Errata and Addenda for Algebraic Geometry I (Edition 2) Show errata for edition 1

Here we post a list of errata and addenda. The name tags refer to the people who found the mistake. We are very grateful to all of them. Further remarks and hints - trivial or not - are very welcome.

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103 errata listed.

PageDescriptionSubmitted byEd.
p. 6,
Preface to 2nd edition
Add Peng Du to the list of names. U. Görtz 2
p. 8,
Replace "field" by "fields". A. Graf 1 2
p. 13,
Line 16
Replace "subset" by "open subset". A. Graf 1 2
p. 14,
Lemma 1.19 (2)
The statement remains true without the assumption that $I$ be finite: Let $\emptyset\ne V\subseteq X$ be open. It is enough to show that $V\cap U_i\ne\emptyset$ for all $i$, because this implies that the closure of $V$ in $X$ contains $U_i$, hence that $V$ is dense in $X$.
Let $I_1 = \{ i\in I;\ V\cap U_i=\emptyset\}$, $I_2=I\setminus I_1$. Then $X = \bigcup_{i\in I_1} U_i\cup \bigcup_{i\in I_2} U_i$, and this union is disjoint. In fact, assume $U_{i_1}\cap U_{i_2}\ne \emptyset$, $i_\nu\in I_\nu$. Since $V\cap U_{i_2}\ne \emptyset$ and any two non-empty opens in the irreducible space $U_{i_2}$ intersect, we find $U_{i_1}\cap U_{i_2}\cap V \ne \emptyset$, a contradiction to $i_1\in I_1$. Since $X$ is connected, $I_1$ must be empty.
M. Schaller 1 2
p. 15,
Line -15
Replace $\mathscr M$ by $\mathcal M$. I. Tselepidis 1 2
p. 17,
Line 20
Replace "function" by "functions". Ehsan Shahoseini 1 2
p. 19,
Line 18
Add "contravariant" (or "opposite category of ..."). Denis Vogel 1 2
p. 28,
Line 21
The reason provided after the "as ..." is not related to this definition giving a topology (you will not need this in a verification of the axioms for a toplogy; it is the final topology wr.t. the inclusion maps for the U_i). Instead, it is the reason that the U_i are open in this topology. Denis Vogel 1 2
p. 29,
Proof of Corollary 1.60
For the proof of the second assertion, i.e. $\mathbb{P}^n$ is a prevariety, one needs to show the connectedness of $\mathbb{P}^n$. But the book use part (2) of Lemma 1.19 (to show irreduciblity) which itself needs the connectedness! If connectedness is trivial, the second assertion of the Corollary 1.60 is proved and there is no need to speak about Lemma 1.19 and irreduciblity; and if connectedness is not trivial, then one cannot use the Lemma 1.19. To resolve this, one should make the reference to Lemma 1.19 more precise: Use part (1) to show that $\mathbb P^n(k)$ is connected (from the construction and previous discussion it is clear that any two of the standard charts have non-empty intersection.). Then irreducibility follows from part (2) (which needs the connectedness), but that was already noted as a general fact in Proposition 1.48. Ehsan Shahoseini 1 2
p. 30,
Line -10
After "are again in a", add reference to Exercise 1.20 (b). Ehsan Shahoseini 1 2
p. 33,
Line -13, -12
Replace $\mathbb{P}_k^n$ by $\mathbb{P}^n(k)$. E. Hong 1 2
p. 35,
Line 5
Replace "function field" with "function fields". Ehsan Shahoseini 1 2
p. 35,
Line -9
Replace "it" by ", the quadric of rank 3". Ehsan Shahoseini 1 2
p. 47,
Line -10
Replace "$R$-module" by "an $R$-module". A. Graf 1 2
p. 52,
Line 8
$\mathscr O(U)$ should be $\mathscr O_{\mathbb C}(U)$ Ehsan Shahoseini 1 2
p. 62,
Line -20
Replace $i^\flat_x$ by $i^\sharp_x$. A. Graf 1 2
p. 62,
Line 6 (in Example 2.37)
Replace "$f\in A$" by "$f\in A\setminus \{0\}$". Harmeet Singh 1 2
p. 66,
Line -13
Replace "subset" by "open subset". A. Graf 1 2
p. 70,
Line 7
Replace $\varphi(s)_{|X_{s}} \in \Gamma(X_{\varphi(s)}, \mathcal{O}_{X})^{\times}$ by $\varphi(s)_{|X_{\varphi(s)}} \in \Gamma(X_{\varphi(s)}, \mathcal{O}_{X})^{\times}$. Harmeet Singh 2
p. 71,
Prop. 3.8
Add a reference to Exer. 2.18. A. Graf 1 2
p. 72,
Line -16
Replace "with an open subscheme of $T_i$" by "with an open subscheme $T_i$ of $T$". E. Hong 1 2
p. 72,
Prop. 3.10
Add a reference to Exer. 2.16. (And/or odd a reference to Prop. 3.10 to the exercise.) A. Graf 1 2
p. 75,
Line -1
Replace the reference to Section (4.14) by a reference to Section (4.13). N.T. 1 2
p. 81,
Proof of Prop. 3.33
Replace the beginning of the first sentence by "If $x\in X$ and if $U={\rm Spec} A$ is an affine open neighborhood of $x$, then $x$ is closed in $U$ and corresponds to ..." A. Graf 1 2
p. 83,
Proof of Thm. 3.37
The point that $X(k)$ is connected should be addressed explicitly. A. Graf 1 2
p. 84,
Line -21
Add index $i$ to the second intersection. A. Graf 1 2
p. 86,
Line 22
Kernels of sheaves have not been defined yet at this point. A. Graf 2
p. 86,
Definition 3.41 (1), second line near the end
Replace $\mathcal{O}$ by $\mathcal{O}_X$. Andreas Blatter 2
p. 92,
Line 23
Replace "component" by "components". A. Graf 1 2
p. 92,
Exercise 3.19
Replace ${\rm Hom}({\rm Spec}(R), \mathbb P^n_R)$ by ${\rm Hom}_{R}({\rm Spec}(R), \mathbb P^n_R)$ to make explicit that we only consider $R$-morphisms here. A. Graf 1 2
p. 93,
Exercise 3.26(a)
The statement is true for $V = X$, but not in general. (E.g., take $Y= \mathbb A^1_k$, $U=\mathbb A^1_k\setminus \{0\}$, $I=\{1, 2\}$, $i=1$, $V=U_2\subseteq X$.) Paulo Lima-Filho 1 2
p. 107,
Remark/Def. 4.24
Conflict of notation: There are too many $f$'s here, $f\in \Gamma(U, \mathscr O_S)$ in line 2, the morphism $f\colon X\to S$, the polynomial $f$ in (1). U. Görtz 1 2
p. 114,
Line -10
The fiber product of the two projective spaces should be $\times_R$ instead of $\times_S$. Jingyi Xu 1 2
p. 116,
Line -15
Replace "as $S$-scheme" by "as an $S$-scheme". A. Graf 1 2
p. 116,
Line -19
Add reference to Cor. 4.7. A. Graf 1 2
p. 129,
Prop. 5.22 (4)
It would be enough to assume that $X$ and $Y$ are locally of finite type over $k$. A. Graf 1 2
p. 131,
Line -12
Since $d$ is never refered to, its definition in the statement of Prop. 5.30 should be removed. A. Graf 1 2
p. 133,
Proof of Cor. 5.33
Since the formulation "vanishes identically" has not been formally defined, it would be clearer to replace "in which case" by "i.e.,". A. Graf 1 2
p. 136,
Line 13
Better: "... where $S$ is the spectrum of a field $k$ and $X$ is of finite type over $k$." A. Graf 1 2
p. 137,
Line -4
It should be explained why $\bar{x}$ is closed in $X_K$. The point $\bar{x}$ is a point of the fiber $p^{-1}(x) = \Spec \kappa(x)\otimes_k K$. We now use Proposition 3.33. Since the extension $\kappa(x)/k$ is finite, $\kappa(x)\otimes_k K$ is a finite-dimensional $K$-vector space. This implies that the residue class field of $\bar{x}$ (which is a quotient of this tensor product) is finite over $K$. Hence $\bar{x}$ is a closed point of $X_K$. (A variant of the argument: Since $\kappa(x)\otimes_k K$ is finite over $K$, it is an Artin ring, hence the fiber $p^{-1}(x)$ has dimension $0$. Therefore $\bar{x}$ is a closed point of the fiber. Since $x$ is closed in $X$, $p^{-1}(x)$ is closed in $X_K$. Altogether we see that $\bar{x}$ is closed in $X_K$.) A. Graf 1 2
p. 137,
Line -15
It might be clearer to move this remark up so that it comes directly after the proof of Corollary 5.45. A. Graf 1 2
p. 140,
Proof of Prop. 5.53
In the proof of (iv) $\Rightarrow$ (iii), further details should be added about the construction of the scheme $Y'$.
To define $Y'$, one applies gluing of schemes (Prop. 3.10) to the family of schemes $U_i := {\rm Spec} (A_i\otimes k')/(f_{ij})$ (this is a different $U_i$ than in the proof of 5.53; I use $U_i$ here to indicate how Prop. 3.10 is applied), and $U_{ij} = U_i \cap {\rm Spec} (A_j \otimes k')$ (scheme-theoretic intersection inside $X\otimes k'$).
To construct a complete gluing datum, one has to show that $U_{ij} = U_{ji}$ as closed subschemes of ${\rm Spec} (A_i\otimes k') \cap {\rm Spec} (A_j\otimes k')$. (Since these identifications are (basically) equalities, it is clear that the cocycle condition of a gluing datum is satisfied, so that Prop. 3.10 can be applied.)
By construction, it is clear that the desired equality holds after base change to $\Omega$, i.e., that $U_{ij}\otimes_{k'}\Omega = U_{ji}\otimes_{k'}\Omega$. It therefore suffices to prove the following general statement:
Let $Z$ be a scheme over a field $k'$, let $\Omega / k'$ be a field extension and let $Z_1, Z_2$ be closed subschemes of $Z$ such that $Z_1 \otimes_{k'} \Omega = Z_2 \otimes_{k'} \Omega$ (as closed subschemes of $Z\otimes \Omega$). Then $Z_1 = Z_2$.
To prove this statement, it is enough to show that $Z_1\subseteq Z_2$ and that $Z_2\subseteq Z_1$, or in other words that the inclusions $Z_1\cap Z_2 \subseteq Z_i$ (for $i=1,2$) are equalities. With this observation, one easily reduces to the case that $Z_2 \subseteq Z_1$ (replace $Z_2$ by $Z_1\cap Z_2$, and apply the statement again with the roles of $Z_1$ and $Z_2$ reversed). Furthermore, one may work locally on $Z$, so that we may assume that $Z$ is affine, say $Z = {\rm Spec} A$.
Then the closed subschemes correspond to ideals $\mathfrak a_1 \subseteq \mathfrak a_2 \subseteq A$ such that $\mathfrak a_1 \otimes \Omega = \mathfrak a_2 \otimes \Omega$ (inside $A\otimes \Omega$). This implies that $(\mathfrak a_2 / \mathfrak a_2) \otimes \Omega = 0$, so $\mathfrak a_2 / \mathfrak a_1 = 0$ (base change by a field extension preserves the vector space dimension), and we get $\mathfrak a_1 = \mathfrak a_2$, as desired.
Jin Yong / U.G. 1 2
p. 142,
Line 15
It is clearer to write $A\otimes_kK = \Gamma(C_K\cap D_K, \mathscr O_{C_K\cap D_K})$ (so to make use of the calculation in the previous line). A. Graf 1 2
p. 143,
Proof of Lemma 5.62
It would be helpful to add a reference to equation (4.12.4). A. Graf 1 2
p. 143,
Line -13
Replace $\bigoplus$ by $\bigoplus_d$. A. Graf 1 2
p. 150,
Line -6
Add subscript $k$ (twice). A. Graf 1 2
p. 150,
Example 6.3 (3)
Maybe explain that the finiteness conditions are imposed to ensure that the construction (and passing to the dual vector space, resp.) is compatible with the tensor product. A. Graf 1 2
p. 150,
Line 12
Label the final arrow with $df_x$. A. Graf 1 2
p. 151,
Line 8
Replace "homomorphisms" by "local homomorphisms". A. Graf 1 2
p. 152,
Line 13
Replace "projection $X(k[\varepsilon])\to X(k)$" by "map $X(k[\varepsilon])\to X(k)$ induced by the projection $k[\varepsilon]\to k$, $\varepsilon\mapsto 0$. A. Graf 1 2
p. 153,
Lines -2, -1
Replace the reference to Section (4.14) by a reference to Section (4.13). Dominik Briganti 1 2
p. 155,
Replace "$k$-scheme" by "$K$-scheme". Dominik Briganti 1 2
p. 155,
Lines 1, 2
It should be added that here ${\rm Spec}(K[\varepsilon])$ is considered as an $S$-scheme via the composition ${\rm Spec}(K[\varepsilon])\to{\rm Spec}(K)\to S$, where the first morphism is the one coresponding to the inclusion $K\subset K[\varepsilon]$. Ho Hai Phung 1 2
p. 156,
Line -17
Replace "a $r\times r$ minor" by "an $(n-d)\times (n-d)$ minor" as $r$ is not defined in Definition 6.14. Jan Willing 1 2
p. 169,
Exercise 6.12 (b) (iii)
Condition (iii) should be changed to: "$X \cong \coprod_i \Spec K_i$, where $K_i$ are finite separable field extensions of $k$." (Alternatively, add the assumption that $X$ is quasi-compact in the beginning.) Rafael M. Saavedra 1 2
p. 171,
The second line of exercise 6.26
Replace ${\rm Spec} R[X](f)$ by ${\rm Spec} R[X]/(f)$. Also, there is a conflict of notation ($X$ is used for the scheme and for the variable). Han Hu 2
p. 173,
Line -9
Somewhere (maybe after Example 7.2) add a sentence like "If $\mathscr A$ is a presheaf of rings and $\mathscr F$ is a presheaf of abelian groups with an $\mathscr A$-module structure (as presheaves), then the sheafification of $\mathscr F$ is a sheaf of modules over the sheafification of $\mathscr A$, because sheafification is compatible with products. A. Graf 1 2
p. 174,
Line 17
Replace "of an $\mathscr O_X$-module" by "of an $\mathscr O_X$-module $\mathscr F$". A. Graf 1 2
p. 177,
Line -1
Add period at the end of the sentence. A. Graf 1 2
p. 178,
Around line -9
Add that the tensor product is associative (up to canonical isomorphism) and that one can similarly form tensor products of more than two factors (with the same result as taking successive tensor products). A. Graf 1 2
p. 182,
Line 6
Add "for all open $U\subseteq X$" at the end of the line. A. Graf 1 2
p. 185,
Line -12
Replace "localization in the prime ideal" by "localization at the prime ideal". A. Graf 1 2
p. 185,
Line 15
Replace "with respect" by "with respect to". U. Görtz 1 2
p. 189,
Proof of Corollary 7.19 (4)
Maybe replace the final sentence by the following, to make this clearer: These isomorphisms show that the presheaf $\mathscr H$ on the basis of the topology given by the $D(f)$ is actually a sheaf, and therefore yield the desired isomorphism (7.10.3). Caiyong Qiu 1 2
p. 190,
Theorem 7.22
Replace $X_s$ by $X_s(\mathscr L)$ in the statement (and in the proof before the reduction to $\mathscr L \cong \mathscr O$. A. Graf 1 2
p. 191,
Add reference to Corollary 3.22 (open subschemes of noetherian schemes are noetherian). A. Graf 1 2
p. 195,
Line -3
Replace "as" by "as an". A. Graf 1 2
p. 202,
Exer. 7.4
Add a comma before $s_r(x)$ in both (a) and (b). A. Graf 1 2
p. 204,
Exer. 7.10 (a)
Replace the first occurrence of "polynomials" by "polynomial". A. Graf 1 2
p. 205,
Exercise 7.16
Replace $\mathscr F$ by $\mathscr E$ (four times). Jan Willing 1 2
p. 209,
Prop. 8.4
It would be helpful to point out explicitly that for an $S$-Scheme $T$ the set ${\rm Hom}_S(T, S)$ is a singleton set (and that therefore in both (1) and (2) of Prop. 8.4 the set $F(T)$ has at most one element). A. Graf 1 2
p. 218,
Line 9, (8.7.1)
Maybe replace $u^*\mathrm{Grass}^e(\mathscr{E})$ by $\mathrm{Grass}^e(\mathscr{E})\times_SS'$ here to emphasize that the Yoneda lemma is used in order to obtain an isomorphism of schemes. The corresponding identification of functors is already obtained in the displayed line above. Xiaolong Liu 1 2
p. 219,
Line 21
Replace $q'$ by $q''$. Xiaolong Liu 1 2
p. 221,
Line 7
Replace $Grass^1(\bigwedge^d\mathscr{O}_{Spec(\mathbb{Z})})$ by $Grass^{\binom{n}{d}-1}(\bigwedge^d\mathscr{O}_{Spec(\mathbb{Z})})$. Xiaolong Liu 1 2
p. 223,
Line -6
Replace $M_{1 \times n}$ by $M_{1\times n}(R)$. Sz-Sheng Wang 1 2
p. 257,
Line 25,26
Replace 'constructible' by 'globally constructible' twice. Xiaolong Liu 1 2
p. 293,
line -6
Replace $f\circ s={\ rm id}_U$ by $h_{|U}\circ s= {\rm id}_U$. Xiaolong Liu 2
p. 295,
Lines -12, -11, -10
Define $\theta$ and $\theta'$ as $\theta := (g_{ij})_[i,j}$ and $\theta' := (g_{ij}')_[i, j}$, respectively. Maybe replace "for all $i$" with "for all $i \in I$". Mention explicitly that the equation (in line 7) holds on $U_i \cap U_j$. N. T. 1 2
p. 302,
line -9
Replace '$f_{|U}$' by '$s_{|U}$'. Xiaolong Liu 2
p. 313,
Lines 8, 11
The expression for $f$ in line 8 (ed. 1: line -4) ignores that there may be denominators. A homogeneous polynomial may be expressed as stated and for those $f$ the expression in line 11 (ed. 1: line -1) is correct. In general, we express an element of $\mathcal R$ as a fraction and extend the map $Z$ in the obvious way so that it becomes a group homomorphism. J.-C. Syu 1 2
p. 321,
Exercise 11.18 (a)
Replace '$Z^1({\rm Spec}\ A)\cong\mathbb{Z}^r$' by '$Z^1({\rm Spec}\ A)\cong\mathbb{Z}$'. Xiaolong Liu 1 2
p. 332,
line 12
Replace 'second equality' by 'equality in (1)'. Xiaolong Liu 2
p. 362,
Proof of Lemma 12.84
All $C$'s should be $A$'s, similarly for the $C_\nu$ (and probably the $c_{ij}$ should be renamed to $a_{ij}$. Xiaolong Liu 2
p. 372,
Prop. 13.2 (2), proof of Prop 13.2 (3)
The set of ideals disjoint from $S$ may be empty. A corrected version of (2) is: "Let $S\subset A_+$ be a non-empty subset such that $s,t\in S$ implies $st\in S$ (hence $S\cup\{1\}$ is a multiplicative subset). Suppose that there exists a homogeneous ideal $I\subsetneq A_+$ such that $S\cap I = \emptyset$. Then every ideal maximal among the set of homogeneous ideals $I$ satisfying $I\subsetneq A_+$ and $S\cap I=\emptyset$ is of the form $\mathfrak{p}_+$ for a relevant prime ideal $\mathfrak{p}$." Maybe also include justification in the proof of (3) why the set of ideals with those properties for $S=\{f,f^2,\dots\}$ is non-empty: "Conversely, if $f\notin\text{rad}(I)$, then let $S=\{f,f^2,\dots\}$ and note $I\cap S=\emptyset$ and $I\subsetneq A_+$. Hence there is a relevant prime ideal disjoint from $S$ by (2), so $f$ is not contained in the intersection of all relevant prime ideals." Matthew Snodgrass 1 2
p. 373,
Proof of Prop 13.2 (3)
To apply (2), $f$ must be in $A_+$. The current argument is sufficient to show (3) (together with a direct proof that the ideal $\operatorname{rad}(I)$ is homogeneous) but not for the statement in the first line of the proof. 1 2
p. 373,
Line -8
We defined $V_+(\mathfrak a)$ only for homogeneous ideals contained in $A_+$. Therefore $V(\mathfrak a^h)$ should be replaced by $V(\mathfrak a^h\cap A_+)$. Note that for $\mathfrak p\in {\rm Proj}(A)$, we have \(\mathfrak a\subseteq \mathfrak p \Leftrightarrow \mathfrak a^h\subseteq\mathfrak p\Leftrightarrow \mathfrak a^h\cap A_+\subseteq\mathfrak p.\) (Alternatively we could define $V_+(-)$ for all homogeneous ideals (or even all subsets of $A$) and explain why this doesn't make a difference.) Dominik Briganti 1 2
p. 373,
Line 2
Replace "set of ideal" by "set of ideals". 2
p. 380,
Line 14
Replace '$(M_{(f)})_n$' by '$(M_f)_n$'. Xiaolong Liu 1 2
p. 386,
Line -12
Replace '$\mathbb{P}^{n+1}=\mathbb{P}(\mathscr{O}_S^{n+1})$' by '$\mathbb{P}_S^{n}=\mathbb{P}(\mathscr{O}_S^{n+1})$'. Xiaolong Liu 1 2
p. 409,
Line -1
Replace '$X$' by '$Z$'. Xiaolong Liu 1 2
p. 412,
line -8
Replace '$A^x=k[x,y,\mu]/(\mu x=y)$' by '$A^x=k[x,y,\mu]/(\mu x-y)$'. Xiaolong Liu 1 2
p. 437,
line 6
Replace '$h=g\circ f$' by '$g=h\circ f$'. Xiaolong Liu 1 2
p. 456,
Line -10
Replace $\varphi$ by $\varphi'$ (twice). Peng DU 1 2
p. 457,
Prop. 14.67
The lower right triangle of the diagram constructed at the end of the proof is not commutative in general, therefore the proof is not valid as given here. Furthermore, in Step (i) of the proof of Theorem 14.66 (ed. 1) / Theorem 14.68 (ed. 2), a morphism $f': T'\rightarrow S'$ is considered where $T'$ is a finite disjoint union of affine open subschemes of $S'$ covering $S'$. If $S'$ is not quasi-separated, then it is not possible to find a quasi-compact such $f'$. Therefore, it seems better to handle individually the two cases of this proposition that are needed in the proof of the theorem: The case where $T'$ is a finite disjoint union of open subschemes of $S'$, and the case where $f' : T'=S \rightarrow S'$ is a section of $p: S' \rightarrow S$. Both cases are easy to deal with (the first one because it is clear that morphisms of this kind satisfy descent). A posteriori, the theorem implies that the proposition is actually true in the form stated. M. Bruneaux, P. Godfard 1 2
p. 468,
Proof of Theorem 14.93
For $n=1$ the proof needs to be modified slightly (this case is easier, but the exception should be stated explicitly).
Furthermore, in the second part of the proof it might be appropriate to give a few more details, see here.
J.-C. Syu / U. Görtz 1 2
p. 475,
Lemma 14.111
It is enough to assume that $f$ is locally of finite type (and in fact in the proof of the following theorem the lemma is applied with that weaker assumption). The same proof works in the general case. Jin yong An 1 2
p. 487,
Exercise 14.8
Replace $X$ by $Y'$. 1 2
p. 513,
Second line in the proof of Prop. 16.11
Add "open subset" after dense. Andreas Blatter 1 2
p. 526,
Line -5
In the expression for $g$, in the third parenthesis there should be $(X_2-1)^2$ instead of $(X_2^2-1)^2$. Alejandro Vargas and Tim Seynnaeve 1 2
p. 543,
Line 16
Add "if" after "only". U. Görtz 1 2
p. 551,
Example A.6.(3)
The category of sets should also be in the list. Andreas Blatter 1 2
p. 571,
Proposition B.73 (6)
What is true (and proved in Matsumura's book) is: If $B$ is normal, then $A$ is normal. If $A$ and $B\otimes_A\kappa(\mathfrak p)$ are normal for all $\mathfrak p\in {\rm Spec}(A)$, then $B$ is normal.
It is not true that normality of $B$ implies normality of the fibers. Consider for instance a discrete valuation ring $A$ with uniformizer $t$ and $B=A[X,Y]/(XY-t)$. Then $B$ is regular and in particular normal, but the special fiber $(A/t)[X, Y]/(XY)$ is not normal.
M. Kerz 1 2
p. 607,
Left col., lines 3, 4
The symbols for direct sum and product should be exchanged. J.-C. Syu 1 2