Possibly false proof in AM

Question

Here is the excerpt of the book where I suspect a mistake (page 66):

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Where they say "The restriction to $A$ of the natural homomorphism $A^\prime \to k^\prime$" I think we don't want a restriction. 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 $B$.

Do you agree with my version and that what is written in Atiyah-Macdonald is not correct? Thank you.

You may want to check out http://mathoverflow.net/questions/42241/errata-for-atiyah-macdonald — Asaf Karagila, Jul 18 '12 at 08:11
@AsafKaragila Thank you, I was aware of the list but this doesn't seem to be mentioned there. — Rudy the Reindeer, Jul 18 '12 at 08:31
You might have a new one for the list. I'm pretty tired and doubting authors of such intelligence is always dangerous, but I don't see how restricting to $A$ helps them here. — Dylan Moreland, Jul 18 '12 at 08:34
@DylanMoreland Thank you for your comment. I agree with you. So I will wait for more feedback. — Rudy the Reindeer, Jul 18 '12 at 09:10
I have the following correction for this penciled in: strike out "the restriction to $A$ of", and change a coup of $A$s to $A'$s. This wasn't in the errata on MO. I checked it in some other book, probably Lang. — Bruce Evans, Jul 18 '12 at 10:02

score 6 · Accepted Answer · answered Jul 18 '12 at 11:03

6

Congratulations for spotting the difficulty and correcting it, Clark: you are absolutely right and I completely agree with your version!

As a slightly different formulation for the proof that $x\notin B$, I would just remark that if we had $x\in B$, we would deduce the absurd conclusion $$1=g(1)=g(x\cdot x^{-1})=g(x)\cdot g(x^{-1})=g(x)\cdot 0=0$$

answered Jul 18 '12 at 11:03

Georges Elencwajg

150,790

Dear Georges, thank you very much! – Rudy the Reindeer Jul 18 '12 at 11:08
1

Dear Clark, you are very welcome: there is nothing more satisfying for a teacher than seeing a student (I'm guessing here!) very critically reading a book, to the point of spotting an, err..., inaccuracy in a famous book by famous authors. By the way, I strongly encourage you to add this erratum to the list on MathOverflow mentioned by Asaf: you would do our community a great favour. – Georges Elencwajg Jul 18 '12 at 11:36
Dear Georges, thank you. Yes, your guess is correct, I am a student. The erratum has already been added: I requested MattN to do it (in the comments to this question). – Rudy the Reindeer Jul 18 '12 at 12:51
Thanks to you and @MattN, then. – Georges Elencwajg Jul 18 '12 at 13:13

Possibly false proof in AM

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