Spectra

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 all posts 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; here we first define it as a set and then put a topology on it. 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 set of all prime ideals of $A$, called the spectrum of $A$.

Now suppose a ring homomorphism $\phi: A \rightarrow B$ is given. Then by [Algebraic Structures] §Field of Fractions, ⁋Proposition 9, the map

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

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.

$\Spec A$ as a topological space

Now we define an appropriate topology on $\Spec A$.

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

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

Several facts can be proved easily. First, $Z$ is inclusion-reversing: if $S_1\subseteq S_2$ are subsets of $A$, then

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

However, this 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 $(S)$ 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 preceding discussion. Thus 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 $\mathfrak{p}$ itself, so we obtain $(S)\subseteq (\mathfrak{p})=\mathfrak{p}$.

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 $\mathfrak{a}\subseteq \sqrt{\mathfrak{a}}$ gives $Z(\sqrt{\mathfrak{a}})\subseteq Z(\mathfrak{a})$ trivially. Conversely, for any $\mathfrak{p}\in Z(\mathfrak{a})$, using [Commutative Algebra] §Properties of Localization, ⁋Corollary 8 we have

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

so $\mathfrak{p}\in Z(\sqrt{\mathfrak{a}})$.

The most important ingredient in defining a topology on $\Spec A$ is the following lemma.

Lemma 6 The following hold.

For any ideals $\mathfrak{a},\mathfrak{b}$ of $A$, we have $Z(\mathfrak{ab})=Z(\mathfrak{a})\cup Z(\mathfrak{b})$.
For any collection of ideals $\{\mathfrak{a}_i\}$ of $A$, we have $Z(\sum \mathfrak{a}_i)=\bigcap Z(\mathfrak{a}_i)$.
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

It is obvious that a prime ideal 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 $b\in\mathfrak{b}$ 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 since $b\not\in \mathfrak{p}$ by the preceding assumption, necessarily $a\in \mathfrak{p}$; thus $\mathfrak{a}\subseteq \mathfrak{p}$.
This is obvious because $\sum \mathfrak{a}_i$ is defined as the smallest ideal containing all the $\mathfrak{a}_i$.
[Commutative Algebra] §Properties of Localization, ⁋Corollary 8.

Then by [Topology] §Interior, Closure, Boundary, ⁋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 consider nice. For example, for any integral domain $A$, $(0)$ is itself a prime ideal, and the only nonempty subset of $A$ contained in it is $(0)$ itself. Hence the only closed set containing $(0)\in\Spec A$ is $Z(0)=\Spec A$, and therefore the singleton $\{(0)\}$ is not closed unless $A$ is a field. In particular, the Zariski topology is generally not Hausdorff. However, as we proceed with the discussion in the posts of 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 Endowing the spectrum $\Spec A$ of a ring $A$ with the topological structure of Definition 7, the functor $\Spec: \cRing^\op \rightarrow \Top$ of Proposition 2 is a functor.

Proof

In addition to Proposition 2, what remains to show is that for any ring homomorphism $\phi: A \rightarrow B$, the map $\Spec \phi: \Spec B \rightarrow \Spec A$ is continuous. Thus it suffices to show that the preimage under $\Spec\phi$ of any closed set in $\Spec A$ is closed in $\Spec B$. ([Topology] §Interior, Closure, Boundary, ⁋Proposition 2)

Now any closed set in $\Spec A$ is of the form $Z(\mathfrak{a})$, and any closed set in $\Spec B$ is of the form $Z(\mathfrak{b})$; hence to show this it suffices to show that for any ideal $\mathfrak{a}$ of $A$, there exists an ideal $\mathfrak{b}$ of $B$ satisfying

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

We claim that in fact

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

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

First, suppose $\mathfrak{q}\in\Spec B$ lies in the left-hand side. That is, $(\Spec\phi)(\mathfrak{q})=\phi^{-1}(\mathfrak{q})\in Z(\mathfrak{a})$. Then from $\mathfrak{a}\subseteq \phi^{-1}(\mathfrak{q})$ we obtain $\phi(\mathfrak{a})\subseteq \mathfrak{q}$, so $\mathfrak{q}\in Z(\phi(\mathfrak{a}))$.

Conversely, suppose $\mathfrak{q}\in\Spec B$ lies in 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., $\mathfrak{q}\in (\Spec\phi)^{-1}(Z(\mathfrak{a}))$.

Two important ring homomorphisms we will use in what follows 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.

$\Spec\pi$ and $\Spec\epsilon$ are both injective, and they define homeomorphisms onto their respective images.
The image of $\Spec\pi$ in $\Spec A$ is a closed set.
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

The first statement follows from the two propositions [Commutative Algebra] §Basic Notions, ⁋Proposition 11 and [Commutative Algebra] §Localization, ⁋Proposition 8 mentioned above. For the second, the image of $\Spec\pi$ in $\Spec A$ is exactly $Z(\mathfrak{a})$. Finally, for the third, the elements of $\Spec\epsilon$ are the prime ideals not containing $f$, and these can be written as $\Spec A\setminus Z(f)$.

It is obvious that the topological space $\Spec A$ has a base, but we also want this base to bear some relation to the algebraic structure of $A$. We make the following definition.

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

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 coincides with $D(f)$ as a set follows from

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

together with [Commutative Algebra] §Localization, ⁋Proposition 8. Moreover, that $D(f)$ carries the same topology as $\Spec A_f$ also follows from the fact that the correspondence in [Commutative Algebra] §Localization, ⁋Proposition 8 preserves inclusion.

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

Proof

For any open set $\Spec A \setminus Z(S)$, using the second statement of Lemma 6 we have the 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 similar computation using this lemma, we can verify that $D(fg)=D(f)\cap D(g)$. However, since this computation uses the first statement of Lemma 6, it cannot in general be extended to infinite index sets.

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

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

Proof

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

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

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

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 lie in $\bigcap_{i\in I} Z(f_i)$; hence the above equality is equivalent to $\sum(f_i)=(1)$. Since $1\in\sum(f_i)$, there exist a finite subset $J\subseteq I$ and a family of elements $(a_j)_{j\in J}$ in $A$ such that $\sum_{j\in J} a_jf_j=1$. Thus

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

and reversing the above computations yields the desired result.

Galois correspondence

By definition, $Z$ is a function that takes an ideal of $A$ and yields a closed subset of $\Spec A$. Conversely, we define the following.

Definition 13 For any subset $T\subseteq \Spec A$,

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

Then $I(T)$ is an ideal, being an intersection of ideals. Moreover, if $T_1\subseteq T_2$ then $I(T_2)\subseteq I(T_1)$ trivially.

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

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

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 all $T\in \mathcal{P}(\Spec A)$.

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

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

$IZ(S)=\sqrt{(S)}$
$ZI(T)=\cl(T)$

Proof

This is obvious from the formula
\[I(Z(S))=I(Z((S)))=\bigcap_\text{\scriptsize$\mathfrak{p}$ a prime in $Z((S))$} \mathfrak{p}=\bigcap_\text{\scriptsize$\mathfrak{p}$ a prime containing $(S)$} \mathfrak{p}=\sqrt{(S)}\]
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 $\Spec A$.

One somewhat confusing point in this correspondence is that since every prime ideal of $A$ is 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 for any irreducible closed subset $Y$, the ideal $I(Y)$ is prime.

First, assume $Z(\mathfrak{p})$ is not irreducible. Then there exist two proper nontrivial closed subsets $Z(\mathfrak{a}), Z(\mathfrak{b})$ of $Z(\mathfrak{p})$ such that $Z(\mathfrak{p})=Z(\mathfrak{a})\cup Z(\mathfrak{b})$. ([Topology] §Dimension, ⁋Definition 6) Now

\[Z(\mathfrak{p})=Z(\mathfrak{a})\cup Z(\mathfrak{b})=Z(\mathfrak{ab})\iff \mathfrak{p}=\sqrt{\mathfrak{p}}=\sqrt{\mathfrak{ab}}=\sqrt{\mathfrak{a}\cap \mathfrak{b}}\]

On the other hand

\[\sqrt{\mathfrak{a}\cap \mathfrak{b}}\supseteq \mathfrak{a}\cap \mathfrak{b}\supseteq \mathfrak{a}\mathfrak{b}\]

so combining the above two formulas we see that $\mathfrak{p}\supseteq \mathfrak{a}$ or $\mathfrak{p}\supseteq \mathfrak{b}$ must hold, and therefore $Z(\mathfrak{p})=Z(\mathfrak{a})$ or $Z(\mathfrak{p})=Z(\mathfrak{b})$, a contradiction.

Conversely, for any irreducible closed subset $Y$, we must show that $I(Y)$ is a prime ideal. Since $Y$ is closed, 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.

To this end, suppose $fg\in \mathfrak{a}$, and consider the two nonempty open subsets $D(f), D(g)$ 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 subsets of the irreducible closed set $Y$, so for this to hold, one of them must be empty. ([Topology] §Dimension, ⁋Proposition 7) Hence $f\in \mathfrak{a}$ 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 satisfying $\mathfrak{q}\subsetneq \mathfrak{p}$—that is, if $\mathfrak{p}$ is a minimal prime ideal—then $Z(\mathfrak{p})$ is maximal among irreducible closed subsets, i.e., 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 $\Spec A$.

In particular, the spectrum of an integral domain—the example considered above—is irreducible.

Classical algebraic geometry

Before defining the structure sheaf in the next post, it is helpful to briefly review classical algebraic geometry. In what follows we fix an algebraically closed field $\mathbb{K}$. Then the (classical) affine $n$-space over $\mathbb{K}$ is the set

\[\mathbb{A}_{\mathbb{K},\text{classical}}^n=\{(x_1,\ldots, x_n)\mid x_1,\ldots, x_n\in \mathbb{K}\}\]

Each element $x=(x_1,\ldots, x_n)$ of $\mathbb{A}_{\mathbb{K},\text{classical}}^n$ is called a point, and each $x_i$ is called the $i$-th coordinate of $x$.

Now consider the polynomial ring $A=\mathbb{K}[\x_1,\ldots,\x_n]$ of polynomials with coefficients in $\mathbb{K}$. Then for any $x=(x_1,\ldots, x_n)\in \mathbb{A}_{\mathbb{K},\text{classical}}^n$, we can define the ideal $\mathfrak{m}_x$ of $A$ by

\[\mathfrak{m}_x=(\x_1-x_1,\ldots, \x_n-x_n)\subseteq \mathbb{K}[\x_1,\ldots,\x_n]\]

and in [Commutative Algebra] §Nullstellensatz we proved that these are all the maximal ideals of $A=\mathbb{K}[\x_1,\ldots,\x_n]$. Therefore, writing $\mSpec A$ for the set of maximal ideals of $A$, we have $\mathbb{A}_{\mathbb{K},\text{classical}}^n= \mSpec A$.

On the other hand, we can regard any $f\in A$ as a function from $\mathbb{A}_{\mathbb{K},\text{classical}}^n$ to $\mathbb{K}$ via

\[\mathbb{A}_{\mathbb{K},\text{classical}}^n \rightarrow \mathbb{K};\qquad (x_1,\ldots, x_n)\mapsto f(x_1,\ldots, x_n)\]

Then, similarly to Definition 3, for any subset $S$ of $A$, defining

\[Z(S)=\{(x_1,\ldots, x_n)\mid \text{$f(x_1,\ldots, x_n)=0$ for all $f\in S$}\}\]

we can think of $Z(S)$ as the common zeros of the collection $S$ of functions on $\mathbb{A}_{\mathbb{K},\text{classical}}^n$, and by a computation similar to Lemma 6 we can define a topology on $\mSpec A$ having the $Z(S)$ as closed sets.

Similarly, in parallel with Definition 13, for a subset $T$ of $\mathbb{A}_{\mathbb{K},\text{classical}}^n$, defining

\[I(T)=\{f\in A\mid \text{$f(x_1,\ldots, x_n)=0$ for all $x\in T$}\}\]

we know, as in Theorem 15, that $I$ and $Z$ define an antitone Galois connection. The proposition in $\mSpec A$ corresponding to Proposition 16 is obtained from [Commutative Algebra] §Nullstellensatz, ⁋Theorem 4 because $\mathbb{K}[\x_1,\ldots,\x_n]$ is Jacobson.

Earlier we saw that the elements of $A$, i.e., polynomials, behave like functions on $\mathbb{A}_{\mathbb{K},\text{classical}}^n$. More generally, on a principal open set $D(g)$, the fraction $f/g$ is also well defined, and in algebraic geometry we are interested in such functions. Functions of this form are not well defined on all of $\mathbb{A}_{\mathbb{K},\text{classical}}^n$, so we will look at functions that can be represented in the form $f/g$ in a suitable neighborhood of each point.

The relationship between rings and spectra

Summarizing the above discussion, we can organize it as follows. The last few rows are results from Classical algebraic geometry; although we have not yet examined how these behave for the spectrum of a general ring, we will verify this in the next post.

$$A$$		$$\Spec A$$
prime ideal		point
$$\mathfrak{p}\subset A$$	$$\longleftrightarrow$$	$$\mathfrak{p}\in \Spec A$$
maximal ideal		closed point
$$\mathfrak{m}\subset A$$	$$\longleftrightarrow$$	$$\mathfrak{m}\in \text{(m)}\Spec A$$
prime ideal		irreducible closed subset
$$\mathfrak{p}$$	$$\longrightarrow$$	$$Z(\mathfrak{p})$$
$$I(Y)$$	$$\longleftarrow$$	$$Y$$
minimal prime ideal		irreducible component
$$\mathfrak{p}$$	$$\longrightarrow$$	$$Z(\mathfrak{p})$$
$$I(Y)$$	$$\longleftarrow$$	$$Y$$
radical ideal		closed set
$$\mathfrak{a}$$	$$\longrightarrow$$	$$Z(\mathfrak{a})$$
$$I(Y)$$	$$\longleftarrow$$	$$Y$$
ideal sum		intersection
$$\mathfrak{a}+\mathfrak{b}$$	$$\longrightarrow$$	$$Z(\mathfrak{a})\cap Z(\mathfrak{b})$$
$$\sqrt{I(V)+I(W)}$$	$$\longleftarrow$$	$$V\cap W$$
ideal product		union
$$\mathfrak{a}\mathfrak{b}$$	$$\longrightarrow$$	$$Z(\mathfrak{a})\cup Z(\mathfrak{b})$$
$$\sqrt{I(V)I(W)}$$	$$\longleftarrow$$	$$V\cup W$$
element		function
$$f\in A$$	$$\longrightarrow$$	$$\mathfrak{p}\mapsto f\pmod{\mathfrak{p}}$$

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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Spectra

\(\Spec A\) as a set

\(\Spec A\) as a topological space

Galois correspondence

Classical algebraic geometry

The relationship between rings and spectra

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