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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142 errata listed.
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p. 6,
¶
Preface to 2nd edition

Add Peng Du to the list of names.  U. Görtz  2 
p. 8,
¶
16

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 nonempty 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. 15,
¶
Second sentence of the proof of Lemma 1.25(3)

It should say "there would exist a minimal element $Z \in \mathscr M$".  Javier de la Bodega  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 nonempty 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. 30,
¶
Line 1

Replace $f_m$ by $f_r$ at the end of the line.  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. 38,
¶
Exercise 1.20 (a)

$x_1, \dots , x_n \in k$ should be $x_1, \dots, x_n \in K$.  Steven Jin  1 2 
p. 47,
¶
Line 25

It should say "nonequivalent"; i.e. the hyphen is missing.  Javier de la Bodega  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. 55,
¶
Line 14

To be more precise, ”by Proposition 2.23" could be "by Proposition 2.23 (2)".  Ehsan Shahoseini  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. 65,
¶
Exer. 2.15

The assumption that $X$ be locally compact is superfluous.  V. Paškūnas  1 2 
p. 66,
¶
Line 13

Replace "subset" by "open subset".  A. Graf  1 2 
p. 70,
¶
Lines 6, 7

Missing closing parenthesis in $\mathscr O_X(f^1(D(s))$ and $\mathscr O_X(f^1(D(s))^\times$.  Jenna Nieminen  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. 79,
¶
Proof of Prop. 3.27 (2)

At this point in the proof, $X$ is not yet assumed to be affine. But $V(f)$ has only been defined in the affine setting until this point in the book. So it might be worth noting that $V(f)$ denotes the set of $x\in X$ where $f(x)=0$ in the sense of the penultimate paragraph of p.57. This is a closed subset of $X$ by Ex. 2.19. (Later in the book, $X_f$ is reintroduced on p.187, without reference to Ex. 2.19).  Denis Vogel  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 20

Replace $\mathscr O$ by $\mathscr O_X$.  Ehsan Shahoseini  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 LimaFilho  1 2 
p. 97,
¶
Section (4.2)

It should be stated explicitly that for the Yoneda lemma we need to assume that $\mathcal C$ is a locally small category, i.e., that all ${\rm Hom}_{\mathcal C}(X, Y)$ are sets.  M. Jagalski  2 
p. 98,
¶
Remark 4.9

In the displayed line, it should say "$x \mapsto x^r$".  Javier de la Bodega  1 2 
p. 102,
¶
Last sentence of part (i) in the proof of Theorem 4.18

It should say "$f = p \circ h$ and $g = q \circ h$"; i.e. it should be = instead of :=  Javier de la Bodega  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,
¶
(4.15.1)

The top morphism of the commutative diagram in the middle should be "${\rm id}_G \times_S e$" instead of "$({\rm id}, e)_S$".  Jan Willing  1 2 
p. 116,
¶
Line 19

Add reference to Cor. 4.7.  A. Graf  1 2 
p. 126,
¶
Proposition 5.11 (i)

Add '.' at the end.  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 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. 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 $p^{1}(x) = \mathop{\rm Spec} \kappa(x)\otimes_k K$. We now use Proposition 3.33. Since the extension $\kappa(x)/k$ is finite, $\kappa(x)\otimes_k K$ is a finitedimensional $K$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. 138,
¶
Line 2

The reference to Proposition B.97 is not required at this point and should be omitted (our definition of separability B.91 (6) is exactly what is needed here).  Jinyong / U.G.  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 $U_{ij} = U_i \cap {\rm Spec} (A_j \otimes k')$ (schemetheoretic 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,
¶
Line 13

Replace $\bigoplus$ by $\bigoplus_d$.  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. 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,
¶
Prop. 6.10 (1)

One should choose a representative $x = (x_0,\dots, x_n)\in \mathbb A^{n+1}_k(k)$ of the point of interest to start with in order to obtain a natural isomorphism between the two sides, because the isomorphism will depend on this choice. See also Section (8.9) (and Section (17.7) in Volume II for a relative version).  U. Görtz  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,
¶
15

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 $(nd)\times (nd)$ minor" as $r$ is not defined in Definition 6.14.  Jan Willing  1 2 
p. 162,
¶
Line 12, proof of Lemma 6.26

For an arbitrary choice of a point $y'$ lying over $y$, the residue class field of $y'$ might be bigger than $\kappa(y)$, so we have to choose $y'$ suitably in order to apply the previous reasoning. In fact, there always exists $y'$ lying over $y$ with $\kappa(y')=\kappa(y)$. Namely, $y$ gives rise to a morphism ${\rm Spec}(\kappa(y)\to X$, and together with the identity morphism of $\kappa(y)$ we obtain a morphism ${\rm Spec}(\kappa(y))\to X\otimes_k\kappa(y)$ whose image point $y'$ has residue class field $\kappa(y)$.  Jin Yong / U.G.  2 
p. 165,
¶
Last sentence of the proof of Lemma 6.38

It should say "every nonempty closed subset".  Javier de la Bodega  1 2 
p. 169,
¶
Exercise 6.12 (b) (iii)

Condition (iii) should be changed to: "$X \cong \coprod_i {\rm Spec}(K_i)$, where $K_i$ are finite separable field extensions of $k$." (Alternatively, add the assumption that $X$ is quasicompact 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. 180,
¶
Line 9ff

In this general setting some stalks of $\mathcal{O}_X$ might be zero. But in order to read off the rank at $x$ from the local rank you need that $\mathcal{O}_{X,x}$ is a nonzero ring.  Denis Vogel  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,
¶
11

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. 204,
¶
Exercise 7.7 (b)

The exercise is not correct as written. Instead, one could say that there is a natural surjective $\mathscr O_X$module homomorphism $\mathscr F^\eta\to \mathscr F^x$ and its kernel $\mathscr G$ has the property that $\mathscr G_{U}$ is not globally generated for any open neighborhood $U$ of $x$.  Mathis Birken  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 ${\rm Grass}^1(\bigwedge^d\mathscr{O}_{Spec(\mathbb{Z})})$ by ${\rm Grass}^{\binom{n}{d}1}(\bigwedge^d\mathscr{O}_{Spec(\mathbb{Z})})$.  Xiaolong Liu  1 2 
p. 223,
¶
Line 6

Replace $M_{1 \times n}$ by $M_{1\times n}(R)$.  SzSheng Wang  1 2 
p. 227,
¶
Exercise 8.7

Replace `$\mathscr N_n$ is the called ...' by `The scheme $\mathscr N_n$ is called'.  Akash N  1 2 
p. 240,
¶
Second equation in Example 9.30

It should say "$(x,y) \mapsto \dots$" rather than "$(x,y)\to\dots$".  Javier de la Bodega  1 2 
p. 257,
¶
Line 25,26

Replace 'constructible' by 'globally constructible' twice.  Xiaolong Liu  1 2 
p. 259,
¶
Line 1ff, Proof of Prop. 10.48

The implicit definition of the $U^i$ is a bit confusing. It might be clearer to write "The set $\mathcal U = \dots$ is totally ordered by inclusion. We define $U^i$ by $\mathcal U = \{ U_0, \dots, U^n\}$ and $U^0\subsetneq U^1\subsetneq\cdots\subsetneq U^n$, so $U^0 = U_0=U$, $U^n = X$.  Jan Willing  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,
¶
Line 2

Write morphism instead of morphisms.  Saskia Kern  1 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. 297,
¶
Line 16

Replace $\check{H}^1(X, G)$ by $\check{H}^1(X, G')$.  Saskia Kern  1 2 
p. 302,
¶
line 9

Replace '$f_{U}$' by '$s_{U}$'.  Xiaolong Liu  2 
p. 312,
¶
Secondtolast line of the proof of Proposition 11.42

It should say "${\rm cyc}(f)$" instead of "${\rm div}(f)$", according to the notation established in the third paragraph of page 310.  Javier de la Bodega  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. 327,
¶
Line 12

Replace "and hence $X' = {\rm Spec}(B' \otimes_B A)$" by "and hence $X' = {\rm Spec}(A' \otimes_A B)$".  Akash N  1 2 
p. 332,
¶
line 12

Replace 'second equality' by 'equality in (1)'.  Xiaolong Liu  2 
p. 340,
¶
Theorem 12.35

The assumption that $X$ be quasiseparated is superfluous. This is currently used to invoke Theorem 7.22 in the proof of (ii) $\Rightarrow$ (i), but it follows from (ii) since $X_{f_i}\cap X_{f_j}$ is a principal open subset of the affine scheme $X_{f_i}$. Alternatively, one could prove (ii) $\Rightarrow$ (i) by saying that for all $i$, the morphism $X_{f_i}\to D(f_i)$ is affine (since $X_{f_i}$ is affine), so that by Prop. 12.1 the morphism $X\to {\rm Spec}(A)$ is affine.  U. Görtz  1 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. 371,
¶
Last paragraph

With the terminology introduced before, the definition of homegeneous ideal should read "A homogeneous ideal I of A is a homogeneous submodule of A." (But maybe the term graded submodule, with the same meaning as homgeneous submodule, should also be introduced.)  Johann Birnick  1 2 
p. 372,
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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 nonempty 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 nonempty: "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,
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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,
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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,
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Line 2

Replace "set of ideal" by "set of ideals".  2  
p. 378,
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Line 2

It is not true in general that the functor considered here commutes with tensor products, see Stacks Project01ML. But this does hold if $A$ is generated by $A_1$ as an $A_0$algebra.  D. Vogel  1 2 
p. 380,
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Line 14

Replace '$(M_{(f)})_n$' by '$(M_f)_n$'.  Xiaolong Liu  1 2 
p. 382,
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Prop. 13.24

The subscheme $Z$ should be assumed to be nonempty, to ensure we can find $I$ as in the proposition which does not contain $A_+$.  U. Görtz  1 2 
p. 386,
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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,
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Line 1

Replace '$X$' by '$Z$'.  Xiaolong Liu  1 2 
p. 412,
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line 8

Replace '$A^x=k[x,y,\mu]/(\mu x=y)$' by '$A^x=k[x,y,\mu]/(\mu xy)$'.  Xiaolong Liu  1 2 
p. 431,
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Proposition 14.14 (ii)

Replace "schematic closure of the generic fiber $f^{1}(\eta)$ in X" by "schematic image of the canonical morphism $f^{1}(\eta) \rightarrow X$". In 10.31, schematic closure is defined only for quasicompact immersion but the morphism above may not be an immersion. An example: $X=Y=Spec\mathbb{Z}$. Then $\{ 0 \}$ is not a subscheme in the sense of 3.43.  1 2  
p. 433,
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Proof of Lemma 14.18

In the first paragraph, one needs to argue more carefully in order to conclude that the homomorphisms $f_x^\sharp$ are bijective. (It is not enough to assume that $f$ is dominant and locally on $X$ an immersion, and $Y$ is reduced.) Note that the morphisms $U_i\to Y$ are dominant since $U_i\subseteq X$ is dense and $f$ is dominant. Thus they are dominant immersions, and $Y$ being reduced are open immersions. From this it is clear that the $f_x^\sharp$ are isomorphisms.  J. Ehrhard  1 2 
p. 437,
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line 6

Replace '$h=g\circ f$' by '$g=h\circ f$'.  Xiaolong Liu  1 2 
p. 456,
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Line 10

Replace $\varphi$ by $\varphi'$ (twice).  Peng DU  1 2 
p. 457,
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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 quasiseparated, then it is not possible to find a quasicompact 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,
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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,
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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,
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Exercise 14.8

Replace $X$ by $Y'$.  1 2  
p. 507,
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Exercise 15.9

Add the assumption that $k$ has characteristic $\ne 2$.  V. Paškūnas  1 2 
p. 513,
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Second line in the proof of Prop. 16.11

Add "open subset" after dense.  Andreas Blatter  1 2 
p. 526,
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Line 5

In the expression for $g$, in the third parenthesis there should be $(X_21)^2$ instead of $(X_2^21)^2$.  Alejandro Vargas and Tim Seynnaeve  1 2 
p. 536,
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First sentence of section 16.26

It should say "the singularities of $Z={\rm Spec}\ k[S,T,U]/(U^3ST)$", i.e. the base field $k$ is missing.  Javier de la Bodega  1 2 
p. 536,
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Third sentence of second paragraph of section 16.28

It should say "If $p(x,y) \in Z(k)$ is a singular point".  Javier de la Bodega  1 2 
p. 537,
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Proposition 16.47

Replace $n$ by $r$ in both statement and proof.  Javier de la Bodega  1 2 
p. 543,
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Line 16

Add "if" after "only".  U. Görtz  1 2 
p. 551,
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Example A.6.(3)

The category of sets should also be in the list.  Andreas Blatter  1 2 
p. 557,
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Line 13

It should say "An $A$module", instead of "Am $A$module".  Javier de la Bodega  1 2 
p. 571,
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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 $B=A[X,Y]/(XYt)$. 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,
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Left col., lines 3, 4

The symbols for direct sum and product should be exchanged.  J.C. Syu  1 2 
p. 609,
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Isol(X/Y)

It should say "Set of points isolated in their fiber".  Javier de la Bodega  2 