Errata for Atiyah–Macdonald

Question

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?

Answer 1

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 Atiyah-MacDonald is, to my taste, the best mathematics book I have ever seen, all subjects considered.

Answer 2

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 $$\phi^n+a_1\,\phi^{n-1}+\cdots+a_n=0$$ 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 $$\phi^n+\psi(a_1)\,\phi^{n-1}+\cdots+\psi(a_n)=0$$ 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 $$ \phi^n+a_1\,\phi^{n-1}+\cdots+a_n\,\phi^0=0. $$

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 Cassels-Froehlich). 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 4-5. 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 $\mathfrak{a}^{\max(0,k-n(i))}$.]

[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 $\mathfrak q^2$.]

Answer 3

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 $\not\supseteq$.

Answer 4

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

"$0\rightarrow\mathbb{Z}\otimes N\stackrel{f\otimes 1}{\longrightarrow} \mathbb{Z}\otimes N$".

Answer 5

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.

A-M 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...

Answer 6

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.

Answer 7

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.

Answer 8

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

Answer 9

page 81, line 5: change $f_i \in A[x]$ to $f_i \in \mathfrak{a}$

Answer 10

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 $\mathfrak{N}^k \supseteq \mathfrak{N}$.

Answer 11

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 non-zero 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 $H=\{ 0 \}$

Answer 12

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".

Answer 13

On page 31, the first line refers to Proposition 2.11, when it should be 2.12.

Answer 14

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').

Answer 15

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 Majidi-Zolbanin. See their answer above from July 17, 2012)

Answer 16

On page 91, the second line in the second Example should refer to Proposition 8.8, not Theorem 8.7.

Answer 17

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);..."

Answer 18

Page 114, Exercise 5, the short exact sequence is missing the middle term.

Answer 19

Page 118, line 16, Example: Poincaré series should be $P(A,t)=(1-t)^{-s}l(A_0)$.

Answer 20

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.

Answer 21

Page 69, Ex5.17: this is not the weak form, and the result is rather trivial.

Answer 22

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.

Answer 23

Proof of Theorem 11.22, page 123, direction iii) $\Rightarrow$ i): The map should be $\alpha : k[t_1,\dotsc,t_d] \rightarrow G_{\mathfrak{m}}(A)$.

Answer 24

p. 107, lines 4-5: change "$A^* = A[x_1,\dots,x_r]$" to "$A^*$ is a quotient of $A[x_1,\dots,x_r]$".

Answer 25

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).

Answer 26

Last line of page 91, exercise 1, should it be hence $0=p_i^{(r)}$ for all large $r$?

Answer 27

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.

Answer 28

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)”;

...

Answer 29

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).

Answer 30

Maybe in page 22, in the proof of corollary 2.7, $=$ should be $\cong$?