# From classical geometry to algebraic geometry

This post is an elementary discussion of understanding points in $k^{n}$ as points in $\mathbb{A}_{k}^{n} = \text{Spec}k[x_{1}, \cdots, x_{n}]$, where $k$ is a (not necessarily algebraically closed) field. The reference is Ravi Vakil’s lecture notes, though the post is not identical to the notes.

Theorem. Given a point $(a_{1}, \cdots, a_{n}) \in k^{n}$, the kernel of the evaluation homomorphism (at the point) $k[x_{1}, \cdots, x_{n}] \rightarrow k$ given by $f \mapsto f(a_{1}, \cdots, a_{n})$ is $(x_{1} - a_{1}, \cdots, x_{n} - a_{n})$. In particular, the ideal $(x_{1} - a_{1}, \cdots, x_{n} - a_{n})$ is maximal, so $(x_{1} - a_{1}, \cdots, x_{n} - a_{n}) \in \text{Spec}k[x_{1}, \cdots, x_{n}]$.

Remark. It is rather hard to compute the kernel in a naive way. Let’s try it. Suppose that $f(a_{1}, \cdots, a_{n}) = 0$. To show that $f \in (x_{1} - a_{1}, \cdots, x_{n} - a_{n})$ directly, we need to find $g_{1}, \cdots, g_{n} \in k[x_{1}, \cdots, x_{n}]$ such that

$f(x_{1}, \cdots, x_{n}) = g_{1}(x_{1}, \cdots, x_{n})(x_{1} - a_{1}) + \cdots + g_{n}(x_{1}, \cdots, x_{n})(x_{n} - a_{n})$.

But there is a way to get around this difficulty: to get help from morphisms!

Proof. Define $\phi : k[x_{1}, \cdots, x_{n}]/(x_{1} - a_{1}, \cdots, x_{n} - a_{n}) \rightarrow k$ by $\bar{f} \mapsto f(a_{1}, \cdots, a_{n})$. This is well-defined because if $\bar{f} = 0$, then $f(x_{1}, \cdots, x_{n}) = \sum_{j=1}^{n}h_{j}(x_{1}, \cdots, x_{n})(x_{j} - a_{j})$ so that $f(a_{1}, \cdots, a_{n}) = 0$.

Now, define $\psi : k \rightarrow k[x_{1}, \cdots, x_{n}]/(x_{1} - a_{1}, \cdots, x_{n} - a_{n})$ by $c \mapsto \bar{c}$. Then

$\psi\phi(\bar{f}) = \overline{f(a_{1}, \cdots, a_{n})} = f(\bar{a_{1}}, \cdots, \bar{a_{n}}) = f(\bar{x_{1}}, \cdots, \bar{x_{n}}) = \bar{f}$,

and $\phi\psi(c) = \phi(\bar{c}) = c$, so $\phi$ is an isomorphism. Let $K$ be the kernel of evaluation. Then we have constructed the following isomorphism

$k[x_{1}, \cdots, x_{n}]/(x_{1} - a_{1}, \cdots, x_{n} - a_{n}) \overset{\sim}{\longrightarrow} k[x_{1}, \cdots, x_{n}]/K$

given by $f \mod (x_{1} - a_{1}, \cdots, x_{n} - a_{n}) \mapsto f \mod K$. Since zeros of each side correspond, we have $(x_{1} - a_{1}, \cdots, x_{n} - a_{n}) = K$. Q.E.D.

Remark. The key of the above proof was to recognize that

$\overline{f(x_{1}, \cdots, x_{n})} = f(\bar{x_{1}}, \cdots, \bar{x_{n}})$

in the quotient ring $k[x_{1}, \cdots, x_{n}]/I$ where $I$ is a given ideal of the polynomial ring. Let $\pi : k[x_{1}, \cdots, x_{n}] \rightarrow k[x_{1}, \cdots, x_{n}]/I$. The above identity can be also understood as

$\pi(k)[\bar{x_{1}}, \cdots, \bar{x_{n}}] = k[x_{1}, \cdots, x_{n}]/I$,

so it is acceptable to write

$\bar{a_{n_{j}}}\bar{x_{j}}^{n} + \cdots + \bar{a_{1}}\bar{x_{j}} + \bar{a_{0}} = a_{n_{j}}\bar{x_{j}}^{n} + \cdots + a_{1}\bar{x_{j}} + a_{0}$,

which is given by the restriction of scalars given by $\pi : k \overset{\sim}{\longrightarrow} \pi(k)$. Notice that our arguments do not need $k$ to be isomorphic to $\pi(k)$! That’s why I believe that this can be generalized as much as we want (see below).

There is another reason that it is legitimate to write $a \bar{x_{j}} = \bar{a}\bar{x_{j}}$: if we look at the quotient $k[x_{1}, \cdots, x_{n}]/I$ as a $k$-algebra, this is how one defines $k$-action on the quotient module: if $N$ is an $A$-submodule of $M$, we define $a\bar{m} = \overline{am}$Both views are valid here!

Question. Can we generalize this to $A[x_{1}, \cdots, x_{n}]$, where $A$ is an arbitrary commutative ring with unity?

Answer (not professional). I think we can, but it is more reasonable to assume that $A$ is an integral domain so that we can conclude that $(x_{1} - a_{1}, \cdots, x_{n} - a_{n})$ are prime ideals given by the isomorphism $A[x_{1}, \cdots, x_{n}]/(x_{1} - a_{1}, \cdots, x_{n} - a_{n}) \simeq A$ similar to above. However, I think that the isomorphism is given over any commutative rings.

Consider an example: $\mathbb{Z}[x, y] \rightarrow \mathbb{Z}$ given by $f \mapsto f(m, n)$. Then we can construct

$\phi : \mathbb{Z}[x, y]/(x - m, y - n) \rightarrow \mathbb{Z}$

by $\bar{f} \mapsto f(m, n)$ and

$\psi : \mathbb{Z} \rightarrow \mathbb{Z}[x, y]/(x - m, y - n)$

by $s \mapsto \bar{s}$. Then we have

$\psi\phi(f(x, y)) = \phi(f(m, n)) = \overline{f(m, n)} = \overline{f(x, y)}$ and

$\phi\psi(s) = \phi(\bar{s}) = s$. This clearly generalizes to the case where $A$ is any commutative ring and we have arbitrarily many finite number of indeterminates. Again, the key is to see $\overline{f(x, y)} = \overline{f(m, n)}$. As we saw before, this is merely because polynomials are “constructed by addition and multiplication”. Just like in Theorem, this argument is not about merely giving an isomorphism, but it also shows that

$f(a_{1}, \cdots, a_{n}) = 0$ if and only if $f(x_{1}, \cdots, x_{n}) \in (x_{1} - a_{1}, \cdots, x_{n} - a_{n})$.

To put it differently, we have

$f(a_{1}, \cdots, a_{n}) = 0$ if and only if $f(x_{1}, \cdots, x_{n}) = 0 \mod (x_{1} - a_{1}, \cdots, x_{n} - a_{n})$.

A more ambitious question. If I have not made any incorrect remark, it is natural to ask what is the notion of ideals $(x_{1} - a_{1}, \cdots, x_{n} - a_{n})$ that is the kernel of $f \mapsto f(a_{1}, \cdots, a_{n})$. If the ground ring is not an integral domain, we necessarily have $(x_{1} - a_{1}, \cdots, x_{n} - a_{n}) \notin \mathbb{A}_{A}^{n}$, so we might want to make $\mathbb{A}_{A}^{n}$ bigger! Let us call such an ideal as a type C ideal. Now we may ask what are the type C ideals in an arbitrary ring and if it enriches the theory! However, this might be just a very shallow question.

Remark. Notice that we have not used Nullstellensatz (weak nor strong) for this discussion. Weak Nullstellensatz is only used to say that over an algebraically closed field, classical points are all the maximal ideals of a polynomial ring over the field.

# Review: Topologification 1

This post is to review some topological concepts. The material is from Topology (3rd edition) by James Munkres and several articles of proofwiki (e.g., the article about a way to create a basis) with some additional thoughts of myself.

Given a set $X$, we are interested in ways to construct a topological structure on $X$.

One natural thought is to pick some collection of subsets $\mathscr{S} \subseteq \mathcal{P}(X)$ and ask if there is the smallest topology on $X$ that contains $X$. If it exists, it must be unique, and let us denote $O_{\mathscr{S}}$ the topology, calling it the topology generated by $\mathscr{S}$.

Existence of the topology generated by a set. We can construct this topology by taking intersection containing $\mathscr{S}$ because there is at least one being intersected, which is the entire power set $\mathcal{P}(X)$.

Remark. Let $\mathscr{T} \subseteq \mathcal{P}(X)$ be a topology on $X$. Then $O_{\mathscr{T}} = \mathscr{T}$. Moreover, we get a functor $\mathscr{S} \mapsto O_{\mathscr{S}}$ from the category of topologies on $X$ to the category of subsets of $X$ (i.e., $\mathcal{P}(X)$).

Remark. Given a collection of subsets $\mathscr{S} \subseteq \mathcal{P}(X)$, the topology $O_{\mathscr{S}}$ is the topological best approximation of $\mathscr{S}$. It deserves a name such as “topologification” and I am pretty sure this is the left-adjoint to the forgetful functor from the category of topologies on $X$ to the powerset $\mathcal{P}(X)$.

We now want to another (less categorical but more set-theoretical) criterion to compute $O_{\mathscr{S}}$. An easier case is given by the concept of “basis”.

The subcollection $\mathscr{B} \subseteq \mathcal{P}(X)$ is called a (topological) basis if it satisfies the following axioms:

1. $\mathscr{B}$ covers $X$.
2. If $x \in B_{1} \cap B_{2}$ where $B_{1}, B_{2} \in \mathscr{B}$, then there is $B_{3} \in \mathscr{B}$ such that $x \in B_{3} \subseteq B_{1} \cap B_{2}$.

We can compute $O_{\mathscr{B}}$ as follows.

Theorem 1.1. Let $\mathscr{B}$ be a topological basis in $\mathcal{P}(X)$. Then

$O_{\mathscr{B}} = \{U \in \mathcal{P}(X) | (\forall x \in U) (\exists B \in \mathscr{B}) (x \in B \subseteq U)\}$.

Remark. Let $O$ denote the right-hand side. It is easy to show that $O$ is a topology on $X$. However, it is hard (at least for me) to show the equality!

We show that $O$ is a topology on $X$. (i) We get $\emptyset \in O$ vacuously and $X \in O$ trivially. Let $\{U_{i}\}_{i \in I} \subseteq O$. (ii) If $x \in \cup_{i \in I}U_{i}$, then $x \in B \subseteq U_{j} \subseteq \cup_{i\in I}U_{i}$ for some $j \in I$ and some $B \in \mathscr{B}$, so $\cup_{i \in I}U_{i} \in O$. (iii) If $x \in U_{i} \cap U_{j}$, then there is $B_{i}, B_{j} \in \mathscr{B}$ such that $x \in B_{i} \subseteq U_{i}$ and $x \in B_{j} \subseteq U_{j}$, so take $B_{k} \in \mathscr{B}$ such that $x \in B_{k} \subseteq B_{i} \cap B_{j} \subseteq U_{i} \cap U_{j}$. Thus $U_{i} \cap U_{j} \in O$.

We have established that $O$ forms a topology on $X$ and clearly, we have $\mathscr{B} \subseteq O$. The difficulty arises when one tries to show $O \subseteq O_{\mathscr{B}}$. To finish showing the theorem, we need to discuss about another way to compute $O_{\mathscr{B}}$ by another description of $O$.

Theorem 1.2. Let $\mathscr{B}$ be a topological basis in $\mathcal{P}(X)$. Then

$O_{\mathscr{B}} = \{U \in \mathcal{P}(X) | U \text{ is a union of some members of }\mathscr{B}\}$.

Proof. Denote $O'$ the right-hand side. It is evident that $O' \subseteq O_{\mathcal{B}}$. If we show that $O' = O$ (see the remark above), then we can conclude at once that $O_{\mathscr{B}} = O' = O$ since $O$ is a topology on $X$ containing $\mathscr{B}$!

Let $U \in O$. For each $x \in U$, (using the axiom of choice) choose $B_{x} \in \mathscr{B}$ such that $x \in B_{x} \subseteq U$. Then $U = \cup_{x \in U}B_{x} \in O'$. It is immediate that $O' \subseteq O$, so we finish the proof (of Theorem 1 and 2 together). Q.E.D.

Remark. Do not forget that the description given in Theorem 1 was not so trivial to prove! As a corollary, we get a nontrivial way to describe open sets of a topological space $(X, O(X))$ by $O_{O(X)} = O(X)$.

Corollary 1.3. Let $X$ be a topological space. Then $U \subseteq X$ is open if and only if for each $x \in U$ there is an open $V \ni x$ such that $x \in V \subseteq U$.

# Subspace and quotient space

A space with an inclusion morphism is a subspace. A space with a projection morphism (mapping each element to its equivalence class) is a quotient space.

# Polynomial rings are UFDs

R is a UFD if and only if the univariate polynomial ring over it is. (See Dummit and Foote p.304.)

# Universal morphisms

Universal morphisms are morphisms induced by an initial or a terminal object in a category of some natural transformations.

# Intersection over the emtpy collection

Any member in the intersection over the empty collection needs to be in every member of the empty collection. Thus any object in the given ambient space satisfies this property.

# Review: Localization 1.

This post is entirely based on Ravi Vakil’s lecture notes (FoAG), Atiyah and Macdonald, and a tiny bit of additional thoughts of mine. It is an informal discussion about localization.

According to Atiyah and MacDonald (p.36) techniques from localization is “the most important technical tools in commutative algebra”. The following is more quote.

“They correspond in the algebro-geometric picture to concentrating attention on an open set or near a point, and the importance of these notions should be self-evident.”

All rings in this post are commutative rings with unity.

Categorical definition (ring). Every ring is a monoid with respect to its multiplication. Let $A$ be a ring and $S$ be a multiplicative submonid (i.e., $1_{A} \in S$ and $S$ is closed under multiplication). Then there is a ring map $\phi : A \rightarrow S^{-1}A$ that is initial among ring maps $g : A \rightarrow B$ such that $g(S)$ has only (multiplicatively) invertible elements in $B$. Note in particular that $\phi(S)$ has only invertible elements in $S^{-1}A$.

Proof of Existence (Sketch). Define $S^{-1}A = \{a/s : a \in A \text{ and } s \in S\}$ where $a/s$ is the equivalence class of $(a, s) \in A \times S$ with the following equivalence relation:

$(a, s) \sim (a', s')$ is defined by $[(\exists u \in S)(u(s'a - sa') = 0)]$.

We give $S^{-1}A$ a ring structure as follows:

$a/s + a'/s' := (s'a + sa')/(ss')$ and $(a/s) \cdot (a'/s') := (aa')/(ss')$.

Then define $\phi : A \rightarrow S^{-1}A$ by $\phi(a) = a/1$. It is mechanical to show that this satisfies the universal property. Q.E.D.

Remark. From the above definition, we see that a ring map $S^{-1}g : S^{-1}A \rightarrow B$ corresponds to a ring map $g : A \rightarrow B$ that sends $S$ to invertible elements in $B$. (After reading the part about localization of a module, one can recognize that $B = S^{-1}B$ in this case.)

We fix $S \subseteq A$ as above throughout.

Proposition 1.1. The following data are equivalent:

• $M$ is an $S^{-1}A$-module;
• $M$ is an $A$-module where actions by elements of $S$ are $A$-module automorphisms.

Sketch of Proof. An $S^{-1}A$-module structure $M$ gives $M$ an $A$-module structure by $a \cdot m := (a/1)m$. Conversely, an $A$-module structure on $M$ with actions of $S$ being module automorphisms give $M$ an $S^{-1}A$-module structure by $(a/s) \cdot m := a {\tilde{s}}^{-1}(m)$, where $\tilde{s}(m) = sm$. Q.E.D.

Remark. $A \rightarrow S^{-1}A$ is injective if and only if $S$ contains no zero-divisors.

Notice that $A \rightarrow S^{-1}A$ is an $A$-module map, so it is not difficult to notice the following generalization.

Categorical definition (module). Given an $A$-module $M$, there is an $A$-module map $\phi : M \rightarrow S^{-1}M$ that is initial among $A$-module maps $\alpha : M \rightarrow N$ such that $S$ has only “action-wise” invertible elements in $N$.

Here, “an invertiable element of $A$ in $N$” means that the action given by the element is an $A$-module automorphism. Notice that being units of a ring is much stronger than this condition (e.g., $\mathbb{Z}$-action on $\mathbb{Z}$).

Remarks.

1. We already know that an $A$-module $N$ with $S$ being invertible is same as saying $N$ being an $S^{-1}A$-module, so in particular, we know $S^{-1}M$ has an $S^{-1}A$-module structure.
2. Moreover, the universal morphism $S^{-1}\alpha : S^{-1}M \rightarrow N$ is required to be $S^{-1}A$-module map with this $S^{-1}A$-module structure. (Admittedly, the morphisms of relevant comma category are $S^{-1}A$-module maps.)
3. Thus, it is easier to think of $M \rightarrow S^{-1}M$ as the initial in the category $A$-module maps from $M$ to an $S^{-1}A$-modules whose morphisms are $S^{-1}A$-module maps between targets making commutative triangles. (But to think of $M \rightarrow S^{-1}M$ as a generalization of the case for rings, it is easier to have the previous definition.)

Proof of Existence (Sketch). Define $S^{-1}M := \{m/s : m \in M \text{ and } s \in S\}$ where $m/s$ is the equivalence class of $(m, s) \in M \times S$ of the following equivalence relation:

$(m, s) \sim (m', s')$ is defined by $u(s'm - sm') = 0$ for some $u \in A$.

We give $S^{-1}M$ an $S^{-1}A$-module structure as follows:

$m/s + m'/s' := (s'm + sm')/(ss')$ and $(a/t) \cdot (m/s) := (am)/(ts)$.

We also get the map $\phi : M \rightarrow S^{-1}M$ by $m \mapsto m/1$ and check the desired universal property.  Q.E.D

Theorem 1.2. Localization is an exact functor $S^{-1} : \text{Mod}_{A} \rightarrow \text{Mod}_{S^{-1}A}$.

Proof. From universal property, it is clear how to define $S^{-1}u : S^{-1}M \rightarrow S^{-1}N$ for an $A$-module $u : M \rightarrow N$. It is also clear from commutative diagrams that $S^{-1}(u + u') = S^{-1}u + S^{-1}u'$ (that is, it is an additive functor).

To show exactness, we fix an exact sequence $M \overset{f}{\longrightarrow} N \overset{g}{\longrightarrow} P$ of $A$-module morphisms and show that $S^{-1}M \overset{S^{-1}f}{\longrightarrow} S^{-1}N \overset{S^{-1}g}{\longrightarrow} S^{-1}P$ is exact. We only need to show that $\ker(S^{-1}g) \leqslant \text{im}(S^{-1}f)$.

If $n/s \in \ker(S^{-1}g)$, then $n/1 \in \ker(S^{-1}g)$, so $g(n)/1 = 0$ and thus $g(un) = ug(n) = 0$ for some $u \in S$. Since $un \in \ker(g) = \text{im}(f)$, we take $x \in M$ such that $f(x) = un$, so $n/s = f(x)/(us) \in \text{im}(S^{-1}f)$ as desired. Q.E.D.

Remark. It seems to be difficult to show the exactness purely categorically, but I might be wrong.

Example. Let $\mathfrak{a} \subseteq A$ be an ideal. Then $S^{-1}\mathfrak{a} \subseteq S^{-1}A$ is an ideal. (We may need a canonical identification to get the last inclusion.)

From here, we only fix the ring $A$ not necessarily $S$.

Primality and localization. Let $\mathfrak{p}$ be an ideal of $A$. Then

$A \setminus \mathfrak{p}$ is a submoid if and only if $\mathfrak{p}$ is prime.

(That is, full information about primality of $\mathfrak{p}$ is in the determination of whether $A \setminus \mathfrak{p}$ is a submonoid!)

For a prime ideal $\mathfrak{p}$, we thus consider the submonoid $S := A \setminus \mathfrak{p}$ and define

$A_{\mathfrak{p}} := S^{-1}A$.

For an $A$-module $M$, we define $M_{\mathfrak{p}} := S^{-1}M$.

Theorem 1.3. Let $\mathfrak{p} \subseteq A$ be prime. Then $A_{\mathfrak{p}}$ is a local ring with the maximal ideal $\mathfrak{p}_{\mathfrak{p}}$.

Proof. Any element $a/s \notin \mathfrak{p}_{\mathfrak{p}}$ can be assumed to be represented by $a \notin \mathfrak{p}$ and $s \notin \mathfrak{p}$, so it is a unit in $A_{\mathfrak{p}}$. Q.E.D.

A pathological remark. If there is a prime ideal $\mathfrak{p} \subseteq A$, then $A \neq 0$ and $A_{\mathfrak{p}} \neq 0$.

Proposition 1.4. For any submonoid $S \subseteq A$, we have

$S^{-1}A = 0$ if and only if $0 \in S$.

Proof. 0/1 = (1/0)(0/1) = 1/1. Q.E.D.

Remark. I think this proposition is the canonical answer for questions about division by zero.

The following is the most important statement of this post! It tells us how to find the prime ideals in $S^{-1}A$.

Theorem 1.5. For any submonoid $S \subseteq A$, the functor $S^{-1}$ “cuts out” a bijection

$S^{-1} : (\text{prime ideals of } A \text{ which do not meet }S) \rightarrow (\text{prime ideals of } S^{-1}A)$,

given by $\mathfrak{p} \mapsto S^{-1}\mathfrak{p}$.

Remark. We need $\mathfrak{p} \cap S = \emptyset$ to ensure that $S^{-1}\mathfrak{p} \subseteq S^{-1}A$ is a prime ideal. Here is why.

Let $(ab)/(st) = (a/s)(b/t) = x/r \in S^{-1}\mathfrak{p}$. To show $S^{-1}\mathfrak{p}$ is a prime ideal, we just need $a \in \mathfrak{p}$ or $b \in \mathfrak{p}$. If not, we have $rab = stx \in \mathfrak{p}$, so one of $r, a, b$ must be in $\mathfrak{p}$, so $r \in \mathfrak{p}$.

If we have the condition $\mathfrak{p} \cap S = \emptyset$, then we contradict the last sentence of the previous paragraph to show that $S^{-1}\mathfrak{p} \subseteq S^{-1}A$ is a prime ideal. However, the theorem says much more. It says the disjointness of $\mathfrak{p}$ and $S$ is a necessary and sufficient condition for $S^{-1}\mathfrak{p}$ to be a prime ideal and all prime ideals of $S^{-1}A$ must be obtained this way.

Proof of Theorem 1.5. (Surjectivity) Let $\mathfrak{q} \subseteq S^{-1}A$ be a prime ideal. Consider

$\mathfrak{p} := \{a \in A : a/1 \in \mathfrak{q}\}$.

We clearly have $S^{-1}\mathfrak{p} = \mathfrak{q}$ and $\mathfrak{p} \cap S = \emptyset$ since $\mathfrak{q}$ cannot contain a unit. It is clearly an ideal and if $ab \in \mathfrak{p}$, then $ab/1 = x/1 \in \mathfrak{q}$ for some $x \in \mathfrak{p}$ and thus $uab = ux \in \mathfrak{p}$ for some $u \in S$. This means that one of $u, a, b$ is $\mathfrak{p}$, so $a \in \mathfrak{p}$ and $b \in \mathfrak{q}$ since $\mathfrak{p} \cap S = \emptyset$. This shows that $\mathfrak{p}$ is a prime ideal.

(Injectivity) Let $S^{-1}\mathfrak{p} = S^{-1}\mathfrak{p}'$ where $\mathfrak{p, p'} \subseteq A$ are prime ideals that do not intersect $S$. To show $\mathfrak{p = p'}$, it is enough to show that $\mathfrak{p \subseteq p'}$. Given $x \in \mathfrak{p}$, we have $x/1 = y/1$ for some $y \in \mathfrak{p'}$ This means that $sx = sy \in \mathfrak{p'}$ for some $s \in S$, so $x \in \mathfrak{p'}$ since $\mathfrak{p'}$ is prime and do not intersect $S$. Q.E.D.