Is there a good list of errata for Atiyah–Macdonald available? A cursory Google search reveals a laughably short list here, with just a few typos. Is there any source available online which lists inaccuracies and gaps?

1$\begingroup$ Please read tea.mathoverflow.net/discussion/154/… $\endgroup$– Willie WongOct 15, 2010 at 20:18

7$\begingroup$ I've just corrected the spelling of Ian G. Macdonald's name, to avoid confusion with the less famous group theorist Ian D. MacDonald. Aside from this, I think it's pointless to use this site to assemble errata for a book. The answer to the question about a Web source of errata is very likely no. If anybody wants to start a special Web site for this purpose, it's fine with me. Virtually all math books do have at least minor errors. $\endgroup$– Jim HumphreysOct 15, 2010 at 23:19

15$\begingroup$ Let me try and "refute" the "highprofile organizer" comment above. To be frank, MO just turned out to be a repository for the CasselsFroehlich errata rather than anything else. The reason I got so many was not because I posted here. It was because I asked for errata in many places rather than just here, all at the same timebut, crucially, I also approached several highprofile people personally (Hendrik Lenstra, Rene Schoof, J.P. Serre, the Conrads [before, I think, they were MOactive] and others) and asked them if they had anything to send me... $\endgroup$– Kevin BuzzardOct 24, 2010 at 9:25

8$\begingroup$ ...and several responded with big lists. Note that almost all of the answers in that thread were posted by me, and are of the form "prof X just sent me this big list". I really pushed to make the errata, and, because I had a deadline myself (the LMS wanted to republish with the errata in) I had to push the people I was asking. I worked very hard to make those errata. So, it's very different to just posting once here and then sitting back and hoping (which, I think, is what is happening here, although I do apologise if I've got this wrong). Also CF was typeset by a company who had very... $\endgroup$– Kevin BuzzardOct 24, 2010 at 9:29

5$\begingroup$ ...limited experience in mathematical typesetting, and they introduced many errors. (OhI should have mentioned Birch and Tate in my list of bigshots I approached directly, and I'm sure there are others I've forgotten). $\endgroup$– Kevin BuzzardOct 24, 2010 at 9:29
30 Answers
Dear Tim, on page 31 they consider a ring $A$ and two $A$ algebras defined by their structural ring morphisms $f:A\to B$ and $g:A\to C$. They then define the tensor product as a ring $D=B\otimes _A C$ and want to make it an $A$ algebra. For that they must define the structural morphism $A\to D$ and they claim that it is given by the formula $a \to f(a)\otimes g(a)$.This is false since that map is not a ring morphism. The correct structural map $A\to D$ is actually $a\mapsto 1_B\otimes g(a) =f(a)\otimes 1_C$.
PS: To prevent misunderstandings, let me add that AtiyahMacDonald is, to my taste, the best mathematics book I have ever seen, all subjects considered.

3$\begingroup$ Do you think we should create an errata here? $\endgroup$– Tim Campion ♦Oct 15, 2010 at 16:13

4$\begingroup$ Dear Tim: no. I think we should stick to the community wiki format: one post per answer. This has the advantage that posts can then be commented (or refuted!) individually. Anyway, given my admiration for this almost perfect book, I conjecture that we will find very few errata ... $\endgroup$ Oct 15, 2010 at 18:19

6$\begingroup$ Dear Tim, let me clarify my point of view. 1) I agree with you that there is probably no errata floating around. 2) I know no other error in the book and thus won't post any more. 3) I encourage everybody else to post new answers. 4) Afterwards, whoever wants to may create a document inspired by this community wiki and spread it as he/she deems fit. $\endgroup$ Oct 15, 2010 at 21:00

33$\begingroup$ @Yuji: Honorary titles in the British system are truly mysterious, but "Sir Atiyah" is actually "Sir Michael". Similarly, SwinnertonDyer is known as "Sir Peter". But lords and ladies are even more troublesome: the fictional detective is "Lord Peter" (not "Lord Wimsey") whereas the philosopher Bertrand Russell was "Lord Russell". Depends on birth order among other things, complicated by lifetime peerages for some. Feel free to call me Lord Jim.... $\endgroup$ Oct 17, 2010 at 12:46

2$\begingroup$ @Vincenzo: compute the image of $a+b$ and realize that it is not the sum of the images of $a$ and $b$. $\endgroup$ Nov 13, 2016 at 18:32
EDIT OF JULY 26, 2017
Proposition 2.4 page 21 reads:
Let $M$ be a finitely generated $A$module, let $\mathfrak a$ be an ideal of $A$, and let $\phi$ be an $A$module endomorphism of $M$ such that $\phi(M)\subseteq\mathfrak a M$. Then $\phi$ satisfies an equation of the form $$\phi^n+a_1\,\phi^{n1}+\cdots+a_n=0$$ where the $a_i$ are in $\mathfrak a$.
Strictly speaking, this makes no sense (it seems to me) because $\phi$ and the $a_i$ belong to different rings. I suggest the following restatement:
Let $M$ be a finitely generated $A$module, let $\mathfrak a$ be an ideal of $A$, let $\phi$ be an $A$module endomorphism of $M$ such that $\phi(M)\subseteq\mathfrak a M$, and let $\psi:A\to\operatorname{End}_A(M)$ be the natural morphism. Then $\phi$ satisfies an equation of the form $$\phi^n+\psi(a_1)\,\phi^{n1}+\cdots+\psi(a_n)=0$$ where the $a_i$ are in $\mathfrak a$.
Another fix would be to equip $\operatorname{End}_A(M)$ with its natural $A$module structure and change the display to $$ \phi^n+a_1\,\phi^{n1}+\cdots+a_n\,\phi^0=0. $$
END OF EDIT OF JULY 26, 2017
EDIT OF JUNE 9, 2011
Page 102, penultimate paragraph:
"... $f$ induces a homomorphism $\widehat{f}:\widehat{G}\to\widehat{H}$, which is continuous."
No topology has been defined on $\widehat{G}$ and $\widehat{H}$.
[July 7, 2011, GMT. The topology on $\widehat{G}$ can be described as follows. For any subset $S$ of $G$, let $\widehat{S}\subset\widehat{G}$ be the set of equivalence classes of Cauchy sequences in $S$, and say that a subset $V$ of $\widehat{G}$ is a neighborhood of $0$ if there is a neighborhood $W$ of $0$ in $G$ such that $\widehat{W}\subset V$.]
By the way, there is (I think) a somewhat similar "mistake" in the article Atiyah wrote with Wall in "Algebraic Number Theory" Ed. Cassels and Froehlich (see Erratum for CasselsFroehlich). Atiyah and Wall forgot to mention the crucial compatibility between change of groups and connecting morphisms. (See p. 99.)
END OF EDIT OF JUNE 9, 2011
Page 25, first line of the proof of (2.13): change (2.11) to (2.12).
Page 29, about two third of the page: change (2.14) to (2.13).
EDIT. Page 39, last line: change $m$ to $m_i$ (three times).
EDIT OF NOV. 22, 2010. Page 63, proof of Lemma 5.14. The current text reads
Conversely, if $x\in r(\mathfrak a^e)$ then $x^n=\sum a_i\,x_i$ for some $n>0$, where the $a_i$ are elements of $\mathfrak a$ and the $x_i$ are elements of $C$. Since each $x_i$ is integral over $A$ it follows from (5.2) that $M=A[x_1,\dots,x_n]\ \dots$
It would be better (I think) to write something like
Conversely, if $x\in r(\mathfrak a^e)$ then $x^n=a_1\,x_1+\cdots+a_m\,x_m$ for some $m,n>0$, where the $a_i$ are elements of $\mathfrak a$ and the $x_i$ are elements of $C$. Since each $x_i$ is integral over $A$ it follows from (5.2) that $M=A[x_1,\dots,x_m]\ \dots$
[July 8, 2011, GMT. Page 90. It seems to me that the second part of the proof of Theorem 8.7 can be simplified. We must check the uniqueness of the decomposition of an Artin ring $A$ as a finite product of Artin local rings $A_i$. To do this it suffices to observe that, for each minimal primary ideal $\mathfrak q$ of $A$, there is a unique $i$ such that $\mathfrak q$ is the kernel of the canonical projection onto $A_i$.]
[July 7, 2011, GMT. Page 107, lines 45. Instead of $A^*=A[x_1,\dots,x_r]$ read $A^*=A[y_1,\dots,y_r]$ where $y_i=(0,x_i,0,\dots)$.]
[July 7, 2011, GMT. Page 112, proof of Proposition (10.24). Instead of $\mathfrak{a}^{k+n(i)}$ read $\mathfrak{a}^{\max(0,kn(i))}$.]
[July 9, 2015. The integer $d(M)$ (and in particular $d(A)$) is defined on p. 117 after the proof of Theorem 11.1. Another definition of $d(A)$ is given on p. 119 after the proof of Proposition 11.6 via the equality $d(A)=d(G_{\mathfrak m}(A))$. But the old meaning of $d(?)$ is used again in the proof of Proposition 11.20 p. 122, where the expression $d(G_{\mathfrak q}(A))$ occurs at the beginning of the last display. To avoid any confusion, let me denote by $D(M)$ the integer given by the first definition, and set $d(A):=D(G_{\mathfrak m}(A))$.
It seems to me the proof of Proposition 11.3 p. 118 is not entirely correct. I suggest to keep the proof, but to weaken slightly the statement, the new statement being: If $P(M/xM,t)\neq0$ and $D(M/xM)\ge1$, then $P(M,t)\neq0$ and $D(M/xM)=D(M)1$.
This new statement applies to the first equality in the last display in the proof of Proposition 11.20 p. 122 if $d:=\dim A\ge1$ (the case $d=0$ being trivial).  On the third line of the proof $\mathfrak q$ should be $\mathfrak q^2$.]

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$\begingroup$ @darijgrinberg  Thanks! I rewrote (in a hopefully correct way) the last edit. $\endgroup$ Jul 9, 2015 at 8:44

$\begingroup$ I just had this doubt regarding Proposition 2.4. Took me a while to understand. Now that I did, I find this. Thanks :) $\endgroup$– MarraMar 6, 2020 at 12:34
On page 8, the proof of part ii of Proposition 1.11 begins "Suppose $\mathfrak{p}\not\subseteq\mathfrak{a}_i$ for all $i$." It should be $\not\supseteq$.

$\begingroup$ I stumbled on this page after catching this same error. However, I wonder if this is a true error? Technically, the proof is by contraposition, so logically it should be $\nsupseteq$. However, the argument is still correct using $\nsubseteq$ since all we need is the existence of $x_i \in \frak a_i$ with $x_i \notin \frak p$. $\endgroup$– user5826Jun 22, 2020 at 17:28

$\begingroup$ @user5826, re, the statement that there exists $x_i \in \mathfrak a_i$ with $x_i \notin \mathfrak p$ is exactly the statement that $\mathfrak p \not\supseteq \mathfrak a_i$. To see that the proposed (dis)containment doesn't allow it, consider, for example, $\mathfrak p = 2\mathbb Z$ and $\mathfrak a_i = 4\mathbb Z$. $\endgroup$– LSpiceOct 15 at 15:14
On page 29, the example at the top has two typos: it says "$(x)=2x$", when it should be "$f(x)=2x$", and the exact sequence at the end of that same line says "$0\rightarrow\mathbb{Z}\otimes \stackrel{f\otimes 1}{\longrightarrow} \mathbb{Z}\otimes N$", when it should be
"$0\rightarrow\mathbb{Z}\otimes N\stackrel{f\otimes 1}{\longrightarrow} \mathbb{Z}\otimes N$".
Nearly all the mistakes pointed out so far were fixed in the Russian translation, which was done by Manin. But not all. I'll list in parentheses the page numbers of the translation where the original error still occurs for the 5 people who might care. (The translation is usually 11 page numbers ahead of the original.) Scan the answers posted before this one to determine which mistakes I am referring to.
p. 29 (> p. 41): on line 8, change (2.14) to (2.13)
p. 55 (> p. 66): exercise 2
p. 71 (> p. 82): exercise 23
p. 88 (> p. 99): exercise 27(v)
There were also completely original mistakes added especially for the translation! On page 30 line 7 and page 31 lines 10 and 14 of the translation, the tensor product signs should be direct sum signs. On page 32 in the statement of Nakayama's Lemma, the ideal a should be in fraktur font.
On p.55, exercise 4.2 reads "If $\mathfrak a = r(\mathfrak a)$, then $\mathfrak a$ has no embedded prime ideals". I believe it should include the assumption that $\mathfrak a$ is decomposable.
AM defines embedded primes for decomposable ideals only. And it doesn't seem that a radical ideal should automatically be decomposable. If you take something like a reduced (nonnoetherian) ring with infinitely many minimal prime ideals, I expect the zero ideal will be radical but not decomposable...

1$\begingroup$ Dear Anna, you are exactly right. Take $A=k[x_1,x_2,...]=k[X_1,X_2,...]/I$ where $k$ is a field, $X_1,X_2,...$ are denumerably many indeterminates and $I=<X_i.X_j>_{i\neq j}$ is the ideal generated by the products of two different indeterminates. Then $<0>=\frak{a}=\sqrt \frak a$ is a reduced ideal which has no decomposition: indeed all the prime ideals $\sqrt {(0:x_i)}=<x_1,...,\hat x_i,...>=\frak p_i$ would have to be associated to a decomposition of $<0>=\frak a $ according to Theorem 4.5 and there are infinitely many such prime ideals: contradiction. $\endgroup$ May 1, 2011 at 11:48

2$\begingroup$ The way I saw that, either a) a is decomposable, and this makes sense or b) a is not decomposable, hence a fortiori has no embedded prime ideals! $\endgroup$ Aug 29, 2012 at 7:59
Minor typos:
p.34, exercise 2.23: Second sentence should start "For each finite subject $J$ of $\Lambda$".
p.48, exercise 3.27(i): The bracketed text should read "Use Exercises 25 and 26".
p.71, exercise 5.23: The hint should start "The only hard part is (iii) => (i). Suppose (i) is false".
p.88, exercise 7.27(v): The last clause should read "the homomorphism $f_{!}$ is a $K_1(A)$module homomorphism".
p.127, index entry for "flat, faithfully": Should cite p. 46, not p. 29.

2$\begingroup$ And a semantic quibble rather than a mistake: I was confused for some time by the wording of exercise 5.2 on p.67 (if $A \subset B$ is an integral extension of rings, then any homomorphism of $A$ to an algebraically closed field extends to $B$). I've since concluded that AM uses "homomorphism into" here to mean just "homomorphism to," i.e., not necessarily injective. $\endgroup$– Anna M.Oct 18, 2010 at 17:52

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7$\begingroup$ Well, that's the darned thing about correcting errata  you just end up introducing new errors...! $\endgroup$– Anna M.Oct 25, 2010 at 2:35

3$\begingroup$ @KConrad, it is a subject that is not equisemantic with any proper subsubject. $\endgroup$– LSpiceJul 26, 2017 at 13:45

5
On page 23, in the third line of the sketch for Proposition 2.9, change "$v \circ u \circ f = 0$ " to "$f \circ v \circ u = 0$".

$\begingroup$ Sorry to resurrect this thread, but this error and its absence from this list confused a friend on Math SE. $\endgroup$ Mar 12, 2012 at 23:15

$\begingroup$ Sorry to resurrect this thread but this also confused me for a second. $\endgroup$ Dec 31, 2017 at 1:26
page 81, line 5: change $f_i \in A[x]$ to $f_i \in \mathfrak{a}$
On page 41 in the proof of proposition 3.10., change
"i) $\implies$ ii) by (3.5) and (2.20)" to "i) $\implies$ ii) by (3.7) and (2.19)"
On page 52 in remark 1) at the bottom of the page, change
"(see Chapter 1, Exercise 25)" to "(see Chapter 1, Exercise 27)"
On page 65 at the end of the proof of proposition 5.18. the black square to denote end of proof is missing.
On page 66 we need to correct the proof of corollary 5.22., one correct version is the following: We start with the quotient map $\pi: A[x^{1}] \to A[x^{1}] /m$ where $m$ is a maximal ideal containing $x^{1}$. We take an algebraic closure $\Omega$ of the field $A[x^{1}] /m$ and consider the map $i \circ \pi: A[x^{1}] \to \Omega$. Then by the previous theorem, (5.21), we can extend $i \circ \pi$ to some valuation ring $B$ of $K$ containing $A[x^{1}]$: $g: B \to \Omega$ such that $g_{A[x^{1}]} = i \circ \pi$. Then $g(x^{1}) = 0$. Hence $x^{1} \in ker(g)$ and since the kernel is a proper ideal of $B$, $x^{1}$ is not a unit in $B$ and hence $x$ is not in $B$. (also see math.SE)
On page 77 in the proof of proposition 6.7., change
"...a composition series, by ii);..." to "...a composition series, by i);..."


$\begingroup$ 3.5 and 2.20 actually work fine to prove 3.10. $\endgroup$ Aug 29, 2012 at 8:29

$\begingroup$ Also, I don't think either of the things you mention on p.66 or p.77 is actually wrong. $\endgroup$ Aug 29, 2012 at 9:33
On p.89, the second to last line of the proof of Proposition 8.4 should say $\mathfrak{N}^k \subseteq \mathfrak{ N}$ instead of $\mathfrak{N}^k \supseteq \mathfrak{N}$.
On page 31, the first line refers to Proposition 2.11, when it should be 2.12.
Also minor: On p. 91, the $a$'s and $\mathfrak a$'s in the proof of Prop 8.8 seems to be a little jumbled.
I guess you want something like "Let $\mathfrak a$ be an ideal of $A$, other than $(0)$ or $(1)$. We have $\mathfrak m = \mathfrak N$, hence $\mathfrak m$ is nilpotent by (8.4) and therefore there exists a positive integer $r$ such that $\mathfrak a \subseteq \mathfrak m^r$ and $\mathfrak a \not\subseteq \mathfrak m^{r + 1}$; hence there exists $y \in \mathfrak a$ and $a \in A$ such that $y = ax^r$ but $y \not\in (x^{r + 1})$," etc.
In p.45, Ex.3.12.iv, one can avoid the tedious argument provided in the hint by noting that $K\otimes_A M\cong(A\{0\})^{1}M$. (I originally did it as hinted ...)
In p.68, Ex.5.10.ii, (b') is actually equivalent to a weaker (c') that asserts only that $f^*:\mathrm{Spec}(B_\mathfrak{q})\to\mathrm{Spec}(A_\mathfrak{p})\cap V(\ker f)$ is surjective. However, (a') does imply the original (c').
Here are a few more small miscellaneous mistakes and typos:
page 18, line 6: $M''$ not defined (it is $M/M'$)
page 20, line 12: in the expression for $A$ as a direct product, it should be $i=1$ not $i=I$.
page 28, line 5: $=$ should read $\cong$
page 29, line 13 (first line of proof that iv) $\Rightarrow$ iii)): $x_{i}$ should read $x_{i}'$
page 91, in the last example, it is not true that ${\mathfrak m}^{2}=0$. It is a nonzero principal ideal. But the following statement is still true, $\dim \left( {\mathfrak m}/{\mathfrak m}^{2} \right)=2$. It is generated by $x^2$ and $x^3$.
page 102, Lemma 10.1(iv) $H=0$ should read $H=\{ 0 \}$

$\begingroup$ No, page 91, in the last example, ${\mathfrak m}^{2}=0$ is true. $\endgroup$– LaotzuFeb 13, 2014 at 12:15

2$\begingroup$ @Laotzu Your comment is wrong: $x^5=x^2x^3\in(x^2,x^3)^2$, and $x^5\notin x^4k[x^2,x^3]$. As it is said, $\mathfrak m^2=(x^5)$. $\endgroup$ Apr 11, 2014 at 16:46
Here are a few more minor errors:
p. 42, proof of Prop. 3.11(iv), last line: change "by i)" to "by ii)".
p. 72, Exercise 29 (clarification): by "a local ring of $A$" they mean "a localization of $A$ at some prime ideal". (See this question.)
p. 72, Exercise 31: for condition (2), change "for all $x,y \in K^*$" to "for all $x,y \in K^*$ such that $x+y \neq 0$". (Similarly on p. 94, unless one uses the convention $v(0)=+\infty$.)
p. 74, Example 2: change "if $a \neq 0$" to "if $a \neq 0$ and $a \neq \pm 1$".
p. 75, proof of Prop. 6.2: change "hence $M_n=M$" to "hence $M_n=N$".
p. 82, definition of irreducible ideal: it is understood (by the first sentence of the section) that an irreducible ideal $\mathfrak{a}$ is also required to be proper (otherwise Lemma 7.12 would be false, for example). Also, there should perhaps be an entry "Ideal, irreducible, 82" in the index on p. 127.
p. 94, paragraph before Prop. 9.2: change "$v \colon K^* \to Z$" to "$v \colon K^* \to \mathbf{Z}$".
p. 111, definition of $G(A)$ in the middle of the page: change $a^n$ to $\mathfrak{a}^n$.
And finally some really important corrections... ;)
p. 30, middle of the page: missing comma in "finite set of elements $x_1,\ldots x_n$".
p. 61, proof of Prop. 5.6(ii): missing space before the brackets in "$x/s \in S^{1}B(x \in B,\, s \in S)$".
p. 83, page header: change "NEOTHERIAN" to "NOETHERIAN".

12$\begingroup$ When I write my first scifi novel, it's going to be about the proud race of Neotherians. $\endgroup$– Tim Campion ♦May 13, 2015 at 13:56

$\begingroup$ Adding to your correction for Exercise 31 on p. 72: Replace "Show that the set of elements $x \in K^\ast$ such that $v\left(x\right) \geq 0$" by "Show that the union of $\left\{0\right\}$ with the set of elements $x \in K^\ast$ such that $v\left(x\right) \geq 0$". (Of course, anyone with a bit of experience glances over issues like this, knowing the convention that $v\left(0\right) = +\infty \geq 0$; but an introductory exercise is a good place to do things right.) $\endgroup$ May 21, 2017 at 15:07

$\begingroup$ Also, Exercise 30 on p. 72 suffers from the same trivial error you found in Exercise 31 (namely, "for all $x, y\in K^\ast$" should be "for all $x, y \in K^\ast$ satisfying $x+y \neq 0$"). $\endgroup$ May 21, 2017 at 15:09
Here is what I found (in LaTeX unlike in the original); only two of the errors mentioned below (both indicated) have been reported in previous answers here.
p.56, line 24 (Exercise 13(iv)): change second ${\mathfrak p}^{(n)}$ to ${\mathfrak p}^n$
p.76, line 11: change "Exercise" to "Example"
p.76, line 14: change "Exercise" to "Example"
p.91, line −11 (line 2 of second example): change (8.7) to (8.8)
(this was already found by Zev Chonoles. See their answer above from Feb 5, 2011)
p.97, line −1: change (9.7) to (9.6)
p.104, line −4: change $\left\{ G'_n \cap G_n \right\}$ to $\left\{ G' \cap G_n \right\}$
p.114, line −12: change $0 \rightarrow {\mathfrak b}^{m} M \rightarrow M/{\mathfrak b}^{m} M \rightarrow 0$ to $0 \rightarrow {\mathfrak b}^{m} M \rightarrow M \rightarrow M/{\mathfrak b}^{m} M \rightarrow 0$
(this was already found by Mahdi MajidiZolbanin. See their answer above from July 17, 2012)
On page 91, the second line in the second Example should refer to Proposition 8.8, not Theorem 8.7.
Page 69, Ex5.17: this is not the weak form, and the result is rather trivial.

1$\begingroup$ The part on page 69 is trivial, but the part on page 70 (still exercise 17) is the weak form of the Nullstellensatz. $\endgroup$ Jun 14, 2012 at 12:43

$\begingroup$ I think in Ex 16 and 17 one should replace $I(X)$ with an arbitrary ideal $I$ that defines $X$, and take $A=k[t_1,\cdots,t_n]/I$. The proof still goes through. This way the second part of Ex 17 can be derived from the first part by taking $I$ to be the maximal ideal. $\endgroup$ Mar 5, 2013 at 3:47
Page 114, Exercise 5, the short exact sequence is missing the middle term.
The following slip on p82 was found by Kenny Lau when he was formalising Prop 7.8 in Lean: In the line "Substituting (1) and making repeated use of (2) shows that each element of C is..." there's an implicit induction proof, but the base case where the element is 1 is not dealt with. This can be fixed in a number of ways, e.g. by adding a new condition (0) 1 = sum_i b_i y_i and using the b_i as further generators of B_0.

7$\begingroup$ Then it's happening! We are entering the time when our mistakes are revealed by formalization! $\endgroup$– Tim Campion ♦Sep 23, 2020 at 13:55
Page 118, line 16, Example: Poincaré series should be $P(A,t)=(1t)^{s}l(A_0)$.
p. 107, lines 45: change "$A^* = A[x_1,\dots,x_r]$" to "$A^*$ is a quotient of $A[x_1,\dots,x_r]$".
Proof of Theorem 11.22, page 123, direction iii) $\Rightarrow$ i): The map should be $\alpha : k[t_1,\dotsc,t_d] \rightarrow G_{\mathfrak{m}}(A)$.
On p. 65, immediately following Proposition 5.18, the authors claim that for any field $K$ and algebraically closed field $\Omega$, if we partially order the set of ring homomorphisms from subrings of $K$ to $\Omega$ by extension then the conditions of Zorn's lemma are satisfied, and thus there exists at least one maximal element. This is false, since the set of ring homomorphisms from subrings of $K$ to $\Omega$ may be empty: for instance taking $K=\mathbb{F}_p$ and $\Omega$ an algebraically closed field of characteristic zero, the only subring of $\mathbb{F}_p$ is $\mathbb{F}_p$ itself and there is no ring homomorphism (i.e. field extension) from $\mathbb{F}_p$ to a field of characteristic zero.
What the authors should be claiming is that if there exists any ring homomorphism from a subring of $K$ to $\Omega$, then there exists a maximal ring homomorphism. This is all they need for the uses they make of this result (in Corollary 5.22 and Proposition 5.23).
Not sure this can be considered an error, but it confused me:
on p. 65 the homomorphisms in "Let $\Sigma$ be the set of all pairs $(A, f)$, where $A$ is a subring of $K$ and $f$ is a homomorphism of $A$ into $\Omega$ " are not supposed to be injective (I'm not sure if this is the universal usage, but I normally understand into = injective).

2$\begingroup$ Although I guess it makes sense as a kind of dual to "onto", I have never heard the convention that "into" means "injective". $\endgroup$– LSpiceNov 1, 2021 at 12:49

1$\begingroup$ @LSpice I think I’ve come across it, although I don’t like it and I don’t think it’s anywhere close to standard. But approximately 666 times worse: a little light Googling reveals that using “into” to mean “not onto” is a thing! $\endgroup$ Nov 1, 2021 at 13:46

$\begingroup$ In fact upon closer looking the authors consistently use into without implying anything about injectivity  so this is clearly not an error. So I probably should delete the answer, right? $\endgroup$ Nov 1, 2021 at 14:03

1$\begingroup$ In a comment by Anna M. to one of her own answers here in 2010 when this MO question first appeared, she noticed the same implicit convention that “homomorphism into” does not mean injective but she thought it did at first. $\endgroup$– KConradNov 1, 2021 at 15:47

$\begingroup$ I agree the book’s convention is potentially confusing but perhaps not strictly an error. I think it is worth keeping your answer here for the benefit of future users of the book. $\endgroup$– KConradNov 1, 2021 at 15:51
Last line of page 91, exercise 1, should it be hence $0=p_i^{(r)}$ for all large $r$?
Here are some mistakes (it is possible that I am wrong):
page 68, line 7, the "$f$" should be "$f_1$";
page 68, line 5, "larger than $g_m$" is already enough;
page 68, ex 10 ii), "(b') $\Rightarrow$ (c')" needs an extra condition that $f$ is injective;
page 89, line 4, "$\mathfrak{N}^k\supseteq\mathfrak{N}$" should be "$\mathfrak{N}^k\subseteq\mathfrak{N}$";
page 97, line 1, “(9.7)” should be “(9.6)”;
...
Maybe in page 22, in the proof of corollary 2.7, $=$ should be $\cong$?
On Chapter 2 (p. 19), for $A$submodules $P,N$ of an $A$module $M$ is defined the quotient $(N:P)$ as a subset of the ring $A$ of scalars. Later, in Corollary 3.15 (p. 43), it is proved that for any multiplicatively closed subset $S$ of $A$ the equality $S^{1}(N:P)=\bigl(S^{1}N:S^{1}P\bigr)$ holds, as long as $P$ is finitely generated. Note that in this case the result deals with an ideal in the ring $S^{1}A$.
Later (p. 96) the notion of fractional ideals is introduced: they are certain $A$submodules of the field of fractions $K$ of $A$, being $A$ a domain. For a fractional ideal $M$ is defined the set $(A:M)=\{x\in K: xM\subseteq A\}$; note that this new definition differs from that given previously. Proposition 9.6 (p. 97) proves a equivalence regarding this new notion, but the proof uses Corollary 3.15, which is incorrect, though the correct reasoning is entirely similar.