Errata for Atiyah-Macdonald - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T18:08:35Z http://mathoverflow.net/feeds/question/42241 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald Errata for Atiyah-Macdonald Tim Campion 2010-10-15T00:56:31Z 2013-03-08T02:58:54Z <p>Is there a good errata for Atiyah-Macdonald available? A cursory Google search reveals a laughably short list <a href="http://math.mit.edu/~ssam/soln/commalg.errata" rel="nofollow">here</a>, with just a few typos. Is there any source available online which lists inaccuracies and gaps?</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/42271#42271 Answer by Georges Elencwajg for Errata for Atiyah-Macdonald Georges Elencwajg 2010-10-15T10:50:17Z 2010-10-15T10:50:17Z <p>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 $a\mapsto 1_B\otimes g(a) =f(a)\otimes 1_C$.</p> <p><strong>PS:</strong> To prevent misunderstandings, let me add that Atiyah-MacDonald is, <em>to my taste</em>, the best mathematics book I have ever seen, all subjects considered.</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/42427#42427 Answer by Zev Chonoles for Errata for Atiyah-Macdonald Zev Chonoles 2010-10-16T22:28:20Z 2010-10-16T22:28:20Z <p>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 $\not\supseteq$.</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/42433#42433 Answer by Zev Chonoles for Errata for Atiyah-Macdonald Zev Chonoles 2010-10-16T22:48:02Z 2010-10-16T22:48:02Z <p>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 </p> <p>"$0\rightarrow\mathbb{Z}\otimes N\stackrel{f\otimes 1}{\longrightarrow} \mathbb{Z}\otimes N$".</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/42436#42436 Answer by Zev Chonoles for Errata for Atiyah-Macdonald Zev Chonoles 2010-10-16T22:53:15Z 2010-10-16T22:53:15Z <p>On page 31, the first line refers to Proposition 2.11, when it should be 2.12.</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/42671#42671 Answer by Anna M. for Errata for Atiyah-Macdonald Anna M. 2010-10-18T17:18:38Z 2010-10-18T17:18:38Z <p>Minor typos: </p> <p>p.34, exercise 2.23: Second sentence should start "For each finite subject $J$ of $\Lambda$".</p> <p>p.48, exercise 3.27(i): The bracketed text should read "Use Exercises 25 and 26".</p> <p>p.71, exercise 5.23: The hint should start "The only hard part is (iii) => (i). Suppose (i) is false".</p> <p>p.88, exercise 7.27(v): The last clause should read "the homomorphism $f_{!}$ is a $K_1(A)$-module homomorphism".</p> <p>p.127, index entry for "flat, faithfully": Should cite p. 46, not p. 29.</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/42674#42674 Answer by Anna M. for Errata for Atiyah-Macdonald Anna M. 2010-10-18T17:31:51Z 2010-10-18T17:31:51Z <p>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.</p> <p>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...</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/42675#42675 Answer by Anna M. for Errata for Atiyah-Macdonald Anna M. 2010-10-18T17:42:32Z 2010-10-18T17:42:32Z <p>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. </p> <p>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.</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/43364#43364 Answer by Pierre-Yves Gaillard for Errata for Atiyah-Macdonald Pierre-Yves Gaillard 2010-10-24T08:27:49Z 2011-07-10T04:15:42Z <p><strong>EDIT OF JUNE 9, 2011</strong></p> <p>Page 102, penultimate paragraph:</p> <p>"... $f$ induces a homomorphism $\widehat{f}:\widehat{G}\to\widehat{H}$, which is continuous."</p> <p>No topology has been defined on $\widehat{G}$ and $\widehat{H}$. </p> <p>[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 $\widehat{W}\subset V$.] </p> <p>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 <a href="http://mathoverflow.net/questions/11437/erratum-for-cassels-froehlich" rel="nofollow">http://mathoverflow.net/questions/11437/erratum-for-cassels-froehlich</a>). Atiyah and Wall forgot to mention the crucial compatibility between change of groups and connecting morphisms. (See p. 99.)</p> <p><strong>END OF EDIT OF JUNE 9, 2011</strong> </p> <p>Page 25, first line of the proof of (2.13): change (2.11) to (2.12).</p> <p>Page 29, about two third of the page: change (2.14) to (2.13). </p> <p><strong>EDIT.</strong> Page 39, last line: change $m$ to $m_i$ (three times).</p> <p><strong>EDIT OF NOV. 22, 2010.</strong> Page 63, proof of Lemma 5.14. The current text reads </p> <p>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 $M=A[x_1,\dots,x_n]\ \dots$ </p> <p>It would be better (I think) to write something like </p> <p>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 $M=A[x_1,\dots,x_m]\ \dots$ </p> <p>[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 $A_i$.] </p> <p>[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 $y_i=(0,x_i,0,\dots)$.]</p> <p>[July 7, 2011, GMT. Page 112, proof of Proposition (10.24). Instead of $\mathfrak{a}^{k+n(i)}$ read $\mathfrak{a}^{\max(0,k-n(i))}$.] </p> <p>[July 9, 2011, GMT. Page 122, proof of Proposition 11.20. There is a minor typo in the third line of the proof: read $\mathfrak{q}^2$ instead of $\mathfrak{q}$. Another problem is that the notation $d(A)$ is used with two different meanings: the one given on p. 117 for graded modules, and the one given on p. 119 for noetherian local rings. I'll use the notation $D(A)$ for the first meaning. The definition of $D(M)$ for a graded module $M$ makes sense only if the Poincaré series $P(M,t)$ is nonzero. In particular Proposition 11.3 makes sense and holds (with the proof given in the book) only if </p> <ul> <li><p>$P(M,t)\not=0\not=P(M/xM,t)$, </p></li> <li><p>$D(M)\ge1$, </p></li> <li><p>$D(M/xM)\ge0$ if $D(M)=1$, </p></li> </ul> <p>and we must take this into account when using Proposition 11.3 to prove Proposition 11.20. The simplest way to do that is (in my opinion) to treat separately the case $s=0$. Indeed, when $s\ge1$, Proposition 11.3 (as amended above) applies.]</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/43376#43376 Answer by KConrad for Errata for Atiyah-Macdonald KConrad 2010-10-24T12:44:48Z 2010-10-24T12:44:48Z <p>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> <p>p. 29 (---> p. 41): on line 8, change (2.14) to (2.13)</p> <p>p. 55 (---> p. 66): exercise 2</p> <p>p. 71 (---> p. 82): exercise 23</p> <p>p. 88 (---> p. 99): exercise 27(v)</p> <p>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. </p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/54387#54387 Answer by Zev Chonoles for Errata for Atiyah-Macdonald Zev Chonoles 2011-02-05T03:10:16Z 2011-02-05T03:10:16Z <p>On page 91, the second line in the second Example should refer to Proposition 8.8, not Theorem 8.7.</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/54465#54465 Answer by Erik Aas for Errata for Atiyah-Macdonald Erik Aas 2011-02-05T21:17:16Z 2011-02-05T21:17:16Z <p>page 81, line 5: change $f_i \in A[x]$ to $f_i \in \mathfrak{a}$</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/73753#73753 Answer by Chan Yat-Fay for Errata for Atiyah-Macdonald Chan Yat-Fay 2011-08-26T10:11:05Z 2011-08-26T10:11:05Z <p>Page 69, Ex5.17: this is not the weak form, and the result is rather trivial.</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/91031#91031 Answer by Dylan Moreland for Errata for Atiyah-Macdonald Dylan Moreland 2012-03-12T23:12:39Z 2012-03-12T23:12:39Z <p>On page 23, in the third line of the sketch for Proposition 2.9, change "$v \circ u \circ f = 0$ " to "$f \circ v \circ u = 0$".</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/99601#99601 Answer by Matt for Errata for Atiyah-Macdonald Matt 2012-06-14T12:14:52Z 2012-07-22T16:11:01Z <p>On page 41 in the proof of proposition 3.10., change</p> <p>"i) $\implies$ ii) by (3.5) and (2.20)" to "i) $\implies$ ii) by (3.7) and (2.19)"</p> <p>On page 52 in remark 1) at the bottom of the page, change</p> <p>"(see Chapter 1, Exercise 25)" to "(see Chapter 1, Exercise 27)"</p> <p>On page 65 at the end of the proof of proposition 5.18. the black square to denote end of proof is missing.</p> <p>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 <a href="http://math.stackexchange.com/questions/172307/possibly-false-proof-in-am" rel="nofollow">math.SE</a>)</p> <p>On page 77 in the proof of proposition 6.7., change </p> <p>"...a composition series, by ii);..." to "...a composition series, by i);..."</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/102443#102443 Answer by Mahdi Majidi-Zolbanin for Errata for Atiyah-Macdonald Mahdi Majidi-Zolbanin 2012-07-17T12:44:06Z 2012-07-17T12:44:06Z <p>Page 114, Exercise 5, the short exact sequence is missing the middle term.</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/105808#105808 Answer by David Corwin for Errata for Atiyah-Macdonald David Corwin 2012-08-29T09:32:09Z 2012-08-29T09:32:09Z <p>On p.89, the second to last line of the proof of Proposition 8.4 should say $\mathfrak{R}^k \subseteq \mathfrak{R}$ instead of $\mathfrak{R}^k \supseteq \mathfrak{R}$.</p> http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald/123590#123590 Answer by Junyan Xu for Errata for Atiyah-Macdonald Junyan Xu 2013-03-05T04:00:54Z 2013-03-08T02:58:54Z <p>In p.45, Ex.3.12.iv, one can avoid the tedious argument provided in the hint by noting that <code>$K\otimes_A M\cong(A-\{0\})^{-1}M$</code>. (I originally did it as hinted ...)</p> <p>In p.68, Ex.5.10.ii, (b') is actually equivalent to a weaker (c') that asserts only that <code>$f^*:\mathrm{Spec}(B_\mathfrak{q})\to\mathrm{Spec}(A_\mathfrak{p})\cap V(\ker f)$</code> is surjective. However, (a') does imply the original (c').</p>