bijection.

Theorem 1.

The above gives a bijection between

$\{ \mathfrak p \in \mathrm{Spec} A : \mathfrak p \cap S = \emptyset \} \leftrightarrow \mathrm{Spec} (S^{-1}A)$

Useful trick

If $\mathfrak a \subseteq A$ is an ideal, then $\frac a s \in \mathfrak a (S^{-1}A)$ implies $s'a \in \mathfrak a$ for some $s'\in S$. Proof: exercise.

Proof

First we show $\mathfrak q = \mathfrak p (S^{-1}A)$ is prime in $S^{-1}A$. Suppose $a,b \in A$, $s, t\in S$ satisfy $\frac a s \frac b t \in \mathfrak q$. Then by the above trick, there exists $s'\in S$ such that $s'ab \in \mathfrak p$. But $\mathfrak p \cap S = \emptyset \implies s'\not\in \mathfrak p$ so either $a\in \mathfrak p$ or $b \in \mathfrak p$ and hence $\frac a s \in \mathfrak q$ or $\frac b t \in \mathfrak q$.

Note that $\mathfrak q$ is not the whole ring since if it contains 1, then for some $a\in\mathfrak p$, $s\in S$,

$\frac a s = \frac 1 1 \implies \exists t \in S, t(s-a) = 0 \implies ts = ta \in S \cap \mathfrak p,$ a contradiction.

Continuing the previous paragraph, it remains to show $\mathfrak p = \mathfrak q \cap A$.

(⊆) Suppose $a\in \mathfrak p$ so $\frac a 1 \in \mathfrak p (S^{-1}A) = \mathfrak q$. Thus $a \in \mathfrak q \cap A$.

(⊇) Suppose $a \in \mathfrak q \cap A$ so that $\frac a 1 \in \mathfrak q = \mathfrak p (S^{-1}A)$. Then $sa \in \mathfrak p$ for some $s\in S$. Since $s\not\in \mathfrak p$ we have $a\in \mathfrak p$. ♦

Proposition 1.

The above map identifies $\mathrm{Spec} S^{-1}A$ as a subspace of $\mathrm{Spec} A$.

Proof

Let $\phi : A\to S^{-1}A, a\mapsto \frac a 1$. Then $\phi^* : \mathrm{Spec} S^{-1}A \to \mathrm{Spec} A$ is injective and continuous. On the other hand, suppose $V(\mathfrak b)$ is a closed subset of $\mathrm{Spec} S^{-1}A$ for an ideal $\mathfrak b \subseteq S^{-1}A$. Let $\mathfrak a = \mathfrak b \cap A$, an ideal of A. It remains to show $V(\mathfrak b) = (\phi^*)^{-1} (V(\mathfrak a))$.

(⊆) Suppose $\mathfrak q \supseteq \mathfrak b$, a prime ideal of $S^{-1}A$. Then $\phi^*(\mathfrak q) = \mathfrak q \cap A \supseteq \mathfrak b \cap A = \mathfrak a$ so $\phi^*(\mathfrak q) \in V(\mathfrak a)$.

(⊇) Suppose $\mathfrak p := \phi^*(\mathfrak q) = \mathfrak q \cap A$ is a prime of A containing $\mathfrak a$. By theorem 1, $\mathfrak q = \mathfrak p (S^{-1}A) \supseteq \mathfrak a (S^{-1}A) =\mathfrak b$ from the previous article. ♦

Philosophically, localization removes prime ideals which are “too big”.

# Basic Open Sets

Let $f\in A$. By the above result, $\mathrm{Spec} A_f$ is identified with the subspace $\{ \mathfrak p \in \mathrm{Spec} A : \mathfrak p \not\ni f\}$. But this is exactly the basic open set $D(f)$. Hence,

$\mathrm{Spec} A_f \cong D(f) \subseteq \mathrm{Spec} A$

as a homeomorphism.

### Geometric Example

Let $A = k[V]$ be the coordinate ring of variety V over an algebraically closed field k. Pick $f\in A$ and let $B = A_f \cong A[Y]/(f\cdot Y - 1)$.

Concretely, if $V\subseteq \mathbb A^n$ is a closed subspace, then $B = k[W]$ where

\begin{aligned} W &= \{ (x_1, \ldots, x_n, y) \in \mathbb A^{n+1} : (x_1, \ldots, x_n) \in V,\ f(x_1, \ldots, x_n) =\frac 1 y \}, \\ &\cong \{(x_1, \ldots, x_n) \in V : f(x_1, \ldots, x_n) \ne 0\}, \end{aligned}

because of the following result.

Exercise A

Prove that if A is a reduced ring, so is $S^{-1}A$.

As a consequence, we see that if $f \in k[V]$ is a regular function on affine variety V, the open set $D(f) := \{\mathbf v \in V : f(\mathbf v) \ne 0\}$ is also an affine variety. In summary, a basic open subset of an affine variety is an affine variety.

The simplest example is $\mathbb A^1(k) - \{0\}$, which is isomorphic to the “hyperbola” $\{ (x,y) \in \mathbb A^2(k) : xy = 1\}$.

# Local Rings

Consider the case $S = A - \mathfrak p$ for a prime ideal $\mathfrak p \subset A$ so $S^{-1}A = A_{\mathfrak p}$. The first result implies:

Corollary 1.

The prime ideals of $A_{\mathfrak p}$ correspond bijectively with the prime ideals of A contained in $\mathfrak p$.

In particular, $A_{\mathfrak p}$ has exactly one maximal ideal $\mathfrak p A_{\mathfrak p}$.

Thus the spectrum of $A_{\mathfrak p}$ looks something like this.

Example

If $A=\mathbb Z$ and $\mathfrak p = 2\mathbb Z$, then

$A_{\mathfrak p} = \mathbb Z_{(2)} = \{\frac a b\in \mathbb Q : b \text{ odd} \}.$

Definition.

A ring is called a local ring if it has a unique maximal ideal.

We often write $(A, \mathfrak m)$ for the local ring, where $\mathfrak m$ is the maximal ideal of A.

The following is all you need to know about local rings.

Proposition 2.

A ring $A$ is local if and only if its set of non-units forms an ideal, in which case this ideal is the unique maximal ideal of $A$.

Proof

We may assume $A \ne 0$.

(⇒) Let $(A,\mathfrak m)$ be local. If $a\in A$ is not a unit it generates a proper ideal $(a) \subsetneq A$, which must be contained in a maximal ideal of A. But $\mathfrak m$ is the only maximal ideal so $a\in\mathfrak m$. Thus $\mathfrak m$ is the set of non-units of $A$.

(⇐) Suppose the set of non-units is an ideal $\mathfrak a \subsetneq A$. Then any maximal ideal $\mathfrak m\subsetneq A$ cannot contain any units so $\mathfrak m \subseteq \mathfrak a$. Equality must hold by maximality. ♦

# Local Rings in Geometry

Let $A = k[V]$ again, where V is an affine k-variety with k algebraically closed. Fix a point $P\in V$ and consider its associated maximal ideal $\mathfrak m = \mathfrak m_P = \{f \in A : f(P) = 0\}$. If $\frac f g \in A_{\mathfrak m}$ where $f\in A$, $g\in A - \mathfrak m$, then $g(P) \ne 0$ so $D(g)$ (defined above) contains P. Hence $\frac f g$ defines a function

$\frac f g : U \longrightarrow k, \quad Q \mapsto \frac{f(Q)}{g(Q)}$,

for some open neighbourhood U of P.

A function of this form is said to be regular at P. Unfortunately this is dependent on our choice of representatives.

Exercise B

Let $\frac f g, \frac {f'}{g'} \in A_{\mathfrak m}$, corresponding to $\frac f g : U\to k$ and $\frac {f'}{g'} : U' \to k$. Prove that $\frac f g = \frac {f'}{g'}$ if and only if there is an open set W,

$P\in W\subseteq U\cap U',\quad \left.\frac f g\right|_W = \left.\frac {f'}{g'}\right|_W$.

Hence we define the following notion.

Definition.

Let $P$ be a point on the topological space $X$. For a set S, an S-valued local function at P is a function $f : U\to S$ for some open neighbourhood U of P.

If $f:U \to S$, $g:U' \to S$ are local functions at P, we say they have the same germ at P if there is an open W,

$P \in W \subseteq U \cap U', \quad f|_W = g|_W$.

Note that this is an equivalence relation.

Hence $A_{\mathfrak m}$ classifies the germs of regular local functions at P.

### Example

Let $V = V(f) \subset \mathbb A^2(\mathbb C)$ where $f = (Y - X^2 + 1)(Y + X^2 - 1) = 0$. Take $P = (0, 1) \in V$.

We have

$k[V] = k[X, Y]/(f), \quad \mathfrak m = \mathfrak m_P = (X, Y - 1) \subset k[V],$

where $X,Y$ represent their images in $k[X, Y]/(f)$. We take the functions:

$g_1 = \frac{X+1}{Y},\ g_2 = \frac 1 {1-X} \in A_{\mathfrak m},$

which are well-defined on $U = V - \{(-1,0), (+1, 0)\}$. Now $g_1 \ne g_2$ since $g_1(0, -1) = -1$ and $g_2(0, -1) = 1$ but they define the same germ at P:

$h = Y - X^2 + 1 \in k[V] - \mathfrak m \implies h\cdot [(1-X)(X+1) - Y\cdot 1] = -h (Y+X^2 - 1) = 0$

in $k[V]$. Indeed, if we take out the component $W = V(Y - X^2 + 1)$, then $V-W$ is an open subset such that $g_1|_{V-W} = g_2|_{V-W}$.

Exercise C

Prove the preceding statement.

# Irreducible Varieties

When V is an irreducible variety, $A = k[V]$ is an integral domain and this case is particularly nice since S has no zero-divisors (as long as it does not contain 0).

Definition.

rational function on V is a regular local function f at any point of V. Thus a rational function is of the form $f : U \to k$ for some non-empty open $U\subseteq V$.

We identify two rational functions

$f : U\to k, \quad g : U'\to k$

if there exists a non-empty open $W \subseteq U \cap U'$ such that $f|_W = g|_W$.

Exercise D

1. Prove that this is an equivalence relation. [Hint: use the fact that in an irreducible space, any two non-empty open subsets intersect.]

2. Prove that in fact, if there exists $\emptyset \ne W \subseteq U \cap U'$ such that $f|_W = g|_W$, then it holds for $W = U \cap U'$.

Algebraically, a rational function is of the form $\frac f g$ with $f, g\in k[V]$ and $g(P) \ne 0$ for some point $P\in V$. The latter condition just means $g\ne 0$ so we have:

Lemma 1.

The set of rational functions on V is exactly the field of fractions of $k[V]$.

We call this the rational function field of V and denote it by $k(V)$.

Exercise E

Take the curve $Y^2 = X^3 - X$ with coordinate ring $k[V] = k[X, Y]/(Y^2 - X^3 + X)$. Prove that $k(V) \cong k(X)[Y]/(Y^2 - X^3 + X)$, where $k(X)$ is the field of rational functions in X.

Question to Ponder

Let $\mathfrak p\subset A = k[V]$ be a prime ideal (not maximal in general) corresponding to the irreducible closed subset $W\subseteq V$. Explain $A_{\mathfrak p}$ in geometric terms.

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### 12 Responses to Commutative Algebra 23

1. Vanya says:

Did you mean $U = D(g)$ in the definition $\frac f g : U \longrightarrow k, \quad Q \mapsto \frac{f(Q)}{g(Q)}.$ in the section “Local rings in Geometry” . Also, can you describe what S is in $S$-valued local function at $P$.

A typo in the lemma in the section “Irreducible varieties” : “We denote call this the ….”

• limsup says:

Thanks, corrected the parts. On your first comment, one can set $U = D(g)$, but the point is that one really cares about what goes on “close to P”. Hopefully, this edit is clearer.

• Vanya says:

Would you please share your personal email I’d with me? I have a few questions regarding measure theory and I believe you may answer them. Thank you.

• limsup says:

Maybe post the question here? I’ll try to help but can’t guarantee. My measure theory is a little rusty.

2. Vanya says:

I wonder why there are many theories of integration; I feel that such a state of affairs seems contradictory to the spirit of Mathematics, namely exactitude, proceeding from a set of axioms through logical deduction.

Progress in algebra is in isolating important subsets of sets and study their behaviour; this approach eventually leads one to answer a question related to number theory. But, in analysis, the approach seems otherwise.Lebesgue’s theory, unlike Riemann’s, can integrate larger class of functions. What does that mean? That I change my theory of integration as and when I like in order to integrate a special class of functions of my interest; for instance Dirichlet’s function on [0,1] in the case of Lebesgue’s theory. Please clarify these issues. Is it the case that adopting Lebesgue’s theory enables one to make sense of several phenomenon, like in Fourier analysis, which eventually helps one to explain many physical phenomenon. Or is it just that Lebesgue’s theory was adopted by mathematicians, because of its simplicity, generality, and applicability as a theory of integration.

• limsup says:

I’m no expert in analysis but my understanding is that Lebesgue integration is definitely preferred to Riemann integration because it can be generalized to integration on measure spaces (e.g. Haar measure on locally compact groups, including p-adic groups and Lie groups). Once you learn Lebesgue integration, you should probably forget Riemann integration.

That being said, Riemann integration has the advantage of being easier to understand so it is not leaving the university curriculum.

3. Vanya says:

What is the need to show that “show $V(\mathfrak b) = (\phi^*)^{-1} (V(\mathfrak a)).$ since we already knew that $\phi^*$ is continuous?

• Vanya says:

Sorry, it is in Proposition 1.

• limsup says:

Hi. What we’re showing is that $\phi^*$ identifies $\mathrm{Spec} S^{-1}A$ as a subspace of $\mathrm{Spec} A$.

For an injective map $f:X\to Y$, it is not enough to show that $f$ is continuous (i.e. if $C\subseteq Y$ is closed, then so is $f^{-1}(C)$). We also need to show that any closed subset of $X$ is of the form $f^{-1}(C)$ for some closed $C\subseteq Y$.

4. Manas Jana says:

I can’t understand why we need the “useful trick”. Because in previous post by a lemma we know that what are the elements of P(S-1A), i mean a/s € S-1A, where a€ P and s€S. Please clarify this!!!!

• Manas Jana says:

Sorry, a/s € P(S-1A) where a € P and s€ S

• limsup says:

It’s a little subtle since $\frac a s = \frac b s$ does not imply $a = b$ here. The “useful trick” is really easy to prove but will be repeatedly used again and again that’s why we mention it here.