The Stacks project

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

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 $D(a) = \emptyset$. 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.

Proof. It is basic notion (41) that $\mathfrak p := \varphi ^{-1}(\mathfrak p')$ is indeed a prime ideal of $R$. The last assertion of the lemma follows directly from the definitions, and implies the first. $\square$

Proof. Denote the right hand side of the arrow of the lemma by $D$. Choose a prime $\mathfrak p' \subset S^{-1}R$ and let $\mathfrak p$ the inverse image of $\mathfrak p'$ in $R$. Since $\mathfrak p'$ does not contain $1$ we see that $\mathfrak p$ does not contain any element of $S$. Hence $\mathfrak p \in D$ and we see that the image is contained in $D$. Let $\mathfrak p \in D$. By assumption the image $\overline{S}$ does not contain $0$. By basic notion (54) $\overline{S}^{-1}(R/\mathfrak p)$ is not the zero ring. By basic notion (62) we see $S^{-1}R / S^{-1}\mathfrak p = \overline{S}^{-1}(R/\mathfrak p)$ is a domain, and hence $S^{-1}\mathfrak p$ is a prime. The equality of rings also shows that the inverse image of $S^{-1}\mathfrak p$ in $R$ is equal to $\mathfrak p$, because $R/\mathfrak p \to \overline{S}^{-1}(R/\mathfrak p)$ is injective by basic notion (55). This proves that the map $\mathop{\mathrm{Spec}}(S^{-1}R) \to \mathop{\mathrm{Spec}}(R)$ is bijective onto $D$ with inverse as given. It is continuous by Lemma 10.17.4. Finally, let $D(g) \subset \mathop{\mathrm{Spec}}(S^{-1}R)$ be a standard open. Write $g = h/s$ for some $h\in R$ and $s\in S$. Since $g$ and $h/1$ differ by a unit we have $D(g) = D(h/1)$ in $\mathop{\mathrm{Spec}}(S^{-1}R)$. Hence by Lemma 10.17.4 and the bijectivity above the image of $D(g) = D(h/1)$ is $D \cap D(h)$. This proves the map is open as well. $\square$

Proof. This is a special case of Lemma 10.17.5. $\square$

Proof. It is immediate that the image is contained in $V(I)$. On the other hand, if $\mathfrak p \in V(I)$ then $\mathfrak p \supset I$ and we may consider the ideal $\mathfrak p /I \subset R/I$. Using basic notion (51) we see that $(R/I)/(\mathfrak p/I) = R/\mathfrak p$ is a domain and hence $\mathfrak p/I$ is a prime ideal. From this it is immediately clear that the image of $D(f + I)$ is $D(f) \cap V(I)$, and hence the map is a homeomorphism. $\square$

Proof. It suffices to prove that any covering of $\mathop{\mathrm{Spec}}(R)$ by standard opens can be refined by a finite covering. Thus suppose that $\mathop{\mathrm{Spec}}(R) = \cup D(f_ i)$ for a set of elements $\{ f_ i\} _{i\in I}$ of $R$. This means that $\cap V(f_ i) = \emptyset$. According to Lemma 10.17.2 this means that $V(\{ f_ i \} ) = \emptyset$. According to the same lemma this means that the ideal generated by the $f_ i$ is the unit ideal of $R$. This means that we can write $1$ as a finite sum: $1 = \sum _{i \in J} r_ i f_ i$ with $J \subset I$ finite. And then it follows that $\mathop{\mathrm{Spec}}(R) = \cup _{i \in J} D(f_ i)$. $\square$

Proof. The spectrum of a ring is quasi-compact, see Lemma 10.17.8. It has a basis for the topology consisting of the standard opens $D(f) = \mathop{\mathrm{Spec}}(R_ f)$ (Lemma 10.17.6) which are quasi-compact by the first remark. The intersection of two standard opens is quasi-compact as $D(f) \cap D(g) = D(fg)$. Given any two quasi-compact opens $U, V \subset X$ we may write $U = D(f_1) \cup \ldots \cup D(f_ n)$ and $V = D(g_1) \cup \ldots \cup D(g_ m)$. Then $U \cap V = \bigcup D(f_ ig_ j)$ which is quasi-compact. $\square$