This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Since a scheme is a geometric as well as an algebraic object, in order to understand it well we must simultaneously consider not only the topological structure of schemes examined in previous posts but also their algebraic structure, and in the previous post we briefly saw how this philosophy is reflected. In this post we develop this philosophy further.
Reduced Schemes and Integral Schemes
Definition 1 A scheme \(X\) is a reduced scheme if for every open set \(U\), \(\mathscr{O}_X(U)\) is reduced. ([Algebraic Structures] §Field of Fractions, ⁋Definition 11) Similarly, \(X\) is integral if for every open set \(U\), \(\mathscr{O}_X(U)\) is an integral domain. ([Algebraic Structures] §Field of Fractions, ⁋Definition 5)
Then the following holds.
Lemma 2 A scheme \(X\) is a reduced scheme if and only if for every \(x\in X\), \(\mathscr{O}_{X,x}\) is a reduced ring.
Proof
First, for any point \(x\in X\) of a reduced scheme \(X\), consider an affine open subscheme \(U=\Spec A\) containing \(x\). If we let \(\mathfrak{p}\) be the prime ideal corresponding to \(x\) in \(\Spec A\), then
\[\mathscr{O}_{X,x}=(\mathscr{O}_X\vert_U)_x\cong \mathscr{O}_{\Spec A, \mathfrak{p}}\cong A_\mathfrak{p}\]and since \(\mathscr{O}_X(U)\cong A\) is reduced by assumption, \(A_\mathfrak{p}\) is also reduced.
Conversely, if \(\mathscr{O}_{X,x}\) is reduced for every \(x\in X\), then for any open set \(U\), considering the inclusion
\[\mathscr{O}_X(U)\hookrightarrow\prod_{x\in U} \mathscr{O}_{X,x}\]we can verify that \(\mathscr{O}_X(U)\) is reduced.
From this we see that reducedness is a stalk-local property. (§Topology of Schemes, ⁋Proposition 16) Also, if a ring \(A\) is reduced then its localization is easily shown to be reduced, so we can show that the spectrum of a reduced ring is reduced.
Similarly, the spectrum of an integral domain is an integral scheme. This is not difficult to show directly, but in Proposition 4 we prove that a scheme \(X\) is integral if and only if it is an irreducible, reduced scheme. Then the spectrum \(\Spec A\) of an integral domain \(A\) is
- a reduced scheme, since \(A\) is a reduced ring, and
- irreducible, since \(A\) has a unique minimal prime ideal \(\{0\}\).
That is, accepting this, we can see that the spectrum of an integral domain is an integral scheme.
To prove Proposition 4, it is useful to rewrite irreducibility in the following algebraic language.
Lemma 3 An affine scheme \(\Spec A\) is irreducible if and only if its nilradical \(\mathfrak{N}(A)\) is a prime ideal.
Proof
That \(\Spec A\) is irreducible is equivalent to the condition that for any two basic open sets \(D(f),D(g)\neq\emptyset\) of this space, \(D(fg)\neq\emptyset\). However, from the equivalence
\[D(f)\neq\emptyset\iff f\not\in \mathfrak{p}\text{ for some $\mathfrak{p}$}\iff f\not\in \mathfrak{N}(A)\]([Algebraic Structures] §Field of Fractions, ⁋Proposition 14) we know that the statement \(D(f),D(g)\neq\emptyset\implies D(fg)\not\in\emptyset\) is equivalent to the statement
\[f,g\not\in \mathfrak{N}(A)\implies fg\not\in \mathfrak{N}(A).\]Now we obtain the following.
Proposition 4 \(X\) is integral if and only if \(X\) is reduced and irreducible.
Proof
First, suppose \(X\) is integral. Since any integral domain is always reduced, \(X\) is a reduced scheme. If \(X\) were not an irreducible scheme, there would exist two disjoint nonempty open sets \(U_1,U_2\neq\emptyset\). Then for the open set \(U_1\cup U_2\),
\[\mathscr{O}_X(U_1\cup U_2)=\mathscr{O}_X(U_1)\times \mathscr{O}_X(U_2)\]and the right-hand side is not an integral domain, which contradicts the assumption that \(X\) is integral.
Conversely, let an irreducible reduced scheme \(X\) be given, and let us show that \(X\) is an integral scheme. That is, given any open set \(U\) of \(X\), we must show that \(\mathscr{O}_X(U)\) is an integral domain. First we prove the following claim.
Claim For any affine open subset \(\Spec A\cong V\subseteq X\), \(\mathscr{O}_X(V)\cong A\) is always an integral domain.
From the assumption that \(X\) is reduced we know that \(A\) must be a reduced ring. On the other hand, \(X\) is an irreducible closed subset of itself, so \(V\) is also irreducible ([Topology] §Dimension, ⁋Proposition 13) and therefore by Lemma 3, \(\mathfrak{N}(A)=0\) is a prime ideal, so \(A\) is an integral domain.
Now we show in general that for any open set \(U\) of \(X\), \(\mathscr{O}_X(U)\) is an integral domain. To this end, let \(f,g\in \mathscr{O}_X(U)\) be two elements satisfying \(fg=0\). Then for the two open sets of \(U\)
\[D_U(f)=\{x\in U\mid f_x\not\in \mathfrak{m}_x\},\qquad D_U(g)=\{x\in U\mid g_x\not\in \mathfrak{m}_x\}\]and their complements \(Z_U(f), Z_U(g)\), we have \(U=Z_U(f)\cup Z_U(g)\). Since \(X\) is irreducible, from [Topology] §Dimension, ⁋Proposition 13 we know that its open set \(U\) is likewise, and therefore either \(Z_U(f)=U\) or \(Z_U(g)=U\). Without loss of generality, let \(Z_U(f)=U\). Then for any open affine subset \(V\) of \(U\), defining
\[D_V(f)=\{x\in V\mid f_x\not\in \mathfrak{m}_x\}\]in \(V\), we have \(D_V(f)=D_U(f)\cap V=D(f\vert_{U\cap D_U(f)})\subseteq V\), and for this to be empty, \(f\vert_{U\cap D_U(f)}\) must be a nilpotent element of \(\mathscr{O}_X(V)\). However, since \(\mathscr{O}_X(V)\) is an integral domain by the claim above, it follows that \(f\vert_{U\cap D_U(f)}=0\), and since this holds for any open affine subset \(V\) of \(U\), we must have \(f=0\).
On the other hand, from §Topology of Schemes, ⁋Example 6 we know that the irreducibility of an arbitrary scheme \(X\) cannot be determined by looking only at stalks. For example, \(Z(\x(\x-1))\) splits into two components, so points in each component know nothing about points in the other component. Therefore integrality also cannot be determined by looking only at stalks.
However, if \(X\) were a connected scheme, the irreducible components would necessarily meet at some point, and by looking at the stalk at this point we might be able to determine irreducibility. The next proposition records this idea precisely.
Proposition 5 A Noetherian scheme \(X\) is integral if and only if \(X\) is nonempty and connected and each \(\mathscr{O}_{X,x}\) is an integral domain.
Proof
First, if \(X\) is integral then \(X\) is irreducible and hence connected, and since the localization of an integral domain is an integral domain, one direction is obvious.
For the opposite direction, that the scheme \(X\) is reduced is obvious because any integral domain is reduced and reducedness is a stalk-local property. Thus if we show that \(X\) is irreducible using the given conditions, the rest follows from Proposition 4.
First, since \(X\) is a Noetherian scheme, there exist suitable Noetherian rings \(A_1,\ldots, A_r\) such that \(X=\bigcup \Spec A_i\). Also, \(X\) is Noetherian as a topological space, and therefore by [Topology] §Dimension, ⁋Proposition 12, \(X\) has finitely many irreducible components. Now if
\[X=\bigcup_{j=1}^s X_j\tag{$\ast$}\]is the decomposition of \(X\) into irreducible components, then for fixed \(i\), among the sets
\[X_1\cap \Spec A_i,\quad X_2\cap \Spec A_i,\quad\ldots,\quad X_s\cap \Spec A_i\]the nonempty ones are the irreducible components of \(\Spec A_i\). Now by §Spectrums, ⁋Proposition 16, each of these defines a minimal prime ideal \(\mathfrak{q}_j=I(X_j)\), and conversely any minimal prime ideal of \(A_i\) uniquely determines an irreducible component \(X_j\cap \Spec A_i\).
On the other hand, since \(X\) is connected, considering in the irreducible decomposition ($$\ast$) the intersection
\[X_1\cap \bigcup_{j=2}^s X_j\]this is the intersection of two nonempty open sets of \(X\), and since \(X\) is connected they must meet at some point \(x\). That is, there exists a suitable \(j\) such that \(x\in X_1\cap X_j\). Now let \(\Spec A_i\) be an affine cover of \(X\) containing the point \(x\), and suppose that at this time \(x\) corresponds to the prime ideal \(\mathfrak{p}\). That is,
\[x\in \Spec A_i\cap X_1\cap X_j=(\Spec A_i\cap X_1)\cap (\Spec A_i\cap X_j).\]From the preceding argument, \(\Spec A_i\cap X_1\) has generic point \(\mathfrak{q}_1\) and \(\Spec A_i\cap X_j\) has generic point \(\mathfrak{q}_j\), and these are minimal prime ideals of \(A_i\). Now considering the stalk at \(x\), \(\mathscr{O}_{X,x}\cong (A_i)_\mathfrak{p}\), from the minimality of \(\mathfrak{q}_1,\mathfrak{q}_2\) we have \(\mathfrak{q}_1,\mathfrak{q}_2\subseteq \mathfrak{p}\) and therefore \(\mathfrak{q}_1 A_i, \mathfrak{q}_2 A_i\) become minimal prime ideals of \(A_i\). ([Commutative Algebra] §Localization, ⁋Proposition 8) However, an integral domain has a unique minimal prime ideal \((0)\), so this contradicts the assumption that \(\mathscr{O}_{X,x}\) is an integral domain.
The key logic in the above proof can be summarized as follows:
- Since \(X\) is connected, if we decompose \(X\) into irreducible components then each irreducible component must meet another irreducible component,1
- letting \(x\) be a point where two irreducible components meet and taking any open neighborhood of \(x\), this open set will contain the generic point of each irreducible component (§Spectrums, ⁋Proposition 16),
- therefore these generic points survive in the stalk \(\mathscr{O}_{X,x}\) at \(x\), but this is impossible because \(\mathscr{O}_{X,x}\) is an integral domain.
We will examine this property of generic points at the end of this post.
Normal Schemes
Similarly to integral schemes, we can define the following.
Definition 6 A scheme \(X\) is normal if for every \(x\in X\), \(\mathscr{O}_{X,x}\) is a normal domain. ([Commutative Algebra] §Integral Extensions, ⁋Definition 3)
In general, the localization of a normal domain is always a normal domain. ([Commutative Algebra] §Integral Extensions, ⁋Proposition 12) From this we see that the spectrum \(\Spec A\) of a normal domain \(A\) is a normal scheme.
Any integral domain is always reduced, and since by Lemma 2 reducedness can be checked on stalks, any normal scheme is reduced. On the other hand, being an integral scheme is not a stalk-local property, so in general a normal scheme need not be an integral scheme. However, if \(X\) is a connected, nonempty Noetherian scheme then by Proposition 5 we know that normality implies integrality.
On the other hand, we know that a unique factorization domain is always a normal domain. ([Commutative Algebra] §Integral Extensions, ⁋Proposition 9) From this we define the following.
Definition 7 A scheme \(X\) is factorial if for every \(x\in X\), \(\mathscr{O}_{X,x}\) is a unique factorization domain.
Therefore any factorial scheme is a normal scheme. Also, since the localization of a unique factorization domain is a unique factorization domain, the spectrum \(\Spec A\) of a unique factorization domain \(A\) is factorial.
Associated Primes
By §Spectrums, ⁋Corollary 17, we know that there is a one-to-one correspondence between the irreducible components of a scheme \(X=\Spec A\) and the minimal prime ideals of the ring \(A\). This was used importantly in Proposition 5 above.
On the other hand, algebraically a minimal prime ideal is always an associated prime ideal. This can be checked by viewing the ring \(A\) as a module over itself; since \(\ann A=\{0\}\), we apply [Commutative Algebra] §Associated Primes, ⁋Theorem 7. Therefore we define the associated point of a (locally Noetherian) scheme as follows.
Definition 8 For a locally Noetherian scheme \(X\), a point \(x\) and an affine open neighborhood \(U\cong \Spec A\) of \(x\), we say that \(x\) is an associated point of \(X\) if the prime ideal \(\mathfrak{p}_x\subset A\) corresponding to \(x\) is an associated prime ideal of \(A\).
Then this definition does not depend on the choice of \(U\), and moreover can be written stalk-locally. This is because for an affine open neighborhood \(\Spec A\) containing \(x\), assuming that \(A\) is a Noetherian ring from the condition that \(X\) is a locally Noetherian scheme, from the third condition of [Commutative Algebra] §Associated Primes, ⁋Theorem 7 we know that there is a one-to-one correspondence between the associated prime ideals of \(A\) contained in \(\mathfrak{p}_x\) and the associated prime ideals of \(A_{\mathfrak{p}_x}\), and from this one-to-one correspondence we can rewrite Definition 8 as
For a locally Noetherian scheme \(X\) and a point \(x\), we say that \(x\) is an associated point of \(X\) if \(\mathfrak{m}_x\) is an associated prime ideal of \(\mathscr{O}_{X,x}\).
Now the first condition of [Commutative Algebra] §Associated Primes, ⁋Theorem 7 also guarantees the finiteness of associated points when \(X\) is a quasicompact locally Noetherian scheme, that is, when \(X\) is a Noetherian scheme.
Also, by this condition any minimal prime ideal is an associated prime ideal, so this notion generalizes the concept of the generic point in \(\Spec A\). By Definition 8 we may restrict our attention to the spectrum \(\Spec A\) of a Noetherian ring.
Definition 9 Among the associated points of the spectrum \(\Spec A\) of a Noetherian ring, those that are not the generic points of the irreducible components of \(\Spec A\) are called embedded points.
Meanwhile, the following holds by definition.
Proposition 10 The associated points of the spectrum \(\Spec A\) of a Noetherian ring are the generic points of the irreducible components of \(\supp(f)\) for suitable \(f\in A\), and conversely.
Proof
First, for any \(g\in A\) and prime ideal \(\mathfrak{q}\in \Spec A\),
\[\mathfrak{q}\in \supp(g)\iff g_\mathfrak{q}\neq 0\text{ in $A_\mathfrak{q}$}\iff \ann(g_\mathfrak{q})\neq A_\mathfrak{q}\]holds. However, by [Commutative Algebra] §Localization, ⁋Proposition 5,
\[\ann(g_\mathfrak{q})=\ann(g)A_\mathfrak{q}\]so the last condition is equivalent to \(\ann(g)\setminus \mathfrak{q}=\emptyset\), that is, \(\mathfrak{q}\in Z(\ann(g))\). From this we know that for any \(g\in A\),
\[\supp(g)=Z(\ann(g))\]holds. Therefore there is a one-to-one correspondence between the irreducible components of \(\supp(g)\) and the minimal prime ideals containing \(\ann(g)\).
Now given any associated point \(\mathfrak{p}\) of \(\Spec A\), by definition there exists a suitable \(f\in A\) such that \(\mathfrak{p}=\ann(f)\). Now
\[\supp(f)=Z(\ann(f))=Z(\mathfrak{p})\]and it is obvious that \(\mathfrak{p}\) is a minimal prime of \(\ann(f)=\mathfrak{p}\), so \(\mathfrak{p}\) is the generic point of \(\supp f\). This argument also works in reverse based on the above observation.
Examining the equation
\[\supp(f)=Z(\ann(f))\]used in the above proof, if \(A\) were an integral domain then \(\ann(f)\) is all of \(A\) only when \(f=0\), and is \(0\) otherwise. That is, in this case \(\supp(f)\) is empty only when \(f=0\), and is all of \(\Spec A\) otherwise, and since we know from Proposition 4 that \(\Spec A\) is irreducible, we see that the only associated point of \(\Spec A\) is the generic point \((0)\).
More generally, if \(A\) is not an integral domain then there exist cases where \(\ann(f)\) is neither \(0\) nor \(A\), so embedded points may exist.
Example 11 Consider the affine scheme \(X=\Spec \mathbb{K}[\x_1,\x_2]/(\x_2^2, \x_1\x_2)\). Then by §Spectrums, ⁋Lemma 6 and §Spectrums, ⁋Proposition 9, as a set
\[X= Z(\x_2^2,\x_1\x_2)=Z(\x_2^2)\cap Z(\x_1\x_2)=\{(0,0)\}\]Rational Functions
Now we define rational functions defined on a scheme. First, by the second result of [Commutative Algebra] §Associated Primes, ⁋Corollary 4, the map
\[A \rightarrow \prod_\text{\scriptsize $\mathfrak{p}$ associated prime} A_\mathfrak{p}\]is injective. Therefore for any open set \(U\) of a locally Noetherian scheme \(X\), the map
\[\Gamma(U, \mathscr{O}_X) \rightarrow \prod_\text{\scriptsize $x$ associated in $U$} \mathscr{O}_{X,x}\tag{$\ast$}\]is injective.
Definition 12 For a locally Noetherian scheme \(X\) and an open set \(U\) containing all associated points of \(X\), the image of \(\Gamma(U, \mathscr{O}_X)\) under (\(\ast\)) is called a rational function defined on \(X\).
Thus, by definition a rational function defined on \(X\) consists of the data of (1) a domain of definition \(U\) containing all associated points of \(X\), and (2) a function \(f\in \Gamma(U, \mathscr{O}_X)\) on it, and two such pairs \((U, f)\) and \((U',f')\) are the same function if \(f\) and \(f'\) define the same function on \(U\cap U'\).
Example 13 Consider the affine scheme \(X=\Spec \mathbb{K}[\x_1,\x_2]/(\x_2-\x_1^2)\). Then \(X\) has a unique associated prime \((0)\), and any open set of \(X\) contains this point, so a rational function on \(X\) consists of any nonempty open set \(U\) and a function \(f\in\Gamma(U, \mathscr{O}_X)\) on it.
On the other hand, we know that any (nonempty) open set of an affine scheme \(X=\Spec A\) is of the form \(\Spec A_f\) for a suitable nonzero \(f\in A\), and a function on it is given by \(A_f\). For example, in this example if we choose \(f\) to be the image of \(\x_1\) in \(A\), then by the isomorphism
\[\left(\frac{\mathbb{K}[\x_1,\x_2]}{(\x_2-\x_1^2)}\right)_{\x_1}\cong\frac{\mathbb{K}[\x_1,\x_2]_{\x_1}}{(\x_2-\x_1^2)_{\x_1}}\]functions such as \(1/\x_1\) defined on the open set \(\Spec A_{\x_1}\) become rational functions on \(X\). From this we see that all rational expressions that do not have the factor \(\x_2-\x_1^2\) in the denominator become rational functions on \(X\) (on a suitable open set).
The collection of rational functions defined on a locally Noetherian scheme \(X\) defines the total quotient ring \(K(X)\). Now suppose, as above, that \(X\) is an integral scheme. Then in particular \(X\) is irreducible, so it has a unique generic point \(x\). This point must correspond to the unique minimal prime ideal \((0)\) of the integral domain \(A\) for any affine open subset \(U\cong\Spec A\). The localization at this point is obtained by adding all nonzero elements of \(A\) to the denominator, that is, it equals \(\Frac(A)\).
References
[Har] R. Hartshorne, Algebraic geometry. Graduate texts in mathematics. Springer, 1977.
[Vak] R. Vakil, The rising sea: Foundation of algebraic geometry. Available online.
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In this step we used that each irreducible component is an open set, which holds because there are only finitely many irreducible components of \(X\). ↩
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