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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13 errata listed.
Page  Description  Submitted by  Ed. 

p. 15,
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Line 15

Replace $\mathscr M$ by $\mathcal M$.  I. Tselepidis  1 2 
p. 33,
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Line 13, 12

Replace $\mathbb{P}_k^n$ by $\mathbb{P}^n(k)$.  E. Hong  1 2 
p. 72,
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Line 16

Replace "with an open subscheme of $T_i$" by "with an open subscheme $T_i$ of $T$".  E. Hong  1 2 
p. 75,
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Line 1

Replace the reference to Section (4.14) by a reference to Section (4.13).  N.T.  1 2 
p. 92,
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Exercise 3.19

Replace ${\rm Hom}(\Spec(R), \mathbb P^n_R)$ by ${\rm Hom}_{R}(\Spec(R), \mathbb P^n_R)$ to make explicit that we only consider $R$morphisms here.  A. Graf  1 2 
p. 93,
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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. 153,
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Lines 2, 1

Replace the reference to Section (4.14) by a reference to Section (4.13).  Dominik Briganti  1 2 
p. 155,
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15

Replace "$k$scheme" by "$K$scheme".  Dominik Briganti  1 2 
p. 189,
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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. 223,
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Line 6

Replace $M_{1 \times n}$ by $M_{1\times n}(R)$.  SzSheng Wang  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. 551,
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Example A.6.(3)

The category of sets should also be in the list.  Andreas Blatter  1 2 