Quasi-separated Schemes

Defn. A topological space is quasi-separated if the intersection of any two quasi-compact open sets is quasi-compact.  (In a separated space, the intersection of any two affine open sets is affine).

Claim. Affine schemes are quasi-separated.
Proof.  The quasi-compact open subsets of an affine schemes are all finite unions of distinguished open sets. So their intersection will again be a finite union of distinguished open sets.

(Note: quasicompact scheme $\leftrightarrow$ has a finite cover by affine opens).

Lemma. A scheme is quasi-separated iff the intersection of any two affine open subsets is a finite union of affine open subsets.

Proof. Suppose a scheme $X$ is quasi-separated. Take any two affine open subsets of $X$. Since they are quasi-compact, their intersection is also quasi-compact. As their intersection can be covered by affine open sets (e.g. by distinguished open sets), it can be covered by finitely many distinguished open sets.

Conversely, suppose the intersection of any two affine open subsets of $X$ is a finite union of affine open subsets. Take two quasi-compact open sets of $X$. Then each of this open set, being a quasi-compact scheme, is a finite union of affine opens. Their intersection will be the union of intersection of their affine open sets, which by hypothesis is a finite union of affine opens.
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In practice, we will see the hypothesis "$X$ is quasi-compact and quasi-separated" appear often. Usually, we would transfer such a hypothesis to the following statement, by using the above Lemma.

Proposition. A scheme $X$ is quasi-compact and quasi-separated iff $X$ can be covered by a finite number of affine open subsets, any two of which have intersection also covered by a finite number of affine open subsets.

Proof. $X$ is quasi-compact iff $X$ has a finite cover by affine opens. Then combine with Lemma above.
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Claim. Projective $A$-schemes are quasi-compact and quasi-separated.
Proof.  Let $S$ be a finitely generated graded ring over $A$ (i.e. $S_0 = A$) (by definition of projective $A$-schemes). Suppose $f_1, \ldots, f_n$ are generators of $S$ over $A$. Let $X = \text{Proj} A$. Then $D_{+}(f_i)$ cover $X$.  Notice $D_+{f_i} = \text{Spec} ((S_{f_i})_0)$ where $(S_{f_i})_0$ denotes the $0$-th graded piece of $S_{f_i}$, and $D_+(f_i) \cap D_+(f_j) = D_+(f_i f_j)$.
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Example. (Non-quasi-separated Scheme).

Let $X = \text{Spec}k[x_1, x_2, \ldots]$. Let $\mathfrak{m}$ be the maximal ideal $(x_1, x_2, \ldots).$. Let $U = X - [\mathfrak{m}].$ Let $Y $ be the result of gluing two copies of $X$ along $U$. Then $Y$ is not quasi-separated.
Proof. The two copies of $X$ are both quasi-compact, being affine schemes. However, their intersection is $U$, which is not quasi-compact.

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Claim. Locally Noetherian schemes are quasi-separated.

Open Subset of Affine NOT Quasi-compact

Let $X = \text{Spec} k[x_1, x_2, \ldots].$ Let $\mathfrak{m}$ be the maximal ideal $(x_1, x_2, \ldots).$. Let $U = X - [\mathfrak{m}].$ Then $U = \bigcup_{i} D(x_i)$.

Claim: $U$ is not quasi-compact

Proof.
 Suppose $U$ can be covered by finitely many of the $D(x_i)$'s say $U= \bigcup_{i=1}^n D(x_i)$. Then $(x_1, \ldots, x_n) \in U$ but $(x_1, \ldots, x_n) \not \in D(x_i)$, a contradiction.

Morphisms of Function Fields do NOT induce Rational Maps

We have seen that dominant rational maps between integral schemes induce morphisms of function fields.  The converse however, is not true.  Morphisms of function fields of two integral schemes do not always induce rational maps.

Example. Let $X = \text{Spec} k[x]$ and $Y = \text{Spec} k(x)$. Then their function fields are canonically isomorphic (both being $k(x)$).

Claim. There is no rational map $X \dashrightarrow Y$.

Proof. Suppose we have a rational map $f: U \to Y$ for some open dense subset $U$ of $X$. Shrinking $U$ if necessary (as all nonempty subsets of $X$ are dense), assume that $U = D(f)$. Thus $f$ induces a morphism $\varphi: k(x) \to k[x, 1/f]$.   Note that on stalks, this morphism should be identity $\varphi_{\eta}: k(x) \to k(x)$, so $\varphi$ must be identity on $k[x]$. This is impossible.
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For integral finite type $k$-schemes (in particular, irreducible varieties), morphisms of function fields do indeed determine rational maps between the schemes.

Proposition. Let $X$ be an integral $k$-scheme and $Y$ be an integral finite type $k$-scheme. Let $\varphi: K(Y) \hookrightarrow K(X)$ be a $k$-homomorphism. Then there exists a dominant $k$-rational map $\phi: X \dashrightarrow Y$ inducing $\varphi$.

Proof. The map $\phi$ is equivalent to a map from a non-empty open subset of $X$ to a subset $Y$ sending the generic point $\eta_X$ to $\eta_Y$. Moreover, as function fields of nonempty open subsets of $X$ (resp. $Y$) are canonically isomorphic to $K(X)$ (resp. $K(Y)$), we can assume $Y = \text{Spec} B$ and $X = \text{Spec} A$ are affine. We have a morphism $\varphi: B \hookrightarrow K(Y) \hookrightarrow K(X)$.

Warning: Clearly $\varphi$ induces a morphism $\text{Spec} \varphi(B) \to \text{Spec} B$. However we do NOT have a way to take $U = \text{Spec}\varphi(B)$ as an open subset of $X$.

So instead, we want to find some $A_g \subset K(X)$ such that $\varphi: B \hookrightarrow A_g$. This would induce a map a map $D(g) \to Y$.

Why can we find such an $A_g$? This is because $Y$ is a $k$-scheme of finite type. Thus $B = k[b_1, \ldots, b_n]$ for some $b_i$'s. Let $\varphi(b_i) = a_i/f_i$. Let $g = f_1 \cdots f_n$. Then as $\varphi$ is a $k$-homomorphism, the image of $B$ indeed lies in $A_g$.
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In particular if both $X$ and $Y$ are integral finite type $k$-schemes then a rational map $X \dashrightarrow Y$ that induces isomorphism $K(Y) \cong K(X)$ must be birational.

Function Field of an Integral Scheme

Let $X$ be an integral scheme. Then it is irreducible so it has a generic point $\eta$. Let $\text{Spec} A$ be a nonempty affine open of $X$ (thus must contain $\eta$). Let $K(A)$ denote the fraction field of $A$.

Claim. $(O_X)_{\eta} \cong K$ canonically.

Proof. We have $\text{Spec} A$ is itself irreducible, being a nonempty open subspace of an irreducible space. Thus $\eta$ corresponds to the generic point of $\text{Spec} A$, i.e. the nilradical of $A$ (which must be prime). However, since $A$ is integral, $\sqrt{0_A} = (0)$.
Therefore we have $(O_X)_{\eta} = A_{\sqrt{0_A}} = A_{(0)} = K(A)$.
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Let $V \supset U$ be nonempty open subsets of $X$.

Claim. The restriction map $\text{res}: O_X(U) \to O_X(V)$ is an inclusion.

Proof. Let $s \in O_X(U)$ be such that $s|_V = 0$. Notice that $U$ is irreducible, being a nonempty open subset of an irreducible space, so $V$ is dense in $X$. Thus as $s$ vanishes on $V$, it also vanishes on $U$. (Warning: this is NOT enough to conclude that $s = 0$). However, as $O_X(U)$ is reduced, its sections are uniquely determined by their values at points of $U$, so $s$ must indeed be $0$.
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Claim. The map $O_X(U) \to (O_X)_{\eta}= K(A)$ is an inclusion.

Proof. Suppose $s_{\eta} = 0$ for some $s \in O_X(U)$. Then $s|_V = 0$ for some $\emptyset \neq V \subset U$. Thus by the previous claim, $s = 0$.
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Advantage of Irreducible Varieties. 

Irreducible varieties form an important class of integral schemes. We thus see that for such varieties, all $O_X(U)$'s can be embedded into the same ring $K(X)$. As restriction maps are simply inclusion, sections glue iff they are the same element of $K(X)$.
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Dominant maps of Integral Schemes induce morphism of Function Fields.

Let $f: X\dashrightarrow Y$ be a dominant rational map. Then $f $ sends $\eta_X$ to $\eta_Y$, as dominant rational map between irreducible schemes preserve generic pt.  Thus it induces a map on stalk $(O_Y)_{\eta_Y} = K(Y) \to (O_X)_{\eta_X} = K(X).$

Dominant Maps and Generic Points

Proposition. Let $\varphi: A \to B$ and $f: \text{Spec} B \to \text{Spec} A$ be the induced map. Then $\text{Im} f$ is dense in $\text{Spec} A$ iff $\varphi^{-1}(\sqrt{0_B}) \subset \sqrt{0_A}.$
Equivalently, $\ker \varphi \subset \sqrt{0_A}.$

Proof.  Note that $\text{Im} f = \{ \varphi^{-1}(q) \mid q \in \text{Spec} B \}.$ So
$$\overline{\text{Im} f} = \mathbb{V}\left( \bigcap_{q \in \text{Spec} B} \varphi^{-1} q \right) $$
But $$\bigcap_{q \in \text{Spec} B} \varphi^{-1} q = \varphi^{-1} \left( \bigcap_{q \in \text{Spec} B} q\right) = \varphi^{-1}(\sqrt{0_B}).$$
Thus $$\overline{\text{Im} f} = \text{Spec} A = \mathbb{V}(\sqrt{0_A}) \iff \varphi^{-1}(\sqrt{0_B}) \subset \sqrt{0_A}.$$
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Note that for a set $S \in \text{Spec} B$, $$\overline{S} = \bigcup_{s \in S} \overline{\{s\}} = \bigcup \mathbb{V}(s) = \mathbb{V}\left( \bigcap \mathfrak{p}_s\right).$$

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If $\text{Spec} A$ and $\text{Spec} B$ are both irreducible, then their nilradicals are their generic points, so the above statement is equivalent to saying that $\text{Im} f$ is dense iff $f$ maps generic pt to generic pt. There is a generalization of this.

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Proposition. A rational map $\pi: X \dashrightarrow Y$ of irreducible schemes is dominant iff $\pi$ sends the genetic point $\eta_X$ of $X$ to the generic point $\eta_Y$ of Y. 

Proof. Let $(U, \pi)$ be a representation of $\pi$. Note that every open set of $X$ contains $\eta_X$.

Suppose $\pi(\eta_X) = \eta_Y$. Then $\pi(U) \ni \eta_Y$ so must be dense in $Y$.

Conversely, suppose that $\pi(U)$ is dense in $Y$.  Let $V\cong \text{Spec} B$ be an affine open of $Y$.   Cover $\pi^{-1}(V)$ by (finitely many) affine opens $U_i\text{Spec} A_i$. Then $\pi$ gives a morphism
$$\pi': \bigsqcup U_i \to V.$$
As $\pi$ is dominant, we must have image of $\pi'$ (which is equal to image of $\pi|_{\pi^{-1}(V)}$) is dense in $V$. On the other hand
$$\bigsqcup U_i = \bigsqcup \text{Spec} A_i = \text{Spec} \left(\prod A_i\right).$$ Let $A = \prod A_i$. Suppose $\varphi: B \to A$ is the induced map on global sections.

As the image of $\pi'$ is dense we know that $\varphi^{-1}(\sqrt{0_A}) \supset \sqrt{0_B}$, which is $\eta_Y$. On the other hand $(\sqrt{0_A}) = \bigcap_{p \in A_1, A_2, \ldots, A_n} p$  as $\text{Spec} A =  \bigsqcup \text{Spec} A_i$. Thus for each $A_i$, as $\bigcap_{p \in A_i} p \supset \bigcap_{p \in A_1, A_2, \ldots, A_n} p$ we have $\varphi^{-1}(\bigcap_{p \in A_i}) \supset \sqrt{0_B})$. But the former is just $\pi(\eta_X).$
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Note: Recall that every open subset of an irreducible space is irreducible.

Note: If $\eta$ is the generic point of an irreducible scheme $X$ and $U = \text{Spec} A$ is a affine open of $X$ then $\eta$ is the generic point of $U$ (i.e., corresponding to the nilradical of $A$).
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Quicker method: Suppose $\pi(\eta_X) \neq \eta_Y$. Then $U = Y - \overline{\pi(\eta_X)} \neq \emptyset$. As $\pi(X)$ is dense in $Y$, $U \cap \pi(X) \neq \emptyset$ so $\pi^{-1}(U)$ is not empty and thus must contain $\eta_X$, a contradiction.

Hypersurfaces have no Embedded Points

In this post, we assume the following properties of associated points of an $A$ module $M$ (for some Noetherian ring $A$).

1. Associated points of $M$ are precisely the generic points of irreducible components of support of some elements in $M$ (viewed as global section on $X = \text{Spec} A$).
2. $M$ has finitely many associated points
3. An element of $A$ annihilates some non-zero element of $M$ iff it vanishes at some associated point of $M$. 

Notice that $m_p = 0$ iff $x m = 0$ for some $x \not \in p$, i.e. $\text{ann}(m) \not \subset p$. In other words, $p \in \text{Supp}(m)$ iff $\text{ann}(m) \subset p$. 

So $\text{Supp}(m)$ consists of exactly the prime ideals of $A$ containing $\text{ann}(m)$. Thus its irreducible components should correspond to the minimal such prime ideals. In other words, the associated primes of $A$ lying inside support of $m$ are exactly minimal prime ideals containing $\text{ann}(m)$. Thus from (1) and (2), we already can deduce that if $f \in A$ annihilates some nonzero $m \in M$ then $f$ lies inside some associated point of $M$ (namely those lying inside $\text{Supp}(m)$).

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Lemma. Let $A = k[x_1, \ldots, x_n]/(f)$. Then $A$ has no embedded points.

Proof. Let $p$ be an associated prime. Then $p$ is a minimal prime ideal containing $\text{ann}(g)$ for some $g \in A^*$. We claim that $p$ is a minimal prime ideal containing $f$. Indeed, if $\text{ann}(g) \neq 0$, then $f \mid hg$ for some $h$ but $f \not \mid g$, so by unique factorization, $f$ and $g$ must have some greatest common divisor $1 \neq f_1 \neq f$. Suppose $f = f_1 f_2$. Then $\text{ann}(g) =  (f_2).$ Thus $p$ is a minimal prime ideal containing $f_2$, and must also be a minimal prime ideal containing $f$.  

Locus of Nonreduced points & Associated Points.

Before, we demonstrated that reduced-ness is not an open condition (though for quasi-compact schemes, we can still check reducedness at closed points). However, for Noetherian schemes, reduced-ness is indeed an open condition.

For this post, we only use two properties of associated points of an $A$-module $M$ for $A$ Noetherian.

• The associated points of $M$ are precisely generic points of irreducible components of supports of some global sections of $\widetilde{M}$.
• $M$ has finitely many associated points.
  
  Warning: we only talk about supports of global sections.

Proposition. Let $X = \text{Spec} A$. Show the the set of nonreduced points of $X$ is the closure of the nonreduced associated points. In other words, let $S$ be the set of non-reduced points of $X$. Then
$$S = \overline{S \cap \text{Ass} X}.$$

Proof.  Let $p$ be in the closure of the set of non-reduced associated points.
Then $p \in \overline{\{q\}}$ for some non-reduced associated point $q$, i.e.$p \supset q$. Thus we have  $A_q = (A_p)_q$. If $A_p$ is reduced then so must $A_q$, therefore $A_p$ is nonreduced.

On the other hand, suppose $p$ has nonreduced stalk, i.e. there is some (not necessarily global) section $s_p \neq 0$ such that $s_p^n = 0$ for some $n$. On a sufficiently small affine open $D(f) \ni p$, we could represent $s$ as $a/f$ for some $a \in A$. Thus we have $(af)$ is nilpotent (by proof of Lemma). Now $(af)_p \neq 0$ so $p \in \text{Supp}(af)$ thus it must be in the closure of some associated point $q$. (In particular, $(af)_q \neq 0$). On the other hand, as $D(f) \ni q$, and $(af)^n = 0$ on $D(f)$. Thus  $(af)^n_q = 0$, so $q$ is a non-reduced point.



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Thus we see that in particular, for an affine scheme $X$, the ``reduced locus'' of $X$ is open.

Recall that reduced-ness in in general not an open condition. However, from the above, we see that, for Noetherian schemes $X$, the "reduced locus" of $X$ is an open subset of $X$.
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In the above proof we have used the following Lemma.

Lemma. Localization of reduced rings is reduced.

Proof. Suppose $A$ is a reduced ring. Let $B = S^{-1} A$.
Suppose $a/s$ is a nilpotent in $B$. This means $a/s \neq 0$ in $B$ and $(a/s)^m = 0$ in $B$. In other words, $x a \neq 0$ for all $x \in S$ and $ya^m = 0$ for some $y \in S$. Thus $ya \neq 0$ but $ya^m = 0$, so $ya$ is a nilpotent in $A$, contradiction.

Lemma. If $s_p \neq 0$ then $p$ is in the support of some global section.

Proof. If $s_p \neq 0$, then $s$ is not $0$ on some neighborhood of $p$. WLOG, assume $s \neq 0$ on some $D(f) \ni p$. Suppose $s$ is represented by $a/f$ on $D(f)$. Then $a_p \neq 0$.

Reduced Schemes have no Embedded Points

In this Post, we only talk about associated points of an affine scheme $X = \text{Spec} A$.

We only use the following axiom of associated points: associated points are generic points of irreducible components of the support of sections.
Note that $X$ is itself is always a possible support (of the function $1$, for example), so the generic points of its irreducible components are associated points. The other associated points are called embedded points.

Example: Let  $A= k[x,y]/(xy, y^2)$.

 Possible supports of sections:
$\emptyset$ = $\text{Supp}(0). $
$X = \text{Supp} f$, for $f \not \in (y)$.
$\{(x,y)\} = \text{Supp}(f)$, $ f \in (y) \backslash (y^2, xy)$.

Thus the associated points of $X$ are $(y)$ and $(x,y)$. Among these, only $(x,y)$ is an embedded point.

Notice that  $A$  has an embedded point corresponding exactly to the unique non-reduced point

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Reduced Schemes have no Embedded Points

Let $A$ be an integral domain. Then as the only possible support of sections of $A$ is $\text{Spec} A$,  the the generic point is the only associate point of $\text{Spec}A$.
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Lemma. If $A$ is reduced then $X = \text{Spec}A$ has no embedded points.

Proof. First consider the case when $\text{Spec} A$ is irreducible, i.e. $A$ is a domain. Then $(0)$ is the only associated point, but it is also the generic point of $X$, so there is no embedded point.

 Now let $X =\text{Spec}A$ be any reduced scheme. Let $f \in A^*$. Then $\text{Supp} (f) = \overline{D(f)}$.We claim that its irreducible components are also irreducible components of $X$ (thus all associated points are generic pts of irreducible components of $X$ and hence there is no embedded points).

Note that the irreducible components of a closed subspace $Y \subset X$ are always closed and irreducible in $X$, but might not be maximal.

Claim: the irreducible components of $\overline{D(f)}$ are exactly the irreducible components of $X$ that meet $D(f)$.

Clearly all irreducible components of $\overline{D(f)}$ are contained in some irreducible components of $X$ that meet $D(f)$. Thus it suffices to show that if $\mathbb{V}(\mathfrak{p})  \cap D(f) \neq \emptyset$ then $$\overline{\{\mathfrak{p}\}} =\mathbb{V}(\mathfrak{p})  \subset \overline{D(f)}, \text{i.e.} $$
$$\mathfrak{p} \in \overline{D(f)}.$$
Suppose $\mathbb{V}(\mathfrak{p})  \cap D(f) \neq \emptyset$, i.e. for some $\mathfrak{q} \supset \mathfrak{p}, \mathfrak{q} \not \ni f$. Then in particular, $\mathfrak{p} \not \ni f$. (Or we can see that $\mathbb{V}(\mathfrak{p}) \not \subset \mathbb{V}(f)$.)
Thus $\mathfrak{p} \in D(f)$.
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Another way to phrase it is that if $D(f) \cap \mathbb{V}(\mathfrak)(p)$ is not empty then it is dense in $\mathbb{V}(\mathfrak{p})$ so its closure in $\mathbb{V}(\mathfrak)(p)$ must be $\mathbb{V}(\mathfrak)(p)$. Hence $\overline{D(f)} \supset \mathbb{V}(\mathfrak)(p)$
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Support of a section

Defn. Let $\mathcal{F}$ be a sheaf (or a separated presheaf) of abelian groups on a scheme $X$. Let $s$ be a global section of $\mathcal{F}$. Define support of $s$ to be the set of points $p \in X$ where $s_p \neq  0 \in \mathcal{F}_p$.

Lemma: $\text{Supp} s$ is closed

Proof.  $s_p = 0$ iff $s$ is $0$ (as a section) on a neighborhood of $p$. Thus the complement of  $\text{Supp} s$ in $X$ is open.

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In general, for an $A$-module $M$ and $p \in \text{Spec} A$, we see that $p \in \text{Supp}(m)$ iff $m \neq 0 $ in $M_p$ $\iff$ $\text{ann}(m) \subset p$.

Claim: If $X = \text{Spec} A$ for a integral domain  $A$ and $f \in A^*$ then $\text{Supp} f = X.$

Proof.  For every $f \in A$, if $f \neq 0$ then $\text{ann}(f) = 0$ so $\text{Supp}(f) = \text{Spec} A$.
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Notice that if $s_p = 0$ iff $s$ vanishes on a neighborhood of $p$, so $s(p) = 0$. In other words, if $s(p) \neq 0$ then $p \in \text{Supp}(s)$, i.e.
$$\overline{D(s)} \subset \text{Supp}(f).$$
 However, the other direction is not true in general. For $X = \text{Spec} k[x,y]/(y^2, xy)$, the global section $s = y$ is $0$ at $p=(x,y)$ but $s_p$ is not zero. (Because there are extra differential operators at $p$).

Claim: If $X = \text{Spec} A$ for a reduced ring $A$ then $\text{Supp} f = \overline{D(f)}.$

Proof. Clearly if $p \in D(f)$ then $f(p) \neq 0$ so $f_p \neq 0$ and thus $\text{Supp} f \supset D(f)$, and thus $\text{Supp} f \supset \overline{D(f)}.$

(Another way to state the same thing: Suppose $p \in \overline{D(f)}$. Then every neighborhood of $p$ contains points of $D(f)$ so $f$ cannot vanish on any neighborhood of $p$.)

Conversely, suppose $p \in \text{Supp} f$. Then $f_p \neq 0$. Thus for all (sufficiently small) neighborhood $W \ni p$, we have $f|_W \neq 0$.

Warning: The common mistake here is to deduce from this that $W \cap D(f) \neq \emptyset$, by arguing that as $f|_W$ is not equal to the zero function, $f$ cannot vanish on all of $W$, i.e, $W \not \supset \mathbb{V}(f)$. However in general schemes, functions are NOT determined by their values at points. That is why we need to reduced hypothesis.  (The difference between two functions that equal pointwise lies in the nilradical, so if the nilradical is $0$, functions are determined by their values at points.)

As $W$ is itself a reduced scheme, and $f|_W$ is uniquely determined by its values at points of $W$, we do have that $W \not \subset \mathbb{V}(f)$ and thus $W \cap D(f) \neq \emptyset$.



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Example. Let $X = \text{Spec} k[x]/(x^2)$ be the fat point.  Let $p = (x)$ be the only point of $X$. Then
$Supp f = \emptyset$ if $f \in (x^2)$ (i.e. if $f \sim 0$ as elements of $k[x]/(x^2)$).
$Supp f = X$, otherwise.

Indeed, if $f$ does not vanish at $p$ to begin with, then $f_p \neq 0$. On the other hand, if $f(p) = 0$ and $f \neq 0$ then $f $ is a constant multiple of $x$. Now $x_p \neq 0$. This can be checked algebraically, but geometrically it is because a neighborhood of $p$ contains the operator: $f \mapsto $ the linear term of $f$. Notice that $x$ does not vanish under this operator.

See What does the Stalk Perceive for more explanation.

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Let $X = \text{Spec} A$. Let $f \in A$. Then $p$ is in $\text{Supp} f$ iff  $f$ is not in the kernel of
$A \to A_p$.

Notice $\text{Supp} (fg) = \text{Supp}(f) \cap \text{Supp} (g)$.


Example. Let $A = \frac{k[x,y]}{(y^2, xy)}$ and $X = \text{Spec} A$. Let $f \in A$. We claim that the support of $f$ is either $\emptyset, (x,y)$ or $X$.

Proof. Note the elements of $\text{Spec} A$ are prime ideals of $k[x,y]$ containing $(y)$. In other words, $(y)$ is the generic point of $X$.

Case 1. $f \not \in (y)$. Then $D(s)\ni (y)$  so $\text{Supp}(s) \supset \overline{D(s)} = X$.

Case 2. $f \in (y)$. Then $f$ vanishes at all points of $X$ since all prime ideals of $A$ contains $(y)$. (Another way to see this is to notice that $D(f)$ is an open subset of $X$ not containing the generic point so must be empty).

For all $p \neq (x,y)$, $p$ is a reduced point of $X$. As $f$ vanishes on a neighborhood of $p$ and $p$ is reduced, $f$ must actually be equal to the zero section on a neighborhood of $p$, i.e. $f_p = 0$. So $p \not \in \text{Supp}(f)$.

Thus we only need to check if $f_{(x, y)} = 0$

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Some other random reasonings that achieve the same thing:
Suppose $p \neq (x, y)$, i.e. $p \in D(x)$.

We have $s_p$ iff $s$ is $0$ on some neighborhood of $p$. Without loss of generality, assume that neighborhood is $D(xf) \subset D(x)$ for some $f$.

We have a morphism $A \to  O_X(D(x)) = A_x = k[x] \to O(D(xf))$ (given by $y \mapsto 0$). As $k[x]$ is an integral domain, the localization map $O_X(D(x)) \to O_X(D(xf))$ is injective,  so $s \in A$ restricts to $0$ in $D(xf)$ iff it restricts to $0$ on $D(x)$, i.e. iff it is in the kernel of
$$A \mapsto A_x$$
i.e.
$$k[x,y]/(y^2, xy) \to k[x]$$ given by $y \mapsto 0$ and $x \mapsto 0$.
The kernel of this map is $(y)$.

So for all $s \in (y)$, $s_p = 0$ at all $p \in D(x)$. We cannot conclude yet for all such $s$, that $\text{Supp} (s) = X - D(x) = (x, y)$. We need to check if $s_{(x,y)} = 0$.
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Let $p =(x,y)$. At the stalk of $X$ at $p$ there are extra differential vectors generated by the $x, y, x^2$. Thus $s_p = 0$ iff the $y$ term, $x$ term and $x^2$ term of $s$ is zero. A way to rephrase it is to think of the ``$y$-operator'', as a point $\text{Spec}k[y]/(y^2) \to X$. As its image is the point $(x, y)$, we can pullback maps on neighborhoods of $(x,y)$ to maps on neighborhood of $(y)$. In other words, we have a local homomorphism of stalks $A_p \to (k[y]/(y^2))_{(y)}$. Thus $s $ is $0$ in $(O_X)_p$ iff its pullback is $0$ in $k[y]/(y^2))_{(y)}$. Among the element $s \in (y)$, only those in $y^2$ pullsback to $0$. So if $s$ is a multiple of $y$, then $s_{(x,y)} \neq 0$, and thus $\text{Supp} (s) = (x,y)$.



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Lemma. Let $M$ be an $A$-module. Then the natural map
$$M \to \prod_{\text{associated }$p$} M_p$$
is an injection.

What does the stalk perceive?

Let $\mathcal{F}$ be a sheaf (or a separated presheaf) of abelian groups on a scheme $X$. Let $s$ be a global section of $\mathcal{F}$.


Notice that even $s(p) = 0$, $s_p$ could still be nonzero.

Example. We give an example to illustrate the fact that two sections become the same element in the stalk at $p$ iff they restrict to the same section on a neighborhood of $p$ (but not that they agree pointwise) (Of course, this is simply by definition of stalk).

 Let $X = \text{Spec} A$. Two global sections $f, g \in A$ satisfies $f_p = g_p$ iff $f_p - g_p$ is $0$ in $A_p$. So let's look at the kernel of $A \to A_p$.

These consist of $s \in A$ such that $h s = 0$ for some $h \not \in p$. Thus $s $ is $0$ on $D(h)$, which is a neighborhood of $p$. Thus if $s_p = 0$ then $s$ vanishes on a neighborhood of $p$.

The converse is not true.

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We see that if $s_p = 0$ then $s$ is $0$ on a neighborhood of $p$ and thus $s(p) = 0$. Is the converse true?

The answer is No. Warning:
 

$$s(p) = 0 \not \implies  s_p = 0.$$

Example. Let $X = \text{Spec} k[x]/(x^2)$. Then $X$ has only one point, but there are extra first derivative vectors at the single point of $X$ also. As in, two global sections of $X$ are identical iff they agree on points of $X$ and also have the same first derivative. Now let $s = x$. At the point $p= (x)$, we have $s(p) $ is $0$ in $ A/(x)$ (or equivalent, in $\text{Frac} A/(x) = A_{(x)}/(x)$). However $s_p$ is not $0$ in $A_{(x)} = (k[x])_{(x)}/(x^2)$. This is because $s= x$ is  $0$ in the latter iff $x f = x^2 g$ for some $f \not \in (x)$, which is impossible as $k[x]$ is a UFD. Thus $\text{Supp} s = X$.

Intuitively, what happened is that for $s_p$ to be $0$, $s$ needs to be $0$ on a neighborhood of $p$. In our case, a neighborhood of $p$ would include the extra derivatives at $p$.   So since $s$ only vanish at $p$ with order $1$, its stalk at $p$ is not $0$.

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The Property of being Reduced

click here

A scheme $X$ is said to be reduced if $O_X(U)$ is reduced for all open $U$. 

Key point

Reduced-ness is stalk-local and affine local. However, it is not an open condition . 

Stalk-local.

Reduced-ness is a stalk-local property: $X$ is reduced iff $(O_X)_x$ is reduced for all $x$.

Paraphrasing this, a scheme is not reduced iff its talk at some point is not reduced. 

proof. (Trivial) Suppose $X$ is not reduced. Then there exists a section nonzero $s \in O_X(U)$ such that $s^n = 0$ for some $n$. As $s$ is not $0$ and $O_X$ is a sheaf, $s_x \neq 0$ for some $x \in U$, but $s_x^n = 0$ in $(O_X)_x$.

On the other hand, suppose $[(s, U)]^n = 0$ in $(O_X)_x$ and $[(s, U)] \neq 0$. Then $s|_W \neq 0$ on some $x \in W \subset U$.

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Affine Local

Affine Communication Lemma.

A property $P$ enjoyed by some affine open sets of a scheme $X$ is called affine-local if

• If $\text{Spec} A \hookrightarrow X$ has property $P$ then so does $\text{Spec} A_f$.
• If $A = (f_1, \ldots, f_n)$ and if $\text{Spec} A_{f_i} \hookrightarrow X$ has property $P$ for all $i$ then so does $\text{Spec} A.$

Suppose $P$ is an affine-local property. Suppose $X = \bigcup_{i} \text{Spec} A_i$ where $\text{Spec} A_i$ all have property $P$. Then every open affine subset of $X$ has property $P$.

Reducedness is affine-local

Suppose $A$ is a ring generated by $f_1, \ldots, f_n$. Then $A$ is reduced iff each $A_i$ is.

$X$ is reduced iff X can be covered by affine open sets $\text{Spec} A$ where $A$ is reduced.

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Example: The scheme $X = \text{Spec} k[x,y]/(y^2, xy)$ is non-reduced. 

The ring $k[x,y]/(y^2, xy)$  contains the element $y \neq 0$ such that $y^2 = 0$. At the stalk of every point $p$, we will still have $y_p^2 = 0$. So $(O_X)_p$ is reduced iff $y_p = 0$. This is possible if we invert $x$, i.e. if $p \in D(x) = $ $x$-axis minus origin.

We show the origin is the only point at which $X$ is non-reduced.

Fuzzy Origin

Proof.  This is equivalent to saying that $\text{Spec} \left(k[x,y]/(y^2, xy)\right)_x$ is reduced, as 

the points of $X$ are prime ideals containing $y^2$ and $xy$, i.e. $(y)$- the $x$-axis.  We think of  $X = \text{Spec} k[x,y]/(y^2, xy)$ then as the $x$ axis with some derivatives at $(0,0)$. 

To say that the origin is the only non-reduced point is equivalent to saying that that on $x$-axis minus the origin, $X$ is reduced. In other words $O_X(D(x))$ is reduced. 

We have $A_x := \left(k[x,y]/(y^2, xy)\right)_x = k[x, y, x^{-1} ]/(y^2, xy)$. As $x$ is a unit in this ring and $xy = 0$, we must have $y = 0$ in this ring. So $A_x = k[x, y, x^{-1}]/(y) = k[x, x^{-1}]$ is reduced. 

At the origin, we have $$A_{(x, y)} = \left(k[x,y]/(y^2, xy)\right)_{(x, y)}  = \frac{k[x,y]_{(x,y)}}{(y^2, xy)}.$$

Thus $y$ is zero in $A_{(x,y)}$ iff the image of $y$ in $k[x,y]_{(x,y)}$ lies in $(y^2, xy)$. Thus for some $h$ outside of $(x, y) we have

$$h y = y^2 f + xy g.$$

Dividing both sides by $y$ gives $h \in (x, y)$ a contradiction. 

More Illuminating Explanation.   Remember that $\text{Spec} k[x,y]/(y^2, xy)$ is just the $x$-axis together with (the span of ) some extra derivative vectors at the origin $\displaystyle \frac{\partial}{\partial_x}, \frac{\partial}{\partial_y}, \frac{\partial^2}{\partial_{x^2}}$ (cf. Visualizing Nilpotents). For $y$ to be $0$ at the stalk of $\mathfrak{p}$ it has to be zero in a neighborhood of $\mathfrak{p}$ (where neighborhood here include the derivative vectors). So for  $y$ to be zero at the stalk of the origin, we not only need $y|_{(0,0)} = 0$ but also that $\frac{\partial}{\partial_x}, \frac{\partial}{\partial_y}$ and $\frac{\partial^2}{\partial_{x^2}}$ applied to $y$ is $0$. The later does not hold as $\frac{\partial}{\partial_y}(y) \neq 0.$  (It gives $y$, the $y$-term)

To say this rigorously, remember that $\frac{\partial}{\partial_y}(y)$ is just the image of $y$ in $k[x, y] \to k[x,y]/(y^2)$ and clearly the image of $y$ is not $0$.

Now how to formalize  "if $y$ vanishes at the stalk of the origin then $\frac{\partial}{\partial_y}(y) =0$"?  To view $\frac{\partial}{\partial_y}$ as an element of $\text{Spec} k[x,y]/(y^2, xy)$ is to have a morphism $k[x,y]/(y^2, xy) \to k[y]/(y^2)$ given by $x \mapsto 0$ and $y \mapsto y$, i.e every $F$ is mapped to its $y$-term.

 (Two polynomials $f, g \in k[x,y]$ are equal in the first ring iff $\displaystyle \frac{\partial}{\partial_x}, \frac{\partial}{\partial_y}, \frac{\partial^2}{\partial_{x^2}}$ applied to them are equal, whereas in the second ring, functions are identified if they agree under $\frac{\partial}{\partial_y}$. As $x$ has no $y$ term, it becomes $0$ in the latter.)

The map $\text{Spec} k[y]/(y^2) \to X$ induces  a pull-back map on stalks, so we have
$$\left(\frac{k[x,y]}{(y^2, xy)}\right)_{(x,y)} \to \left(\frac{k[y]}{(y^2)}\right)_{(y)}$$  

which is a local homomorphism.
If $y$ is $0$ in $\left(\frac{k[x,y]}{(y^2, xy)}\right)_{(x,y)} $ then its image in $\left(\frac{k[y]}{(y^2)}\right)_{(y)}$ must also be $0$. In other words, as the latter is just $\frac{k[y]_{(y)}}{(y^2)}$, we must have the image of  $y$ in $k[y]_{(y)}$ lie in $y^2$ and thus $yh = y^2 f$ for some $h, f \not \in (y)$. This is impossible by unique factorization in $k[y]$. Another way to see this is to see that if $y$ is $0$ in the ring above then $\frac{k[y]_{(y)}}{(y^2)} =  k$, i.e. $y^2$ is kernel of $k \to k[y]_{(y)}$ which is not possible as the latter is injective. 

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Not an open condition:

Reducedness is in general, not an open condition (see example below). However if $X$ is a locally Noetherian scheme then reducedness is open.

Example.
Let $X$ be $\text{Spec} A$ where

$$A = \text{Spec} \frac{\mathbb{C}[x, y_1, \ldots]}{(y_1^2, y_2^2, y_3^2, \ldots, (x-1)y_1, (x-2)y_2,  (x- 3)y_2, \ldots)}$$

Notice that prime ideals of $A$ are exactly prime ideals $\mathfrak{p}$ of $\mathbb{C}[x, y_1, y_2, \ldots]$ containing $ (y_1^2, y_2^2, y_3^2, \ldots, (x-1)y_1, (x-2)y_2,  (x- 3)y_2, \ldots)$. These are exactly prime ideals containing $(y_1, y_2, \ldots,)$.

So $\text{Spec} A$ as a set is homeomorphic to the $$\text{Spec} \frac{\mathbb{C}[x, y_1, \ldots]}{(y_1, y_2, y_3, \ldots } = \text{Spec} \mathbb{C}[x].$$

Claim: the nonreduced points in $X$ correspond to the set $\{\text{x - m}\mid m \in \mathbb{Z}\}$ in $ \text{Spec} \mathbb{C}[x]$. In particular the set of  reduced points in $\text{Spec} A$ is therefore not open.

Proof:  Indeed similar to the above example, $A_{\mathfrak{p}}$ is not reduced iff one of $y_1, y_2 \ldots$ is not zero in the stalk at $\mathfrak{p}$.

Clearly, at points $\mathfrak{p}$ where $(x-m)$ is invertible in $\mathfrak{p}$, we have $y_m = 0$ in $\mathfrak{p}$, so $y_m$ is not nilpotent. Thus $A$ is reduced at $\mathfrak{p}$ if for all $m$, $x- m$ is invertible in $\mathfrak{p}$. This is true if $x-m$ all lie outside $\mathfrak{p}$, i.e. $\mathfrak{p} \neq x-m$ for any $m$. (As $\mathfrak{p}$ is maximal).

Let $\mathfrak{p}$ correspond to $(x-m)$ for some $m$. Then in $A_{\mathfrak{p}}$, we invert $(x - n)$ for every $n\neq m$, so for all such $n$, $y_n = 0$. Thus
$$A_{\mathfrak{p}} = \text{Spec}\frac{\mathbb{C}[x,y_m]}{(y_m^2, (x-m)y_m)},$$
which reduces the problem to the previous example.

Relationship with quasi-compact schemes: Recall that for every open condition $P$, to check if all points of a  quasi-compact scheme $X$ has $P$, it suffices to check that all closed points do. Though reducedness is not an open condition, it is still possible to check the reducedness of a quasi-compact schemes by checking it at closed points.

Closed Points of Quasi-Compact Schemes (Part 2)

Recall that quasi-compact schemes contain closed points- in fact, every closed subset of a quasi-compact scheme has closed point. In particular, every $x \in X$ is the generalization of some closed point, so any covering of the closed points must also be a cover of $X$. Thus to check if an open property holds for all points of $X$, it suffices to check if it holds for all closed points of $X$.

How about for properties $P$ that are not open?
Strategy 1.

Suppose P is  a property such that
for every $x$ not satisfying $P$, $\overline{\{x\}}$ contains another point  not satisfying $P$.
Then P holds for all $x \in X$ iff $P$ holds for all closed points of $X$.

Proof.
If this is true then we have a descending sequence
$$\overline{\{x_1\}} \supsetneq \overline{\{x_2\}} \supsetneq \overline{\{x_3\}} \supsetneq \overline{\{x_4\}} \overline \cdots.$$
On each affine piece, the sequence terminates as affine schemes are Noetherian. As $X$ as finitely many affine pieces, the sequence terminates in $X$ also. Thus there is $\overline{\{x\}}$ which contains no non-$P$ point other than $x$. Thus $x$ must be a closed point. (Contradiction)
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Strategy 2. 

Suppose $P$ is a property (of rings) such that if $A$ satisfies $P$ then $S^{-1}A$ satisfies $P$ (for every multiplicatively closed subset $S\subset A$ not containing $0$). Then $P$ holds for (the stalk of) all $x \in X$ iff $P$ holds for all closed points of $X$.

Proof.  Let $x \in X$. Then $\overline{\{x\}}$ contains a closed point $y$. Let $U= \text{Spec} A$ be an affine neighborhood of $y$. Then $U$ must contains $X$ as $y$ is in the closure of $x$. Thus we have $\mathbb{V}(\mathfrak{p}_x) \ni \mathfrak{p}_y$ so $\mathfrak{p}_y \supset \mathfrak{p}_x$ so we have a localization map
$$A_{\mathfrak{p}_y} \to A_{\mathfrak{p}_x}.$$
As $A_{\mathfrak{p}_y}$ has property $P$ and $P$ is compatible with localization, we have $A_{\mathfrak{p}_x}$ also has $P$.

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Example.

If $X$ is a quasi-compact space that is reduced at closed points, then it is reduced everywhere.

(Note: reducedness is a stalk local property. A scheme is reduced, i.e. its rings of sections are reduced, iff its stalks are all reduced).
Proof. Here we do not need to show that reduced-ness is an open condition (NOT true in general).  Rather we see that reducedness is compatible with localization (inverting elements cannot create nilpotents) so we can use strategy 2. 

Visualizing Nilpotents (Brainstorming)

Warning: this is a free-writing scratch so might be  incoherent. 

Geometric Interpretation of non-radical I

For a ring $A$, radical ideals of $A$ are in bijective correspondence with closed subsets of $\text{Spec} A$. How should we think about a general ideal $I$ geometrically? 

Every ideal $I$ and its power $I^n$ of $A$ cut out the same closed subsets $\mathbb{V}(I)$ of $\text{Spec} A$ as $\sqrt{I}$. Thus we distinguish them instead by the closed scheme structures each puts on $\mathbb{V}(I)$  (i.e. by the structure sheaves). In particular, we think of $I$ instead as the closed subschemes $\text{Spec} (A/I)$. 

Non-reduced rings and Differential Informations.

Let $A$ be a finitely generated $k$-algebra. If $I$ as above is not a radical ideal, then $A/I$ is not a reduced ring. Classically, every reduced finitely generated $k$-algebra is the coordinate ring of some (classical) affine variety. Its elements corresponds to regular functions of the variety. How should we think about the elements of a non-reduced f.g. $k$-algebra geometrically? 

Taylor Expansion.

Classically if $A = A(V)$ is the coordinate ring of a variety $V\subset \mathbb{A}^n(k)$, then its elements are polynomials of $k[x_1, \ldots, x_n]$ up to the relation $f \sim g$ iff $f|_V = g|_V$ as functions.  If $A $ is not reduced then its elements will correspond to polynomials of $k[x_1, \ldots, x_n]$ up to a stricter relation. If $V$ is the variety associated to $A/\sqrt{0_A}$ the reduction of $A$, then for two polynomials $f$ and $g$ correspond to the same element of $A$, they not only need to be equal on $V$, but also need to satisfy some extra differential condition. 

Example.  $\text{Spec}\mathbb{C}[x]/(x)$ and $\text{Spec}\mathbb{C}[x]/(x^2)$ both cut out the origin in $\mathbb{A}^1$. For every function $f$ in $\mathbb{C}[x]$, its image in $\mathbb{C}[x]/(x)$ gives us $f(0)$, whereas its image in $\mathbb{C}[x]/(x^2)$ gives us the linear term of $f$

In other words, $\mathbb{C}[x]/((x-a)^2)$ gives us the linear term of the Taylor explansion of $f$ at $a$, and hence first order differential information about $f$ at $a$. 

 Thus

$$\mathbb{C}[x]/(x^2) = \{ f \in \mathbb{C}[x] \} / \sim$ where

$$f \sim g \iff  f(0) = g(0) \text{ and (the Taylor expansions at $0$) of $f$ and $g$  have the same linear term}  .$$

Example.  Let $f \in \mathbb{C}[x,y]$. When we reduce $f$ mod $(x, y)$, we lose all the term of (the Taylor expansion at $0$) of $f$ except $f(0)$.

In the above list, on the left we list the rings and on the right we list the information preserved by the image of $f$ in those rings, i.e. the monomials of (the Taylor expansion at $(0,0)$)  of $f$ whose coefficients will be preserved.  

• $\mathbb{C}[x,y]/(x,y) \leftrightarrow  x^0, y^0$
• $\mathbb{C}[x,y]/(x^2,y)= \{(0,0)\} \leftrightarrow  x, x^0, y^0$
• $\mathbb{C}[x,y]/(x,y)^2= \{(0,0)\} \leftrightarrow x, y, x^0, y^0$
• $\mathbb{C}[x,y]/(x^2,y^2)= \{(0,0)\} \leftrightarrow xy, x, y, x^0, y^0$
• $\mathbb{C}[x,y]/(y^2)= \{(0,0)\} \leftrightarrow x^2, xy, x, y, x^0, y^0$

Recall the the coefficient of $x^i y^j$ in the Taylor explansion of $f$ at $(0,0)$ is $$\frac{1}{i! j!} \frac{\partial^{i+j}}{\partial_x^i \partial_y^j} f.$$

Thus image of $f$ in the above rings gives differential information about $f$.

Remark: we see that the first derivatives of $f$ at $0$ are just coefficient of $x, y$ in $f$. Thus we can view them as ``functions'' on $x$, $y$, or more sensibly as linear functions, on the $k$-vector space with basis $x, y$. In other words, $\frac{\partial f}{\partial x}(0)$ and $\frac{\partial f}{\partial y}(0)$ are just elements of the $k$-vector space

$$ \left(\frac{(x, y)}{(x, y)^2}\right)^*.$$

Extra-infinitesimal Vectors

We can rephrase the above examples by adding infinitesimal points. We think of the points of $\text{Spec}\mathbb{C}[x]/(x^2)$ as $0$ together with an infinitesimal point $\epsilon_x$, such that $f (\epsilon_x)$ is just the coefficient of $x$ in (the Taylor expansion at $(0,0)$) of $f$. In that case the ring $\mathbb{C}[x]/(x^2)$ is now indeed just the polynomials of $\mathbb{C}[x]$ up to the equivalence relation

$$f \sim g \iff f(0) = g(0) \text{ and } f(\epsilon_x) = g(\epsilon_x).$$

Notice that these new ``points'' are not points in the usual sense since they do not preserve multiplication, i.e. $$f(\epsilon_x) g(\epsilon_x) \neq (fg)(\epsilon_x).$$

Points in the usual sense corresponds to evaluation map $\mathbb{C}[x]/(x^2) \to \mathbb{C}$. The vector $\epsilon_x$ is instead, a derivation $\partial_x: \mathbb{C}[x]/(x^2) \to \mathbb{C}$ as it satisfies the Leibnitz rule (coefficient of $x$ is $fg$ is ....- use this, polynomial multiplication and coefficient, to explain Leibnitz rule to calculus students later).

How many additional points?

Be careful though. Using the above the logic, we could think of the points of $\text{Spec}\left(\mathbb{C}[x,y]/(x^2,y^2)\right)$ as the origin together with an infinitesimal vectors $\epsilon_{xy}$, $\epsilon_{x}$, $\epsilon_{y}$. Intuitively, if our ring is over $\mathbb{R}$, when we know the derivative $$\frac{\partial f}{\partial_{x}}(0) \text{ and } \frac{\partial f}{\partial_{y}}(0) ,$$

we know all the ``directional derivatives'' of $f$ at $0$ since the above two vectors generated the tangent space of $\mathbb{A}^2$ at $0$. In general, the span of the two first order partial derivatives of $f$ at $0$ is canonically isomorphic to the space of all derivations on $\mathbb{C}[x,y]$.

So instead of adding just say $\epsilon_{x}$ and $\epsilon_{y}$, we should add their ``span''. In other words, we should think of  $\epsilon_{x}$ and $\epsilon_y$  as the tangent vectors $(1,0)$ and $(0, 1)$ in the tangent space of $\mathbb{A}^2$ at $(0,0)$ which we identify with $\mathbb{C}^2$. So when we add the points $(1,0)$ and $(0,1)$ to our space, we should also add their span in $\mathbb{C}^2$, i.e. the whole of $\mathbb{C}^2$.

How about  $\epsilon_{xy}$? What are the points we should add if we add $\epsilon_{xy}$? Remember $f(\epsilon_{xy})$ is just the coefficient of $xy$ in (the Taylor expansion at $(0,0)$) of $f$, and thus is (a multiple of ) $\frac{\partial f}{\partial x \partial y} (0, 0)$.  Using the same idea as before, we want to think of all second derivatives in two variables as a vector space, and then take the span of  $\frac{\partial}{\partial x \partial y}$ in that space. Namely we take the $3$-dimensional vector space with basis  $\frac{\partial}{\partial x \partial y}, \frac{\partial}{\partial x^2}, \frac{\partial}{\partial y^2}$. This is exactly the space

$$\frac{(x,y)^2}{(x, y)^3}.$$

Spec of Nilpotent rings has additional Infinitesimal vector

Example.

Let $\mathfrak{m} $ denote the ideal $(x,y)$ in $\mathbb{C}[x,y]$

Thus we can describe the sets of ``points''  of the following affine schemes as follows.

• $\text{Spec}\mathbb{C}[x,y]/(x,y) = \{(0,0)\} $
• $\text{Spec}\mathbb{C}[x,y]/(x^2,y)= \{(0,0)\}$ and span of $\{\frac{\partial }{\partial x}|_{(0,0)} \}$ in the tangent space of $\mathbb{C}[x,y]$ at $0$. The latter is also the space of derivations on $\mathbb{C}[x,y]$, as well as $(\mathfrak{m}/\mathfrak{m}^2)^*$. 
• $\text{Spec}\mathbb{C}[x,y]/(x,y)^2= \{(0,0)\}$ together with all "directional derivatives'' at $(0,0)$, viewed as first order derivations, or points of $(\mathfrak{m}/\mathfrak{m}^2)^*$
• $\text{Spec}\mathbb{C}[x,y]/(x^2,y^2)= \{(0,0)\}$, all points of $(\mathfrak{m}/\mathfrak{m}^2)^*$ and $\text{span} (xy) $ in $(\mathfrak{m}^2/\mathfrak{m}^3)^*$
• $\text{Spec}\mathbb{C}[x,y]/(y^2)= \{(0,0)\} $, (span of ) $\frac{\partial}{\partial_y}$, and for each $i$, (span of) $\frac{\partial^i}{\partial x^i}.$

Closed Points of Quasi-Compact Schemes (Part 1)

Does every nonempty scheme contains a closed point?

Answer: No. There exist nonempty schemes with no closed points.
However, for quasi-compact schemes the statement is true.

Recall the following
Easy Lemma: A scheme $X$ is quasi-compact iff it can be written as a finite union of affine schemes.
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Quasi-Compact Schemes have Closed Points

Let $X$ be a quasi-compact scheme

Proof 1:  (Stack Exchange) As $X$ is quasi-compact, it is a finite union of affine pieces $X = \bigcup_{i=1}^n U_i$. Notice that a subset of $X$ is closed iff it is closed in every $U_i$.

Suppose $X$ does not contain any closed point.

Let $x_1$ be a closed point of $U_1$, then $\overline{\{ x_1\}} \cap U_1 = \{x_1\}$, i.e. $x_1$ is the unique point in $U_1$ of its closure.

As $x$ is not closed, $x$ must not be closed in some $U_i$. WLOG, assume $x_1$ is not closed in $U_2$.  Then $\overline{\{ x_1\}} \cap U_2$, being a closed subset of an affine scheme, must contain some closed point $x_2$ of $U_2$ other than $x_1$.

Repeating the argument, we can find $x_3 \in \overline{\{ x_2\}} \subset \overline{\{ x_1\}} $ such that $x_3$ is closed in $U_3$ but $x_3 \not \in U_2 \cup U_1$. Continuing the process we will get $x_{n+1} \not \in U_{n} \cup U_{n-1}\cup \cdots \cup U_1$, a contradiction.

Remark.  We constructed a sequence $x_1, \ldots, x_n, \ldots$ satisfying:

• $x_i$ is a closed point of $U_i$;
• $x_j$ is a specialization of $x_i$ for every $j > i$;
• $x_i$'s are all distinct
• $x_j \not \in U_i$ for all $j < i$. 

Proof 2: (Akhil) Using Quasi-compactness + Zorn's Lemma, we can show that $X$ has a minimal nonempty closed set $Z$. We can put a scheme structure on $Z$ such that the underlying topology is the subspace topology. As $Z$ is a scheme, it contains an affine piece $U$. As $Z - U$ is a closed subset of $Z$, by minimality of $Z$ we must have $Z- U = \emptyset$, i.e. $Z = U$ is affine so it must contains a closed point.

Proof 3: (Mine)

If $X$ has no closed points, then the closure of every point contains a point other than itself.
Thus we have a nested sequence
$$\overline{\{x_1\}} \supsetneq \overline{\{x_2\}} \supsetneq \overline{\{x_3\}} \supsetneq \overline{\{x_4\}} \overline \cdots.$$
On each affine piece, the sequence terminates as affine schemes are Noetherian. As $X$ as finitely many affine pieces, the sequence terminates in $X$ also. Thus there is some $x$ such that $\overline{\{x\}}$ does not contain any closed point other than $x$, contradiction.

This kind of argument is useful in proving some property (that might not be open) holds for every point of $X$ if it holds for closed point of $X$. For example, we use this argument to check that for a quasi-compact scheme $X$, it suffices to check reducedness at closed points. 

Main Usage

We say that property $P$ of points of a scheme  is open if whenever a point $x$ satisfies $P$, a neighborhood of it must satisfy $P$ also. 

Let $X$ be a quasi-compact scheme. To show that all points of $X$ satisfies $P$. It suffices to show that all \emph{closed} points of $X$ do. 

Explanation: As every closed subset of a quasi-compact scheme is quasi-compact, it must contains a closed point. In particular, for every $x \in X$, $\overline{\{x\}}$ contains a closed point $y$. By definition of closure, we have that every neighborhood of $y$ must contain $x$. So any covering of closed points must also be a cover of $X$. 

Generic Points (Part 1)

Correspondence between points and irreducible closed subsets

Affine Schemes

We have a bijection $X=\text{Spec} A \leftrightarrow \{\text{Irreducible closed subsets of } X\}$ given by

$$x \mapsto \overline{\{x\}};$$

$$ \text{generic point of } Z \leftarrow Z.$$

Why is $\overline{\{x\}}$ irreducible? That is because it is equal to $\mathbb{V}(\mathfrak{p}_x).$

Why does every irreducible closed subset $Z$ of $X$ has a  generic point? Let $Z = \mathbb{V}(\mathfrak{p})$ for some prime ideal $\mathfrak{p} \in A$. Then $[\mathfrak{p}]$ is the generic point of $Z$ (i.e. its closure in $X$ is $Z$). '

Why is this generic point unique? $\mathbb{V}(\mathfrak{p}) = \mathbb{V}(\mathfrak{q})$ iff $\mathfrak{p}= \mathfrak{q}$.

General Schemes

Let $X$ be any scheme. We claim that there is a bijection

$$X \leftrightarrow \{\text{Irreducible closed subsets of } X\},$$

given by the same map as above.

Why is $\overline{\{x\}}$ irreducible? This is true for every topological space. Suppose $\overline{\{x\}} = Y_1 \cup Y_2 $ for some closed $Y_1, Y_2$ of $X$. Then one of the $Y_i$ must contain $x$ and therefore contain $\overline{\{x\}}$ by minimality of closure.

Why does every irreducible closed subset $Z$ of $X$ has a generic point?

Let $U \subset X$ be affine such that $U \cap Z \neq \emptyset$. Then $Z \cap U$ is irreducible in $U$ (by Lemma 1 below). So it has a relative generic point, i.e. a point $x \in U$ such that the closure of $x$ in $U$ is $Z \cap U$. (Note that the closure of $x$ in $U$ is $\overline{\{x\}} \cap U$.)

As $Z$ is closed, we have  the closure of $Z\cap U$ in $X$ is the same as its closure in $Z$, which is $Z$ as $Z$ is irreducible. As $\overline{{\{x\}}}$ is a closed subset of $X$ containing $Z \cap U$, it must contains $\overline{Z\cap U} = Z$.  

Thus $x$ is a generic point of $Z$. 

Lemma 1. Let $Z$ be an irreducible topological space and $U$ be a nonempty open subset of $Z$. Then $U$ is irreducible.

Proof.  Indeed, let $V$ be any nonempty open subset of $U$. Then $V = V' \cap U$ for some open subset  $V'$ of $Z$ so $V$ itself is open in $Z$. As $Z$ is irreducible, $V$ is dense in $Z$, i.e. $\overline{V} = Z$. Thus the closure of $V$ in $U$ is $\overline{V} \cap U = U$, so $V$ is dense in $U$.

Note here we use the following Lemma

Lemma 2. Let $A$ be an arbitrary subset of $X$. For all subset $B \subset A$, we have   $\overline{B} \cap A$ is the closure $B'$ of $B$ in $A$. 

Proof. $$B' = \bigcap_{F \text{ closed in } A, F \supset B } F = \bigcap_{F \text{ closed in }X, F \supset B} (F \cap A) = \left(\bigcap_{F \text{closed in X}, F \supset B} F\right) \cap A = \overline{B} \cap A.$$

-------------------------------

Construction of Projective Space (Part 2)

The Proj construction allows us to define the projective space $\mathbb{P}^n(k)$ in a new way, namely as $\text{Proj}k[x_0, \ldots, x_n]$. One advantage of this approach is that it emphasizes the symmetry of $\mathbb{P}^n(k)$: we glue all basic opens $D_+(f)$ instead of just $D_+(x_i)$, as in the previous approach. In particular, in this new construction, it is easy to see that $D_+(f)$ are affines, whereas in the previous construction, the corresponding sets are not clearly affine (except for $D(x_i)$.

We now show that these two constructions agree.

The Proj Construction

Let $S = S_0 \oplus S_1 \oplus \cdots$ be a $\mathbb{Z}_{\geq 0}$-graded ring. Call $S_{+} = S_1 \oplus S_2 \oplus \cdots$ the irrevelant ideal.  We call a homogeneous prime ideal $\mathfrak{p}$ of $S$  relevant if it does not contain $S_+$.

We will define $\text{Proj} S$.

As a Set

As a set, let $\text{Proj} S$ be the set of relevant homogeneous prime ideals of $S$. 

As a Topological Space

As a subset of $\text{Spec} S$, $\text{Proj}(S)$ inherits the Zarisky topology. In particular, its closed subsets are

$$\mathbb{V}_+(E) = \mathbb{V}(E) \cap \text{Proj} (S)$$. It has a basis of distinguished opens $D_+(f) = D(f) \cap \text{Proj} (S).$

As a Scheme

Clearly the $D_+(f)$'s (where $f$ runs over the homogeneous elements of $S$) form an open cover of $\text{Proj} S$. We will identify each $D_+(f)$ with an affine scheme in a compatible way. Thus gluing these schemes give a scheme structure for $\text{Proj} S$.

Affine structures on Distinguished Opens

Claim: $$D_+(f) \cong \text{Spec} S_{(f)}$$

where $\text{Spec} S_{(f)} $ is the $0$-degree component of the graded ring $S_f$. 

Proof. Define a map $\psi_f: D_+(f) \to \text{Spec} S_{(f)}$ by

$$\mathfrak{p} \mapsto \mathfrak{p} S_f \cap S_{(f)}.$$

(Note that the definition makes sense since if $\mathfrak{p} \not \ni f$ then $\mathfrak{p} S_f $ is a prime ideal of $S_f$ so its contraction to $S_{(f)}$ is a prime ideal of the latter.)

We want to show that this is a homeomorphism of topological spaces.

First, we show that $\psi_f$ is injective.

Indeed, suppose $\psi(\mathfrak{p}) = \psi(\mathfrak{q})$, i.e. the elements of degree $0$ in $\mathfrak{p}S_f$ are the same as the $0$-degree elements in $\mathfrak{q}S_f$. 

Let $d = \deg(f)$ and $h \in \mathfrak{p} $ be of degree $m$. Then $h^d/ f^m$ is of degree $0$ in $\mathfrak{p}S_f$ so must be an element of  $\mathfrak{q}S_f$. This implies $h^d/1 \in \mathfrak{q}S_f$ and thus $h \in \mathfrak{q}$. 

Here we used the obsevation that if $h^m/f^n \in \mathfrak{p}S_f$ for some $m, n$ then $h \in \mathfrak{p}$; and that we can turn every element of $\mathfrak{p}$ into an element of degree $0$ in $\mathfrak{p}S_f$. 

Now we show that $\psi_f$ is surjective. 

From the above discussion, we notice that $h \in \mathfrak{p}_m$ iff $\frac{h^d}{f^m} \in \mathfrak{p} S_f \cap S_{(f)}$. We use this as a characterization of the graded components of $\mathfrak{p}$.

Let $P$ be a prime ideal of $S_{(f)}$. For each $m$, define

$$\mathfrak{p}_m = \left\{h \in S_m \mid \frac{h^d}{f^m} \in P \right\}.$$

We need to check that the $\mathfrak{p}_m $ are subgroups of $S_m$, that they satisfy the grading criterion, so that $\mathfrak{p}:= \oplus \mathfrak{p}_m$ is a graded ideal of $S$. Then we need to check that $\mathfrak{p}$ is a relevant prime ideal.

It's easy to see that $\psi_f$ is a homeomorphism. 

L-valued pt of Projective Space versus Classical Points

In the previous post, we constructed $\mathbb{P}^n(k)$ by gluing affine schemes. Now we will show that if $k$ is algebraically closed, then the closed points of $\mathbb{P}^n(k)$ are in bijective correspondence with points of the classic projective space which we will denote $P^n(k)$. More generally, for any field $k$ and any extension $L$ of $k$, we have that the $K$-points of $\mathbb{P}^n(k)$, i.e. $\hom_k(\text{Spec} L, \mathbb{P}^n(k))$ is in bijective correspondence with $P^n(L)$.

$L$-valued points of $\mathbb{P}^n(k)$ are Classical Points

 Let $\overline{x}: \text{Spec} L \to \mathbb{P}^n$ be an $L$-valued point of $\mathbb{P}^n(k)$ with image $x$. Then $x$ lies in one of the pieces $X_i = \text{Spec} A_i$ where $$A_i = k\left[\frac{t_0}{t_i}, \ldots, \frac{t_n}{t_i}\right].$$ So $\overline{x}$ corresponds to a morphism $\sigma_i: A_i \to L$, so we can associate to it the point $$\left[\sigma_i\left(\frac{t_0}{t_i}\right) : \ldots: 1 :\ldots: \sigma_i\left(\frac{t_n}{t_i}\right)\right]$$ of $P^n(L)$.

Well-defined.

 This assignment is well-defined. Indeed, if $x \in X_j$ also, then $\overline{x}$ factors through $X_i \cap X_j$. This means that $\sigma_i$ and $\sigma_j$ \emph{extends} to a homomorphism $$\sigma_{ij}: A_{ij} = k\left[t_0, \ldots, t_n, \frac{1}{t_i}, \frac{1}{t_j}\right] \to L.$$
Then we can write $$\left[\sigma_j\left(\frac{t_0}{t_j}\right) : \ldots: \sigma_j\left(\frac{t_n}{t_j}\right)\right] $$ as $$\left[\sigma_{ij}\left( \frac{t_i}{t_j}\right)\sigma_i\left(\frac{t_0}{t_i}\right) : \ldots: \sigma_{ij}\left( \frac{t_i}{t_j}\right)\sigma_i\left(\frac{t_n}{t_i}\right)\right]$$ which is the same as $$\left[\sigma_i\left(\frac{t_0}{t_i}\right) : \ldots: \sigma_i\left(\frac{t_n}{t_i}\right)\right].$$

Bijection

For each choice of element of $X_i$ we get a unique morphism $\sigma_i: A_i \to L$ and hence from $\text{Spec} L \to X_i \to X$. 

If $k= \overline{k}$ then closed points of  $\mathbb{P}^n(k)$ are Classical Points

It suffices to show that for $k$ algebraically closed, and $X$ any scheme over $k$ i.e. with a canonical morphism $X \to \text{Spec }k$.
$$ k\text{-points of $X$} \leftrightarrow \text{ closed points of $X$}.$$

Notice that for every point $x \in X$, restricting the canonical morphism to an affine piece piece $\text{Spec} A$ over $X$ gives a homomorphism $k \to A$. Thus this induces a map $k \to \kappa(x)$

Indeed let $\overline{x}: \text{Spec} k \to X$ be a $k$-point with image $x$. Then, as this is a morphism of schemes, we have an induced map on function fields $\kappa(x) \to k$  compatible with the canonical morphism $k \to \kappa(x)$. So we must have $\kappa(x) \cong k$ canonically. On the other hand $\kappa(x) = \text{Frac}(A/\mathfrak{p}_x) \supset A/\mathfrak{p}_x \supset k$

Do I need $A$ to be finitely generated?

n-dim Projective Space is NOT affine if n > 0

In the previous post, we glue together affine spaces $X_i = \text{Spec} A_i = \text{Spec} k\left[x_0, \ldots, x_n, \frac{1}{x_i} \right]$ to get $\mathbb{P}^n(k)$.

To show that $\mathbb{P}^n(k)$ is not affine, we first compute its global sections.

The global sections of $\mathbb{P}^n$ are in bijective correspondence with tuples $(s_0, \ldots, s_n) \in A_0 \times \cdots \times A_n$ that are compatible with respect to restrictions. However as restrictions $A_i \to A_{ij}$ are simply inclusion maps, we have $(s_i)_{X_{ij}} = (s_j)_{X_{ij}} $ iff $s_i = s_j$ as elements of $A_{ij}$. So we have
$$s_0 = s_1 = \ldots = s_n \text{ as elements of } k(x_0, \ldots, x_n).$$
Thus the global sections $\mathbb{P}^n$ is $\bigcap_i A_i = k$. If $\mathbb{P}^n(k) $ is affine, then it must then be $\text{Spec} k$. But if $n >0$, then $\mathbb{P}^n(k)$ consists of more than one point. In fact, its closed points are in bijective correspondence with points of the classical projective  Space.

Construction of Projective Space (Part 1)

We construct the Projective Space $\mathbb{P}^n(k)$ in two ways: via gluing and via the Proj construction. In this first post. We will glue Affine Schemes $\mathbb{A}^n(k)$ to get $\mathbb{P}^n$.
We first present a pedantic construction, then we introduce a simpler way of making the same construction.

Constructing $\mathbb{P}^n$ by gluing.

Classical Construction

Before proceeding to the scheme case, let us look at the the classical $$\mathbb{P}^n(k) = \{[X_0: \ldots: X_n] \mid  X_i \text{ not all $0$} \}/ \sim.$$

We see that $\mathbb{P}^n$ is covered by pieces $U_i = \{ X_i \neq 0\}$. We can identify each $U_i$ with affine space $k^n$ (i.e. put affine coordinates on $U_i$) via the map $\phi_i: U_i \to k^n$ given by

$$[X_0: \ldots: X_n] \mapsto (X_{0}/X_i, \ldots, \widehat{X_i/X_i}, \ldots, X_n/X_i).$$

If $x \in U_i \cap U_j$, what is the relationship between the coordinate $\phi_i(x)$ and $\phi_j(x)$?

Suppose $(x_{0/i}, \ldots, \widehat{x_{i/i}}, \ldots, x_{n/i})$ denotes the coordinate of $\phi_i(x)$. Then we must have $\phi_j(x) = (\frac{x_{0/i}}{x_{j/i}}, \ldots, \widehat{x_{i/i}}, \ldots, \frac{x_{n/i}}{x_{j/i}}).$

In other words, we have a map between coordinate rings of $U_i$ and $U_j$, viewed as affine varieties under identification $\phi_i$ and $\phi_j$ given by

 $$A(U_j) =  k[x_{0/j}, \ldots, \widehat{x_{j/j}}, \ldots, x_{n/j}] \to  A(U_i) =  k[x_{0/i}, \ldots, \widehat{x_{i/i}}, \ldots, x_{n/i}] $$

$$x_{0/j} \mapsto \frac{x_{0/i}}{x_{j/i}}$$ and so on.

Gluing Schemes  

Motivated by the above discussion, for $i = 0$ to $n$, we let $X_i = \mathbb{A}^n = \text{Spec} k[x_{0/i}, \ldots, \widehat{x_{i/i}}, \ldots, x_{n/i}] = \text{Spec} \left(k[x_{0/i},  \ldots, x_{n/i}] /(x_{i/i} - 1)\right).$

Let $$X_{ij} = \text{Spec} \left(k\left[\frac{x_{0/i}}{x_{j/i}}, \ldots, \frac{x_{n/i}}{x_{j/i}} \right]/(x_{i/i}- 1)\right) = \text{Spec}\left(k\left[x_{0/i}, \ldots, x_{n/i}, \frac{1}{x_{j/i}}\right]/(x_{i/i} - 1) \right) =: \text{Spec} A_{ij}$$

be an open subset of $X_i$.

We have the following identification $\varphi_{ij}: X_{ij} \to X_{ji}$ corresponding to the ring homomorphism $A_{ji} \to A_{ij}$ given by

$$ x_{0/j} \mapsto \frac{x_{0/i}}{x_{j/i}}$$

and so on.

Note that this map is well defined since $x_{j/j} - 1$ is mapped to $\frac{x_{j/i}}{x_{j/i}} - 1 = 0$.

It is easy to see that $\varphi_{ij}$ and $X_{ij}$ satisfies the cocycle condition. This is because 

$X_{ij} \cap X_{ik}$ is affine and equal to $\text{Spec}(A_{ijk})$ where

$$A_{ijk} =  k\left[x_{0/i}, \ldots, x_{n/i}, \frac{1}{x_{j/i}}, \frac{1}{x_{k/i}}\right]/(x_{i/i} - 1).$$

As before, the map $\varphi_{ij}|_{X_{ijk}}: X_{ijk} \to X_{jik}$ (restriction of $\varphi_{ij}$) corresponds to the map $A_{jik} \to A_{ijk}$ (extension of the previous map) given by

$$x_{0/i} \mapsto \frac{x_{0/i}}{x_{j/i}}.$$

In particular $\frac{1}{x_{j/i}} \mapsto 1$ and $\frac{1}{x_{k/i}} \mapsto \frac{x_{k_i}}{x_{j/i}}.$

On the other hand the map $\varphi_{kj} \circ \varphi_{ik}|_{X_{ijk}}$ corresponds to the map

$A_{ijk} \to A_{kji} \to A_{jik}$ given by

$$x_{0/i} \mapsto \frac{x_{0/i}}{x_{k/i}} \mapsto \frac{\frac{x_{0/i}}{x_{k/i}}}{\frac{x_{j/i}}{x_{k_i}}} = \frac{x_{0/i}}{x_{j/i}}.$$

Thus we can glue $X_i$ along $X_{ij}$ via $\varphi_{ij}$ to get a unique scheme $\mathbb{P}^n$.

Simpler Method. (Bosch)

In the previous post, we constructed $\mathbb{P}^n(k)$ by gluing $(n+1)$-copies $X_i$'s of $\mathbb{A}^n$. In that post, we think of $X_i=\mathbb{A}^n $ as $\text{Spec}k\left[x_{0/i}, \ldots, x_{n/i}\right]/(x_{i/i} -1).$ A simpler way to rewrite the construction is to use a different coordinate on each $X_i$, namely to think of $X_i$ as $\text{Spec} A_i$ where $$A_i = k\left[\frac{x_0}{x_i}, \ldots, \frac{x_n}{x_i}\right] = k\left[x_0,\ldots, x_n, \frac{1}{x_i}\right],$$ viewed as subrings of $k(x_0, \ldots, x_n).$ Then we see that $X_{ij}:= (X_i)_{x_j}$ and $X_{ji}:= (X_j)_{x_i}$ can be canonically identified with $\text{Spec} A_{ij}$ where $$A_{ij} = k\left[x_0, \ldots, x_n, \frac{1}{x_i}, \frac{1}{x_j}\right].$$ (Notice that the natural injection from $A_i$ and $A_j$ into $A_{ij} = k[x_0, \ldots, x_n, \frac{1}{x_i}, \frac{1}{x_j}]$ corresponds to restriction maps from $X_i \to X_{ij}$ and $X_j \to X_{ji}$, respectively.) This identification is transitive, hence satisfies the cocyle condition.

Elementary Examples of Nonaffine Schemes

If only all schemes were affine!

Today we look at the simplest examples of non-affine schemes. How do we show that they are not affine?

 These schemes $X$ will be obtained by gluing affine pieces. Thus we can compute their global sections $O_X(X)$: these correspond to compatible sections on each affine piece. Using the fact that the functor Spec gives an equivalence of categories, we argue that if $X$ is affine, it must equals $\text{Spec}(O_X(X)).$ By comparing number of points, we argue that this is impossible.

Example 1. The Punctured Plane 

Let $U = \mathbb{A}^2 - \{(0,0)\}$,  be equipped with the structure of open subscheme  of $\mathbb{A}^2=: X$.

We compute its global section. To this end, we think of $U $ as $D(x) \cup D(y)$. So global sections of $U$ correspond to $(s, t) \in O_X(D(x)) \times O_X(D(y))$ that agree on intersection $O_X(D(xy))$.

Notice that the corresponding restriction maps are just inclusion $k[x,y]_x \hookrightarrow k[x,y]_{x,y}$ and $k[x,y]_y \hookrightarrow k[x,y]_{(x, y)}.$ So $s$ and $t$ are compatible iff $s = t$. Thus $O_U(U) = O_X(D(x)) \cap O_X(D(y)) = k[x,y]_x \cap k[x,y]_y = k[x, y].$

If $U$ is affine, it must therefore be $X = \mathbb{A}^2.$

However, the function $x$ and $y$ have common zeroes on $X$, but they not have any common zeroes in $U$.

Another way to phrase it is that we should have a bijection:
$ \{ \text{Prime ideals of $O_U(U)$} \} \leftrightarrow U$ given by
$\mathfrak{p} \mapsto \text{generic point of } \mathbb{V}(\mathfrak{p})$
$\mathbb{I}(x) \leftarrow x.$
So in particular the prime ideal $(x, y)$ should cut out a point in $U$. But it does not.

Example 2. Affine line with doubled origin

We glue $X = \text{Spec} k[t]$ and $Y = \text{Spec}  k[u]$ together along $X \supset U = \text{Spec} k[t, t^{-1}]$ and $Y \supset V = \text{Spec} k[u, u^{-1}]$ via the isomorphism

$ k[t, t^{-1}] \to k[u, u^{-1}]$ given by $t \mapsto u$.

To show that this scheme is not affine, compute its global section in the same way as above.

Example 3. Projective Line 

We glue $X = \text{Spec} k[t]$ and $Y = \text{Spec}  k[u]$ together along $X \supset U = \text{Spec} k[t, t^{-1}]$ and $Y \supset V = \text{Spec} k[u, u^{-1}]$ via the isomorphism

$ k[t, t^{-1}] \to k[u, u^{-1}]$ given by $t \mapsto 1/u$.

We compute the global sections of $\mathbb{P}^1$. These correspond to pairs of sections $(s_1, s_2 ) \in O_X(X) \times O_Y(Y)$ that agree on intersection $U \cong V$.

Notice that corresponding restriction maps  are injections, so $s_1|_U = s_1$ and $s_2|_V = s_2$. We use the map $O_X(U) \to O_Y(V)$ to identify $s_1 = f(t)$ as the section $f(1/u)$ on $V$.  So must have $f(1/u) = s_2$ as elements of $k[u, u^{-1}]$. As $s_2 \in k[u]$, this means $s_2 \in k$, and so is $s_1$, and $s_1 = s_2$.

So $\Gamma(\mathbb{P}^1, O_{\mathbb{P}^1}) = k$. If $\mathbb{P}^1$ were affine, it must equal $\text{Spec} k$ which consists of a single point.

Example 4. $\mathbb{P}^n$ for $n>0$.

Induced morphim on Spec

Classical Case

Finding pullback

Question: Suppose $X \subset k^m$ and $Y \subset k^n$ are irreducible algebraic sets (with Zarisky topology). Let $f: X \to Y$ be given by
      $(x_1, \ldots, x_m) \mapsto (f_1(x), \ldots, f_n(x)).$

What is the induced map on coordinate rings?

Answer: The induced maps $f^{\sharp}: A(Y) = k[y_1, \ldots, y_n]/ I  \to A(X) = k[x_1. \ldots, x_m]/ J$ is given by pre-composition with $f$ so it is
   $y_i \mapsto f_i$.

Recovering from Pullback

Question: On the other hand, suppose we have a map $\varphi: A(Y) \to A(X)$. How do we recover $f$ such that $\varphi = f^{\sharp}$?

Answer: $f: X \to Y$ can be defined by
 $(x_1, \ldots, x_m) \mapsto \left(\varphi (y_1)(x), \ldots, \varphi(y_n)(x)\right).$

Affine Scheme Case.      

In the examples below, we first define the maps on points on $\mathbb{C}^n$. Then we interpret the map as the maps on coordinate rings as before. Afterwards, we recover the map on $\text{Spec} \mathbb{C}^n$.

Example. Parabola

Let $f: \mathbb{C} \to \mathbb{C}$ be given by $x \mapsto y = x^2$. Then the induced map on coordinate rings is $f^{\sharp}: \mathbb{C}[y] \to \mathbb{C}[x]$ given by $y \mapsto x^2$.

We claim that the induced map on Spec is

$ f: \text{Spec} \mathbb{C} \to \text{Spec} \mathbb{C}$
$ (x - a) \mapsto (y - a^2)$. 

Indeed $f(x-a) = (f^{\sharp})^{-1} (x-a)$. Now $f^{\sharp}(y - a^2) = (x^2 - a^2) \in (x-a)$ so $(f^{\sharp})^{-1} (x-a) \ni (y- a^2)$, but the latter is maximal. 

 Generalization

Suppose $f: k^m \to k^n$ is given by $x \mapsto (f_1(x), \ldots, f_n(x))$, where $f_1, \ldots, f_n \in k[x_1, \ldots, x_m]$. Then as before, it induces $f^{\sharp}: k[y_1, \ldots, y_n] \to k[x_1, \ldots, x_m]$ be given by

$y_i \mapsto f_i$. 

Note: We could have also just started with $f^{\sharp}.$

For any ideal $I \subset k[x_1, \ldots, x_m]$ and $J \subset k[y_1, \ldots, y_n]$ such that $I \supset \phi(J)$ we have a morphism 

$f: \text{Spec}  k[x_1, \ldots, x_m]/ I \to k[y_1, \ldots, y_n]/J$. 

Claim: $f$ sends points $(x_1 - a_1, \ldots, x_m - a_m)$ to $(y_1 - f_1(a), \ldots, y_n - f_n(a)).$

Proof: As $(y - f_1(a), \ldots, y - f_n(a))$ is a maximal ideal, it suffices to show that it is in the preimage under $f^{\sharp}$ of $(x_1 - a_1, \ldots, x_m - a_m)$. Indeed, 

$$f^{\sharp}(y_1 - f_1(a), \ldots, y_n - f_n(a)) = (f_1(x) - f_1(a), \ldots, f_n(x) - f_n(a)) \in (x_1 - a_1, \ldots, x_n - a_n) $$  as the latter is the kernel of the evaluation at $a$ map. 

Thus we see that the induced map on Spec generalizes the induced map on classical algebraic varieties.

Morphisms between Locally Ringed Spaces

Locally ringed spaces are generalization of differential manifolds, and classical algebraic varieties. The notion of morphisms of ringed spaces then, should be a generalization of the notion morphisms in the latter categories. Thus, to define morphism between ringed spaces, we need to first be able to rephrase the notion of differentiable maps and morphisms of varieties in a way that only uses the structure sheaf.  (Recall that classically, varieties are irreducible subset of $k^n$ with Zarisky topology, and their morphisms are defined to be restrictions of polynomial maps). The idea is that a continuous map between two differentiable manifolds is differentiable iff it pullbacks differentiable functions to differentiable functions. In the case of classical varieties, a morphism is a continuous map which pullbacks regular functions to regular functions.

In a general ringed space $X$, the sections of $O_X$ cannot be canonically interpreted as functions, so a morphism of ringed spaces $(X, O_X)$ and $(Y, O_Y)$ consists of a pair $(f, f^{\sharp})$ where
$f: X \to Y$ is continuous
and $f^{\sharp}: O_Y \to f_* O_X$.
This $f^{\sharp}$ is an artificial version of pullback of $f$.

On the other hand, if $X$ is a locally ringed space, then given a section $s \in O_X(U)$ and $x \in X$, we can define $s(p) $ to be the image of $s$ in $O_X(U) \to (O_X)_x \to (O_X)_x/\mathfrak{m}_x = \kappa(x)$. Thus we can put restriction of $f^{\sharp}$ to make it resemble a pullback. We will require that

if $t$ vanishes $ f(x)$ then  $f^{\sharp} (t) $ vanishes at $x$.

In other words, if $t \in \mathfrak{m}_{f(x)} $  then $f^{\sharp}(t)$ is in $\mathfrak{m}_x$, or more concisely,

$f^{\sharp}_x: (O_Y)_{f(x)} \to (O_X)_x$ is a local homomorphism.