The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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.

$\square$


Comments (0)

There are also:

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

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00E0. Beware of the difference between the letter 'O' and the digit '0'.