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Postmodern Algebra

Sam Macdonald

Section 11.2 The Spectrum of a Ring

“What is life but a spectrum and what is music but life itself.”
―Billy Cobham

Definition 11.12. Prime Spectrum.

Let \(R\) be a ring. The prime spectrum of \(R\text{,}\) denoted \(\Spec(R)\text{,}\) is the set of prime ideals of \(R\text{.}\) The set of maximal ideals of \(R\) is denoted \(\mSpec(R)\) .

Definition 11.13. \(V(I)\).

For a ring \(R\) and an ideal \(I\text{,}\) we set
\begin{equation*} V(I) := \{ P \in \Spec(R) \ \mid \ I \sse P\}. \end{equation*}

Proof.

Both \((a)\) and \((b)\) are straightforward, so we just prove \((c)\text{.}\)
To see \(V(I)\cup V(J) \subseteq V(I \cap J)\text{,}\) just observe that if \(I \sse P\) or \(J \sse P\text{,}\) then \(I \cap J \sse P\text{.}\)
Since \(IJ \subseteq I \cap J\text{,}\) we have \(V(I \cap J) \subseteq V(IJ)\text{.}\) To show \(V(IJ) \subseteq V(I) \cup V(J)\text{,}\) if \(P\) is a prime and \(P \notin V(I) \cup V(J)\text{,}\) then \(P \not\supseteq I, P \not\supseteq J\text{.}\) Thus we can find \(f \in I\text{,}\) and \(g \in J\) such that \(f, g \notin P\text{.}\) Since \(P\) is prime, \(fg \notin P\text{,}\) while also \(fg \in IJ\text{.}\) Therefore, \(P \nsubseteq IJ\text{.}\)
To show \((d)\text{,}\) since ideals are closed for sums, if \(P \supseteq I_\lambda\) for all \(\lambda\text{,}\) then \(P \supseteq \sum_{\lambda} I_{\lambda}\text{.}\) Moreover, if \(P \supseteq \sum_{\lambda} I_{\lambda}\text{,}\) then in particular \(P \supseteq I_\lambda\text{.}\)

Example 11.15. Poset Structure of Spec.

Note that \(\Spec(R)\) is also a poset under inclusion. Show that the poset structure of \(\Spec(R)\) can be recovered from the topology as follows:
\begin{equation*} P \subseteq Q \quad \Leftrightarrow \quad Q \in \overline{\{ P \}}. \end{equation*}

Example 11.16. Spectrum of Z.

The spectrum of \(\mathbb{Z}\) is, as a poset:
Solution.
#empty
The closed sets are of the form \(V((n))\text{,}\) which are the whole space when \(n=0\text{,}\) the empty set when \(n=1\text{,}\) and any finite union of things in the top row. Any closed set that contains \((0)\) must be all of \(\Spec(\Z)\text{.}\)

Definition 11.17. Radical Ideal.

The radical of an ideal \(I\) in a ring \(R\) is the ideal
\begin{equation*} \sqrt{I} := \{f\in R \ | \ f^n\in I \textrm{ for some } n\}. \end{equation*}
An ideal is a radical ideal if \(I=\sqrt{I}\text{.}\)

Definition 11.19. Reduced Ring.

A ring \(R\) is reduced if it has no nonzero nilpotents.

Example 11.20. Ideal Radical iff R/I Reduced.

An ideal \(I\) in \(R\) is radical if and only if \(R/I\) is reduced.

Proof.

The containment \(I \subseteq \sqrt{I}\) is immediate from the definition of radical, and thus we have \(V(I) \subseteq V(\sqrt{I})\text{.}\) Now let \(P \supseteq I\) be a prime ideal, and let \(f \in \sqrt{I}\text{.}\) By definition, there exists some \(n\) such that \(f^n \in I \subseteq P\text{,}\) but since \(P\) is prime, we conclude that \(f \in P\text{.}\) Therefore, \(V(\sqrt{I}) \subseteq V(I)\text{,}\) and we are done.

Proof.

Let \(P\) be a prime ideal in \(S\text{.}\) Given elements \(f, g \in R\) such that \(fg \in P \cap R\text{,}\) then \(\varphi(f)\varphi(g)=\varphi(fg) \in P\text{,}\) and since \(P\) is prime, we conclude that \(\varphi(f) \in P\) or \(\varphi(g) \in P\text{.}\) Therefore, \(f \in P \cap R\) or \(g \in P \cap R\text{.}\)

Definition 11.23. Induced Map on \(\Spec\).

Each ring homomorphism \(\varphi\!: R\to S\) induces a map on spectra \(\varphi^*\!:\Spec(S)\to\Spec(R)\) given by \(\varphi^*(P)=\varphi^{-1}(P) = P \cap R\text{.}\)

Example 11.24. Induced Map on Spec Continuous.

The induced map on spectra is not only an order-preserving map, but also it is continuous: if \(U\subseteq \Spec(R)\) is open, we have \(U=\Spec(R) \setminus V(I)\) for some ideal \(I\text{;}\) then for a prime \(Q\) of \(S\text{,}\)
\begin{equation*} Q \in (\varphi^*)^{-1}(U) \quad \Longleftrightarrow \quad Q \cap R \not\supseteq I \quad \Longleftrightarrow \quad Q \not\supseteq IS \quad \Longleftrightarrow \quad Q \notin V(IS). \end{equation*}
So \((\varphi^*)^{-1}(U)\) is the complement of \(V(IS)\text{,}\) and thus open.

Definition 11.25. Minimal Prime.

Let \(I\) be an ideal in a ring \(R\text{.}\) A prime \(P\) is a minimal prime of \(I\) if \(P\) is minimal in \(V(I)\text{.}\) A minimal prime of \(R\) is a minimal element in \(\Spec(R)\text{.}\)

Definition 11.26. Minimal Prime.

Let \(I\) be an ideal in a ring \(R\text{.}\) A minimal prime of \(I\) is a minimal element (with respect to containment) in \(V(I)\text{.}\) More precisely, \(P\) is a minimal prime of \(I\) if the following hold: - \(P\) is a prime ideal, - \(P \supseteq I\text{,}\) and - if \(Q\) is also a prime ideal and \(I \subseteq Q \subseteq P\text{,}\) then \(Q = P\text{.}\)
The set of minimal primes of \(I\) is denoted \(\Min(I)\) .

Proof.

Fix an ideal \(I\) and a prime \(P \supseteq I\text{,}\) and consider the set
\begin{equation*} S = \lbrace Q \in V(I) \mid P \supseteq Q \rbrace, \end{equation*}
which is partially ordered with \(\supseteq\text{.}\) On the one hand, \(P \in S\text{,}\) so \(S\) is nonempty. On the other hand, given any chain \(\{ Q_i \}_i\) in \(S\text{,}\) \(Q := \bigcap_i Q_i\) is a prime ideal in \(R\) ( #prove !). Moreover, \(Q\) contains \(I\text{,}\) since every \(Q_i\) contain \(I\text{,}\) and \(Q\) is contained in \(P\text{,}\) since every \(P_i \subseteq Q\text{.}\) Therefore, \(Q \in S\text{,}\) and Zorn’s Lemma applies to \(S\text{.}\) By [provisional cross-reference: cite] [[Mathematics/Foundations/Results/Theorem - Zorn’s Lemma]], \(S\) contains a maximal element for \(\supseteq\text{,}\) say \(Q\text{.}\)
Notice that \(Q\) is equivalently a minimal element for \(\subseteq\text{.}\) Now if \(Q'\) is a prime ideal with \(I \subseteq Q' \subseteq Q\text{,}\) then \(Q' \subseteq Q \subseteq P\text{,}\) and thus \(Q' \in S\text{.}\) Therefore, we must have \(Q' = Q\text{,}\) by maximality of \(Q\) with respect to \(\supseteq\text{.}\) We conclude that \(Q\) is a minimal prime of \(I\text{,}\) and by definition \(Q\) is contained in \(P\text{.}\)

Definition 11.28. Multiplicatively Closed Set.

Suppose \(R\) is a commutative ring and \(S \subseteq R\) is a subset such that 1. \(1 \in S\text{,}\) 2. \(S\) is closed under multiplication (i.e., if \(x,y \in S\text{,}\) then \(xy \in S\)), and 3. \(S\) does not contain \(0\) nor any zerodivisors. Such a subset \(S\) is called a multiplicatively closed subset of non zero divisors of \(R\text{.}\)

Proof.

Consider the family of ideals \(\mathcal{F} := \{ J \ | \ J \supseteq I, J \cap W=\varnothing\}\) ordered with inclusion. This is nonempty, since it contains \(I\text{,}\) and any chain \(J_1 \subseteq J_2 \subseteq \cdots\) has an upper bound \(\cup_i J_i\text{.}\) Therefore, \(\mathcal{F}\) has some maximal element \(\A\) by a basic application of [provisional cross-reference: cite] [[Mathematics/Foundations/Results/Theorem - Zorn’s Lemma|Zorn’s Lemma]]. We claim \(\A\) is prime. Suppose \(f,g\notin \A\text{.}\) By maximality, \(\A+(f)\) and \(\A+(g)\) both have nonempty intersection with \(W\text{,}\) so there exist \(r_1f+a_1\text{,}\) \(r_2 g+a_2 \in W\text{,}\) with \(a_1,a_2\in \A\text{.}\) If \(fg\in \A\text{,}\) then
\begin{equation*} \underset{\in W}{(r_1f+a_1)}\underset{\in W}{(r_2g+a_2)} = r_1 r_2 fg + r_1 f \underset{\in \mathbb{A}}{a_2} + r_2 g \underset{\in \mathbb{A}}{a_1} + \underset{\in \mathbb{A}}{a_1 a_2} \in W \cap \A, \end{equation*}

Proof.

First to justify the equivalence of the two statements we observe:
\begin{equation*} V(I) \subseteq V(f) \Longleftrightarrow f\in P \ \text{for all} \ P \in V(I) \Longleftrightarrow f \in \bigcap_{P \in V(I)} P. \end{equation*}
Now we will show that \(\displaystyle\bigcap_{P \in V(I)} P = \sqrt{I}\text{:}\)
\((\supseteq)\text{:}\) It suffices to show that \(P \supseteq I\) implies \(P \supseteq \sqrt{I}\text{,}\) and indeed
\begin{equation*} f \in \sqrt{I} \implies f^n\in I \subseteq P \implies f\in P. \end{equation*}
\((\subseteq)\text{:}\) If \(f\notin \sqrt{I}\text{,}\) consider the multiplicatively closed set \(W=\{1,f,f^2,f^3,\ldots\}\text{.}\) We have \(W\cap I=\varnothing\) by hypothesis. There is a prime \(P\) in \(V(I)\) that does not intersect \(W\text{,}\) and hence \(P\) does not contain \(f.\)
Finally, \(\Min(I) \subseteq V(I)\text{,}\) and since every prime in \(V(I)\) contains a minimal prime of \(I\text{,}\) we conclude that
\begin{equation*} \bigcap_{P \in V(I)} P = \bigcap_{P \in \Min(I)} P. \end{equation*}

Example 11.31. Radical of \((x^2)\) is \((x)\).

Let \(k\) be a field, \(R = k[x]\text{,}\) and \(I = (x^2)\text{.}\) On the one hand, it is immediate form the definition that \(x \in \sqrt{I}\text{,}\) and thus \((x) \subseteq \sqrt{I}\text{.}\) On the other hand, \((x)\) is a prime ideal that contains \(I\text{,}\) and thus by [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Theorem - Spectrum Analogue of Strong Nullstellensatz|Theorem]] we must have \(\sqrt{I} = (x)\text{.}\)

Proof.

The closed sets of \(\Spec(R)\) are precisely the sets of the form \(V(I)\) for some ideal \(I\text{.}\) As \(V(I) = V(\sqrt{I})\) the closed sets of \(\Spec(R)\) are given by \(V(I)\) where \(I\) ranges over all radical ideals. We showed in [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Proposition - Properties of V|Proposition]] that the map
\begin{equation*} V: \{\text{radical ideals } I\subseteq R\} \longrightarrow \{ \text{closed subsets of } \Spec(R) \} \end{equation*}
is order-reversing. Finally, suppose that \(I\) and \(J\) are ideals such that \(V(I) = V(J)\text{.}\) By [provisional cross-reference: cite] [[Mathematics/Commutative Algebra/Results/Theorem - Spectrum Analogue of Strong Nullstellensatz|Theorem]],
\begin{equation*} \sqrt{I} = \bigcap_{P \in V(I)} P = \bigcap_{P \in V(J)} P = \sqrt{J}. \end{equation*}
Conversely, suppose that \(\sqrt{I} = \sqrt{J}\text{.}\) Given a prime \(P \in V(I)\text{,}\) we also have \(P \in V(\sqrt{I})\text{,}\) and thus
\begin{equation*} P \supseteq \sqrt{I} = \sqrt{J} \supseteq J, \end{equation*}
so \(P \in V(I)\text{.}\) Since the same argument applies to show that \(V(J) \supseteq V(I)\text{,}\) we conclude that \(V(I) = V(J)\text{.}\)

Example 11.33. Properties of Radical Ideals.

Let \(I\) and \(J\) be ideals in a ring \(R\text{.}\) (a) Show that \(\sqrt{\sqrt{I}} = \sqrt{I}\text{.}\) (b) Show that if \(I \subseteq J\text{,}\) then \(\sqrt{I} \subseteq \sqrt{J}\text{.}\) (c) Show that \(\sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J}\text{.}\) (d) Show that \(\sqrt{I^n} = \sqrt{I}\) for all \(n \geqslant 1\text{.}\) (e) Show that if \(P\) is a prime ideal, then \(\sqrt{P^n} = P\) for all \(n \geqslant 1\text{.}\)
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