
Lemma 10.16.2. Let $R$ be a ring.

1. The spectrum of a ring $R$ is empty if and only if $R$ is the zero ring.

2. Every nonzero ring has a maximal ideal.

3. Every nonzero ring has a minimal prime ideal.

4. Given an ideal $I \subset R$ and a prime ideal $I \subset \mathfrak p$ there exists a prime $I \subset \mathfrak q \subset \mathfrak p$ such that $\mathfrak q$ is minimal over $I$.

5. If $T \subset R$, and if $(T)$ is the ideal generated by $T$ in $R$, then $V((T)) = V(T)$.

6. If $I$ is an ideal and $\sqrt{I}$ is its radical, see basic notion (27), then $V(I) = V(\sqrt{I})$.

7. Given an ideal $I$ of $R$ we have $\sqrt{I} = \bigcap _{I \subset \mathfrak p} \mathfrak p$.

8. If $I$ is an ideal then $V(I) = \emptyset$ if and only if $I$ is the unit ideal.

9. If $I$, $J$ are ideals of $R$ then $V(I) \cup V(J) = V(I \cap J)$.

10. If $(I_ a)_{a\in A}$ is a set of ideals of $R$ then $\cap _{a\in A} V(I_ a) = V(\cup _{a\in A} I_ a)$.

11. If $f \in R$, then $D(f) \amalg V(f) = \mathop{\mathrm{Spec}}(R)$.

12. If $f \in R$ then $D(f) = \emptyset$ if and only if $f$ is nilpotent.

13. If $f = u f'$ for some unit $u \in R$, then $D(f) = D(f')$.

14. If $I \subset R$ is an ideal, and $\mathfrak p$ is a prime of $R$ with $\mathfrak p \not\in V(I)$, then there exists an $f \in R$ such that $\mathfrak p \in D(f)$, and $D(f) \cap V(I) = \emptyset$.

15. If $f, g \in R$, then $D(fg) = D(f) \cap D(g)$.

16. If $f_ i \in R$ for $i \in I$, then $\bigcup _{i\in I} D(f_ i)$ is the complement of $V(\{ f_ i \} _{i\in I})$ in $\mathop{\mathrm{Spec}}(R)$.

17. If $f \in R$ and $D(f) = \mathop{\mathrm{Spec}}(R)$, then $f$ is a unit.

Proof. We address each part in the corresponding item below.

1. This is a direct consequence of (2) or (3).

2. Let $\mathfrak {A}$ be the set of all proper ideals of $R$. This set is ordered by inclusion and is non-empty, since $(0) \in \mathfrak {A}$ is a proper ideal. Let $A$ be a totally ordered subset of $\mathfrak A$. Then $\bigcup _{I \in A} I$ is in fact an ideal. Since 1 $\notin I$ for all $I \in A$, the union does not contain 1 and thus is proper. Hence $\bigcup _{I \in A} I$ is in $\mathfrak {A}$ and is an upper bound for the set $A$. Thus by Zorn's lemma $\mathfrak {A}$ has a maximal element, which is the sought-after maximal ideal.

3. Since $R$ is nonzero, it contains a maximal ideal which is a prime ideal. Thus the set $\mathfrak {A}$ of all prime ideals of $R$ is nonempty. $\mathfrak {A}$ is ordered by reverse-inclusion. Let $A$ be a totally ordered subset of $\mathfrak {A}$. It's pretty clear that $J = \bigcap _{I \in A} I$ is in fact an ideal. Not so clear, however, is that it is prime. Let $xy \in J$. Then $xy \in I$ for all $I \in A$. Now let $B = \{ I \in A | y \in I\}$. Let $K = \bigcap _{I \in B} I$. Since $A$ is totally ordered, either $K = J$ (and we're done, since then $y \in J$) or $K \supset J$ and for all $I \in A$ such that $I$ is properly contained in $K$, we have $y \notin I$. But that means that for all those $I, x \in I$, since they are prime. Hence $x \in J$. In either case, $J$ is prime as desired. Hence by Zorn's lemma we get a maximal element which in this case is a minimal prime ideal.

4. This is the same exact argument as (3) except you only consider prime ideals contained in $\mathfrak {p}$ and containing $I$.

5. $(T)$ is the smallest ideal containing $T$. Hence if $T \subset I$, some ideal, then $(T) \subset I$ as well. Hence if $I \in V(T)$, then $I \in V((T))$ as well. The other inclusion is obvious.

6. Since $I \subset \sqrt{I}, V(\sqrt{I}) \subset V(I)$. Now let $\mathfrak {p} \in V(I)$. Let $x \in \sqrt{I}$. Then $x^ n \in I$ for some $n$. Hence $x^ n \in \mathfrak {p}$. But since $\mathfrak {p}$ is prime, a boring induction argument gets you that $x \in \mathfrak {p}$. Hence $\sqrt{I} \subset \mathfrak {p}$ and $\mathfrak {p} \in V(\sqrt{I})$.

7. Let $f \in R \setminus \sqrt{I}$. Then $f^ n \notin I$ for all $n$. Hence $S = \{ 1, f, f^2, \ldots \}$ is a multiplicative subset, not containing $0$. Take a prime ideal $\bar{\mathfrak {p}} \subset S^{-1}R$ containing $S^{-1}I$. Then the pull-back $\mathfrak {p}$ in $R$ of $\bar{\mathfrak {p}}$ is a prime ideal containing $I$ that does not intersect $S$. This shows that $\bigcap _{I \subset \mathfrak p} \mathfrak p \subset \sqrt{I}$. Now if $a \in \sqrt{I}$, then $a^ n \in I$ for some $n$. Hence if $I \subset \mathfrak {p}$, then $a^ n \in \mathfrak {p}$. But since $\mathfrak {p}$ is prime, we have $a \in \mathfrak {p}$. Thus the equality is shown.

8. $I$ is not the unit ideal if and only if $I$ is contained in some maximal ideal (to see this, apply (2) to the ring $R/I$) which is therefore prime.

9. If $\mathfrak {p} \in V(I) \cup V(J)$, then $I \subset \mathfrak {p}$ or $J \subset \mathfrak {p}$ which means that $I \cap J \subset \mathfrak {p}$. Now if $I \cap J \subset \mathfrak {p}$, then $IJ \subset \mathfrak {p}$ and hence either $I$ of $J$ is in $\mathfrak {p}$, since $\mathfrak {p}$ is prime.

10. $\mathfrak {p} \in \bigcap _{a \in A} V(I_ a) \Leftrightarrow I_ a \subset \mathfrak {p}, \forall a \in A \Leftrightarrow \mathfrak {p} \in V(\cup _{a\in A} I_ a)$

11. If $\mathfrak {p}$ is a prime ideal and $f \in R$, then either $f \in \mathfrak {p}$ or $f \notin \mathfrak {p}$ (strictly) which is what the disjoint union says.

12. If $a \in R$ is nilpotent, then $a^ n = 0$ for some $n$. Hence $a^ n \in \mathfrak {p}$ for any prime ideal. Thus $a \in \mathfrak {p}$ as can be shown by induction and $D(f) = \emptyset$. Now, as shown in (7), if $a \in R$ is not nilpotent, then there is a prime ideal that does not contain it.

13. $f \in \mathfrak {p} \Leftrightarrow uf \in \mathfrak {p}$, since $u$ is invertible.

14. If $\mathfrak {p} \notin V(I)$, then $\exists f \in I \setminus \mathfrak {p}$. Then $f \notin \mathfrak {p}$ so $\mathfrak {p} \in D(f)$. Also if $\mathfrak {q} \in D(f)$, then $f \notin \mathfrak {q}$ and thus $I$ is not contained in $\mathfrak {q}$. Thus $D(f) \cap V(I) = \emptyset$.

15. If $fg \in \mathfrak {p}$, then $f \in \mathfrak {p}$ or $g \in \mathfrak {p}$. Hence if $f \notin \mathfrak {p}$ and $g \notin \mathfrak {p}$, then $fg \notin \mathfrak {p}$. Since $\mathfrak {p}$ is an ideal, if $fg \notin \mathfrak {p}$, then $f \notin \mathfrak {p}$ and $g \notin \mathfrak {p}$.

16. $\mathfrak {p} \in \bigcup _{i \in I} D(f_ i) \Leftrightarrow \exists i \in I, f_ i \notin \mathfrak {p} \Leftrightarrow \mathfrak {p} \in \mathop{\mathrm{Spec}}(R) \setminus V(\{ f_ i\} _{i \in I})$

17. If $D(f) = \mathop{\mathrm{Spec}}(R)$, then $V(f) = \emptyset$ and hence $fR = R$, so $f$ is a unit.

$\square$

There are also:

• 2 comment(s) on Section 10.16: The spectrum of a ring

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).