The Spectrum

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Remark In every post in this category, a ring means a commutative ring (with unity).

In this post we define the spectrum, the most fundamental object in algebraic geometry. The spectrum is a topological space equipped with a suitable structure sheaf; for now we define it as a set and put a topology on it. Afterwards, in the next post, we will define the structure sheaf on \(\Spec A\).

\(\Spec A\) as a set

Definition 1 For a ring \(A\), \(\Spec A\) is the collection of all prime ideals of \(A\), and we call it the spectrum of \(A\).

Now suppose a ring homomorphism \(\phi: A \rightarrow B\) is given. Then by [Algebraic Structures] §Field of Fractions, ⁋Proposition 10, the function

\[\Spec\phi: \Spec B \rightarrow \Spec A;\qquad \mathfrak{q}\mapsto \phi^{-1}(\mathfrak{q})\]

is well-defined.

Proposition 2 The \(\Spec: \cRing^\op \rightarrow \Set\) defined above is a functor.

That is, \(\Spec(\phi\circ\psi)=(\Spec\psi)\circ(\Spec\phi)\) and \(\Spec(\id_A)=\id_{\Spec A}\), and the proof is not difficult either.

\(\Spec A\) as a topological space

Now we define an appropriate topological structure on \(\Spec A\).

Definition 3 Fix a ring \(A\) and its spectrum \(\Spec A\). For any subset \(S\) of \(A\), we define the subset \(Z(S)\) of \(\Spec A\) by the formula

\[Z(S)=\{\mathfrak{p}\in\Spec A\mid S\subseteq \mathfrak{p}\}.\]

Then several facts can be proved easily. First, \(Z\) is inclusion-reversing. That is, if two subsets \(S_1\subseteq S_2\) of \(A\) are given,

\[Z(S_1)=\{\mathfrak{p}\in\Spec A\mid S_1\subseteq \mathfrak{p}\}\supseteq \{\mathfrak{p}\in\Spec A\mid S_2\subseteq \mathfrak{p}\}=Z(S_2)\]

holds. However, the above inclusion need not be strict.

Proposition 4 Fix a ring \(A\) and its spectrum \(\Spec A\). For any subset \(S\) of \(A\) and the ideal \((S)\) of \(A\) generated by \(S\), we have \(Z(S)=Z((S))\).

Proof

Since \(S\subseteq (S)\) trivially, the inclusion \(Z((S))\subseteq Z(S)\) is obvious from the discussion above. Hence it suffices to prove the reverse direction. Let \(\mathfrak{p}\) be an arbitrary element of \(Z(S)\). That is, \(S\subseteq \mathfrak{p}\). But the ideal generated by the subset \(\mathfrak{p}\) of \(A\) is itself, so from this we know \((S)\subseteq (\mathfrak{p})=\mathfrak{p}\).

Moreover, for two ideals \(\mathfrak{a}_1\subseteq \mathfrak{a}_2\) of \(A\), the inclusion \(Z(\mathfrak{a}_1)\supseteq Z(\mathfrak{a}_2)\) may also be an equality in general.

Proposition 5 Fix a ring \(A\) and its spectrum \(\Spec A\). For any ideal \(\mathfrak{a}\) of \(A\) and its radical \(\sqrt{\mathfrak{a}}\), we have \(Z(\mathfrak{a})=Z(\sqrt{\mathfrak{a}})\).

Proof

Again, the inclusion \(Z(\sqrt{\mathfrak{a}})\subseteq Z(\mathfrak{a})\) is obvious from \(\mathfrak{a}\subseteq \sqrt{\mathfrak{a}}\). Conversely, for any \(\mathfrak{p}\in Z(\mathfrak{a})\), using [Commutative Algebra] §Properties of Localization, ⁋Corollary 8 we have

\[\sqrt{\mathfrak{a}}=\bigcap_\text{\scriptsize$\mathfrak{q}$ a prime containing $\mathfrak{a}$}\mathfrak{q}\subseteq \mathfrak{p}\]

and thus we know \(\mathfrak{p}\in Z(\sqrt{\mathfrak{a}})\).

The most important thing when defining a topological structure on \(\Spec A\) is the following lemma.

Lemma 6 The following hold.

1. For any ideals \(\mathfrak{a},\mathfrak{b}\) of \(A\), we have \(Z(\mathfrak{ab})=Z(\mathfrak{a})\cup Z(\mathfrak{b})\).
2. For a collection of ideals \(\{\mathfrak{a}_i\}\) of \(A\), we have \(Z(\sum \mathfrak{a}_i)=\bigcap Z(\mathfrak{a}_i)\).
3. For any ideals \(\mathfrak{a},\mathfrak{b}\) of \(A\), we have \(Z(\mathfrak{a})\subseteq Z(\mathfrak{b})\iff \sqrt{\mathfrak{a}}\supseteq \sqrt{\mathfrak{b}}\).

Proof

1. It is obvious that a prime ideal \(\mathfrak{p}\) containing \(\mathfrak{a}\) or \(\mathfrak{b}\) also contains the smaller ideal \(\mathfrak{ab}\), so it suffices to show the reverse inclusion. Assume \(\mathfrak{p}\supset \mathfrak{ab}\). If \(\mathfrak{p}\not\supseteq \mathfrak{b}\), then we can find an element \(b\) of \(\mathfrak{b}\) with \(b\not\in \mathfrak{p}\). On the other hand, for any \(a\in \mathfrak{a}\), we have \(ab\in \mathfrak{ab}\subseteq \mathfrak{p}\), and by the preceding assumption \(b\not\in \mathfrak{p}\), so necessarily \(a\in \mathfrak{p}\) and therefore \(\mathfrak{a}\subseteq \mathfrak{p}\) holds.
2. This is obvious because \(\sum \mathfrak{a}_i\) is defined as the smallest ideal containing all the ideals \(\mathfrak{a}_i\).
3. [Commutative Algebra] §Properties of Localization, ⁋Corollary 8.

Then by [Topology] §Interior, Closure, Boundary of Sets, ⁋Proposition 2, there exists a unique topology \(\mathcal{T}\) having the \(Z(\mathfrak{a})\) as closed sets, which makes \(\Spec A\) a topological space.

Definition 7 The topology on \(\Spec A\) defined above is called the Zariski topology.

The Zariski topology is generally not a topology we would have thought nice. For example, for any integral domain \(A\), \((0)\) is itself a prime ideal, and the only non-empty subset of \(A\) contained in it is also \((0)\). That is, the only closed set containing \((0)\in\Spec A\) is \(Z(0)=\Spec A\), and therefore unless \(A\) is a field the singleton \(\{(0)\}\) is not a closed set. In particular, the Zariski topology is generally not a Hausdorff space. However, as we proceed through the posts in this category, we will discover familiar geometric concepts in this structure.

Earlier, in Proposition 2, we saw that \(\Spec\) can be regarded as a functor \(\Spec: \cRing^\op \rightarrow \Set\). Moreover, \(\Spec\) is also a functor from \(\cRing^\op\) to \(\Top\).

Proposition 8 If we equip the spectrum \(\Spec A\) of a ring \(A\) with the topological structure of Definition 7, then the functor \(\Spec: \cRing^\op \rightarrow \Top\) of Proposition 2 is a functor.

Proof

What remains to be shown in addition to Proposition 2 is that for any ring homomorphism \(\phi: A \rightarrow B\), the map \(\Spec \phi: \Spec B \rightarrow \Spec A\) is a continuous function. Hence it suffices to show that for any closed set of \(\Spec A\), its preimage under \(\Spec\phi\) is also a closed set in \(\Spec B\). ([Topology] §Interior, Closure, Boundary of Sets, ⁋Proposition 2)

On the other hand, since every closed set of \(\Spec A\) is of the form \(Z(\mathfrak{a})\) and every closed set of \(\Spec B\) is of the form \(Z(\mathfrak{b})\), to show this it suffices to show that for any ideal \(\mathfrak{a}\) of \(A\), there exists an ideal \(\mathfrak{b}\) of \(B\) satisfying the formula

\[(\Spec\phi)^{-1}(Z(\mathfrak{a}))=Z(\mathfrak{b}).\]

Our claim is that the formula

\[(\Spec\phi)^{-1}(Z(\mathfrak{a}))=Z(\phi(\mathfrak{a}))\]

holds. Then the ideal generated by \(\phi(\mathfrak{a})\) satisfies the above formula, so the proof is complete.

First, let \(\mathfrak{q}\in\Spec B\) belong to the left-hand side. That is, \((\Spec\phi)(\mathfrak{q})=\phi^{-1}(\mathfrak{q})\in Z(\mathfrak{a})\) holds. Then from \(\mathfrak{a}\subseteq \phi^{-1}(\mathfrak{q})\) we have \(\phi(\mathfrak{a})\subseteq \mathfrak{q}\), so \(\mathfrak{q}\in Z(\phi(\mathfrak{a}))\) holds.

Conversely, let \(\mathfrak{q}\in\Spec B\) belong to the right-hand side. Then from \(\phi(\mathfrak{a})\subseteq \mathfrak{q}\), we obtain the inclusion

\[\mathfrak{a}\subseteq \phi^{-1}(\phi(\mathfrak{a}))\subseteq\phi^{-1}(\mathfrak{q})=(\Spec\phi)(\mathfrak{q})\]

and this proves that \((\Spec\phi)(\mathfrak{q})\in Z(\mathfrak{a})\), i.e. \(\mathfrak{q}\in (\Spec\phi)^{-1}(Z(\mathfrak{a}))\).

The two important ring homomorphisms we will use from now on are the quotient \(\pi:A \rightarrow A/\mathfrak{a}\) and the localization \(\epsilon: A \rightarrow S^{-1}A\). ([Commutative Algebra] §Basic Notions, ⁋Proposition 11 and [Commutative Algebra] §Localization, ⁋Proposition 8)

Proposition 9 For the \(\pi:A \rightarrow A/\mathfrak{a}\) and \(\epsilon: A \rightarrow S^{-1}A\) defined above, the following hold.

1. \(\Spec\pi\) and \(\Spec\epsilon\) are both injective, and they define homeomorphisms onto their respective images.
2. The image of \(\Spec\pi\) in \(\Spec A\) is a closed set.
3. If \(S=\{1,f,f^2,\ldots\}\) for some \(f\in A\), then the image of \(\Spec \epsilon\) in \(\Spec A\) is an open set.

Proof

That \(\Spec\pi\) and \(\Spec\epsilon\) are injective in the first result follows from the two propositions [Commutative Algebra] §Basic Notions, ⁋Proposition 11 and [Commutative Algebra] §Localization, ⁋Proposition 8 mentioned above. Since the two maps are continuous by Proposition 8, what remains is that their inverses are continuous, i.e. that the two maps carry closed sets to closed sets onto their images. This is obtained from the following two formulas for any ideal \(\mathfrak{b}\supseteq \mathfrak{a}\) and any ideal \(J\) of \(S^{-1}A\):

\[(\Spec\pi)\left(Z_{A/\mathfrak{a}}(\mathfrak{b}/\mathfrak{a})\right)=Z_A(\mathfrak{b}),\qquad (\Spec\epsilon)\left(Z_{S^{-1}A}(J)\right)=Z_A(\epsilon^{-1}J)\cap \im(\Spec\epsilon)\]

The first formula follows from the equivalence of \(\mathfrak{q}\supseteq \mathfrak{b}/\mathfrak{a}\) and \(\pi^{-1}(\mathfrak{q})\supseteq \mathfrak{b}\), and the second from the fact that every ideal of \(S^{-1}A\) satisfies \(J=\epsilon(\epsilon^{-1}J)\cdot S^{-1}A\). In the second result, the image of \(\Spec\pi\) in \(\Spec A\) is exactly \(Z(\mathfrak{a})\). Finally, in the third result, the elements of \(\Spec\epsilon\) are the collection of prime ideals not containing \(f\), and these can be written as \(\Spec A\setminus Z(f)\).

That a base for the topological space \(\Spec A\) exists is of course natural, but we also hope that this base has some relation to the algebraic structure of \(A\). Let us define the following.

Definition 10 For any element \(f\in A\) of a ring \(A\), we write \(D(f)\) for the complement of \(Z(f)\) in \(\Spec A\). An open set of this form is called a principal open set.

Now for any \(f\in A\), let \(S_f=\{1,f,f^2,\ldots\}\) and define \(A_f=S_f^{-1}A\). Then by Proposition 9, the image of \(\Spec A_f\) under \(\Spec \epsilon\) in \(\Spec A\) is an open set. That this open set is exactly the same as \(D(f)\) as a set follows from

\[\mathfrak{p}\not\in Z(f)\iff (f)\not\subseteq \mathfrak{p} \iff f\not\in \mathfrak{p}\iff f^k\not\in \mathfrak{p}\text{ for all $k\geq 0$}\iff S_f\cap \mathfrak{p}=\emptyset\]

and [Commutative Algebra] §Localization, ⁋Proposition 8. Moreover, that \(D(f)\) has the same topological structure as \(\Spec A_f\) also follows from the first result of Proposition 9.

Lemma 11 The collection of principal open sets forms a base for \(\Spec A\). ([Topology] §Base of a Topological Space, ⁋Definition 1)

Proof

For any open set \(\Spec A \setminus Z(S)\), using the second result of Lemma 6 we have the following computation:

\[\Spec A\setminus Z(S)=\Spec A\setminus Z\left(\sum_{f\in S} (f)\right)=\Spec A\setminus\left(\bigcap_{f\in S}Z(f)\right)=\bigcup_{f\in S} (\Spec A\setminus Z(f))=\bigcup_{f\in S} D(f).\]

Then by a computation similar to this lemma we can verify that \(D(fg)=D(f)\cap D(g)\). However, since this computation uses the first result of Lemma 6, it cannot in general be extended to infinite indices.

Separately, we can show that \(\Spec A\) always satisfies the condition of [Topology] §Compact Spaces, ⁋Definition 1. However, many properties we usually expect from a compact space often also require the Hausdorff condition ([Topology] §Compact Spaces, §§Compact Hausdorff Spaces), and since the Zariski topology is generally not a Hausdorff space, we call this quasi-compact.

Lemma 12 \(\Spec A\) is a quasi-compact space as a topological space.

Proof

By Lemma 11, it suffices to show that whenever a collection of principal open sets \(\{D(f_i)\}_{i\in I}\) covering \(\Spec A\) is given, there exists a finite subset \(J\) of \(I\) such that still \(\Spec A=\bigcup_{j\in J} D(f_j)\).

From the given assumption \(\Spec A=\bigcup_{i\in I} D(f_i)\) we know

\[\emptyset=\Spec A\setminus\bigcup_{i\in I} D(f_i)=\bigcap_{i\in I}(\Spec A\setminus D(f_i))=\bigcap_{i\in I} Z(f_i).\]

On the other hand, by Lemma 6

\[\emptyset=\bigcap_{i\in I} Z(f_i)=Z\left(\sum_{i\in I}(f_i)\right)\]

and if \(\sum (f_i)\) were a proper ideal of \(A\), then a maximal ideal containing it would be an element of \(\bigcap_{i\in I} Z(f_i)\), so the above formula is equivalent to \(\sum(f_i)=(1)\). Now since \(1\in\sum(f_i)\), we can find a finite subset \(J\subseteq I\) and a family of elements \((a_j)_{j\in J}\) of \(A\) such that \(\sum_{j\in J} a_jf_j=1\). That is,

\[Z\left(\sum_{j\in J} (f_j)\right)=\emptyset\]

and reversing the computations above yields the desired result.

Galois correspondence

By definition \(Z\) is a function taking an ideal of \(A\) to a closed subset of \(\Spec A\). We define the reverse as follows.

Definition 13 For any subset \(T\subseteq \Spec A\), we define

\[I(T)=\{f\in A\mid\text{$f\in \mathfrak{p}$ for all $\mathfrak{p}\in T$}\}=\bigcap_\text{\scriptsize$\mathfrak{p}$ a prime in $T$} \mathfrak{p}.\]

Then \(I(T)\) is an ideal because it is an intersection of ideals. Moreover, if \(T_1\subseteq T_2\), then \(I(T_2)\subseteq I(T_1)\) is obvious.

Now combining Definition 3 and Definition 13, we have defined two functions between the two ordered sets \(\mathcal{P}(A)\), \(\mathcal{P}(\Spec A)\):

\[Z: \mathcal{P}(A) \rightarrow \mathcal{P}(\Spec A);\quad S\mapsto Z(S),\qquad I: \mathcal{P}(\Spec A) \rightarrow \mathcal{P}(A);\quad T\mapsto I(T).\]

Then for any \(S\in \mathcal{P}(A)\) and any \(T\in \mathcal{P}(\Spec A)\),

\[T\subseteq Z(S)\iff\text{$\mathfrak{p}\in Z(S)$ for all $\mathfrak{p}\in T$}\iff\text{$f\in \mathfrak{p}$ for all $f\in S$ and all $\mathfrak{p}\in T$}\iff S\subseteq I(T)\]

so \((Z, I)\) defines an antitone Galois connection between \(\mathcal{P}(A)\) and \(\mathcal{P}(\Spec A)\). ([Set Theory] §Filters and Ideals, Galois Correspondence, ⁋Definition 6) Therefore, the two formulas

\[Z(I(Z(S)))=Z(S),\qquad I(Z(I(T)))=I(T)\]

hold for all \(S\in \mathcal{P}(A)\) and any \(T\in \mathcal{P}(\Spec A)\).

Now consider the two closure operators \(IZ: \mathcal{P}(A) \rightarrow \mathcal{P}(A)\), \(ZI: \mathcal{P}(\Spec A) \rightarrow \mathcal{P}(\Spec A)\).

Proposition 14 For the closure operators \(IZ: \mathcal{P}(A) \rightarrow \mathcal{P}(A)\), \(ZI: \mathcal{P}(\Spec A) \rightarrow \mathcal{P}(\Spec A)\), the following hold.

1. \(IZ(S)=\sqrt{(S)}\)
2. \(ZI(T)=\cl(T)\)

Proof

1. This is obvious from the formula

   \[I(Z(S))=I(Z((S)))=\bigcap_\text{\scriptsize$\mathfrak{p}$ a prime in $Z((S))$} \mathfrak{p}=\bigcap_\text{\scriptsize$\mathfrak{p}$ a prime containing $(S)$} \mathfrak{p}=\sqrt{(S)}.\]

2. This is obvious from the formula

   \[\cl(T) =\bigcap_\text{\scriptsize $Z(S)\supseteq T$} Z(S) =\bigcap_\text{\scriptsize $S\subseteq I(T)$} Z(S) =Z\left(\sum_\text{\scriptsize $S\subseteq I(T)$}(S)\right) =Z(I(T)).\]

Thus we obtain the following.

Theorem 15 There is a Galois correspondence between the radical ideals of a ring \(A\) and the closed subsets of \(\Spec A\).

What can be somewhat confusing in this correspondence is that since any prime ideal of \(A\) is always a radical ideal, the collection of radical ideals of \(A\) always contains (as a set) \(\Spec A\).

Proposition 16 There is a Galois correspondence between the prime ideals of a ring \(A\) and the irreducible closed subsets of \(\Spec A\).

Proof

That is, we must show that for any prime ideal \(\mathfrak{p}\), the set \(Z(\mathfrak{p})\) is irreducible, and that for any irreducible closed subset \(Y\), the set \(I(Y)\) is a prime ideal.

First, since \(I(\{\mathfrak{p}\})=\mathfrak{p}\), the second result of Proposition 14 gives the formula

\[Z(\mathfrak{p})=Z(I(\{\mathfrak{p}\}))=\cl(\{\mathfrak{p}\}).\]

But a singleton is always irreducible and the closure of an irreducible subset is again irreducible, so \(Z(\mathfrak{p})\) is irreducible. That is, \(\mathfrak{p}\) is the generic point of \(Z(\mathfrak{p})\).

Conversely, we must show that for any irreducible closed subset \(Y\), \(I(Y)\) is a prime ideal. First, since \(Y\) is a closed set, by Theorem 15 and Proposition 5 there exists a radical ideal \(\mathfrak{a}\) such that \(Y=Z(\mathfrak{a})\). Then it suffices to show that \(\mathfrak{a}=IZ(\mathfrak{a})=I(Y)\) is prime. Here \(Y\) is irreducible so it is non-empty, and therefore \(\mathfrak{a}=I(Y)\) is contained in a prime ideal that is an element of \(Y\), hence a proper ideal. ([Topology] §Dimension, ⁋Definition 6)

For this, suppose \(fg\in \mathfrak{a}\), and consider the two open sets \(D(f), D(g)\) of \(\Spec A\); then

\[(D(f)\cap Y)\cap (D(g)\cap Y)=D(f)\cap D(g)\cap Y=D(fg)\cap Y\]

must be empty. But \(D(f)\cap Y\) and \(D(g)\cap Y\) are two open sets of the irreducible closed set \(Y\), so for this to hold one of them must be empty. ([Topology] §Dimension, ⁋Proposition 7) On the other hand, for any \(h\in A\), that \(D(h)\cap Y=\emptyset\) is equivalent to every element of \(Y\) containing \(h\), i.e. \(h\in I(Y)=\mathfrak{a}\). From this we know that \(f\in \mathfrak{a}\) or \(g\in \mathfrak{a}\) must hold, and therefore \(\mathfrak{a}\) is a prime ideal.

In particular, for a prime ideal \(\mathfrak{p}\), if there is no prime ideal \(\mathfrak{q}\subsetneq \mathfrak{p}\), i.e. if \(\mathfrak{p}\) is a minimal prime ideal, then \(Z(\mathfrak{p})\) becomes maximal among irreducible closed subsets with respect to inclusion, that is, an irreducible component. From this we obtain the following.

Corollary 17 There is a Galois correspondence between the minimal prime ideals of a ring \(A\) and the irreducible components of \(\Spec A\).

Proof

By Proposition 14, for any prime ideal \(\mathfrak{p}\) we have \(IZ(\mathfrak{p})=\sqrt{\mathfrak{p}}=\mathfrak{p}\), and for any closed subset \(Y\) of \(\Spec A\) we have \(ZI(Y)=\cl(Y)=Y\), so \(Z\) and \(I\) are mutually inverse correspondences. Therefore, by Proposition 16, the map \(\mathfrak{p}\mapsto Z(\mathfrak{p})\) is a bijection between the prime ideals of \(A\) and the irreducible closed subsets of \(\Spec A\), and its inverse is \(Y\mapsto I(Y)\). Moreover, applying the third result of Lemma 6 to two prime ideals \(\mathfrak{p},\mathfrak{q}\) gives

\[Z(\mathfrak{q})\subseteq Z(\mathfrak{p})\iff \sqrt{\mathfrak{q}}\supseteq \sqrt{\mathfrak{p}}\iff \mathfrak{q}\supseteq \mathfrak{p}\]

so this bijection reverses inclusion.

On the other hand, by [Topology] §Dimension, ⁋Definition 9, an irreducible component of \(\Spec A\) is a maximal irreducible subset, and since the closure of an irreducible set is again irreducible, these are always closed sets. Conversely, suppose a maximal \(Y\) among irreducible closed subsets is given, and consider any irreducible subset \(S\) containing \(Y\). Then \(\cl(S)\) is an irreducible closed subset containing \(Y\), so by the maximality of \(Y\) we have \(\cl(S)=Y\), and therefore \(S\subseteq Y\), i.e. \(S=Y\). From the above, the irreducible components of \(\Spec A\) are exactly the maximal ones among irreducible closed subsets with respect to inclusion.

Now since the above bijection reverses inclusion, it pairs the minimal ones among prime ideals, i.e. the minimal prime ideals, with the maximal ones among irreducible closed subsets, i.e. the irreducible components of \(\Spec A\).

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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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