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I have some questions of the book Introduction to commutative algebra by M. F. Atiyah and I. G. Macdonald.

On Line 8-9 of Page 42, it is said that $(xs-a)t=0$ for some $t\in S$ iff $xst\in \mathfrak{a}$. If $(xs-a)t=0$, then $xst=at \in \mathfrak{a}$. But if $xst \in \mathfrak{a}$, could we conclude that $(xs-a)t=0$ for some $a\in \mathfrak{a}$?

On Line 10-11 of Page 42, it is said that $\mathfrak{a} \in C$ iff $\mathfrak{a}^{ec} \subseteq \mathfrak{a}$. But on Page 10, Proposition 1.17(iii), it is said that $\mathfrak{a} \in C$ iff $\mathfrak{a}^{ec} = \mathfrak{a}$.

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LJR
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2 Answers2

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  1. I don't think the implication you are asking about is very clear either (but being unpracticed at commutative algebra, I could be overlooking something.) I think they are being a bit terse at readers' expense. At the very least, we could conclude that the last item implies the third item in the chain of implications. That is, $x\in\cup_{s\in S}(\mathfrak{a}:s)$ implies $xs=a\in\mathfrak{a}$ for some $s\in S$, whence $\frac{x}{1}=\frac{a}{s}$. This would make the line of implications into a circle, so that equivalence is guaranteed everywhere.

  2. In 1.17 i), they show that $\mathfrak{a}\subseteq\mathfrak{a}^{ec}$ in all cases. Then if $\mathfrak{a}\in C$, $\mathfrak{a}=\mathfrak{a}^{ec}$ would be equivalent with $\mathfrak{a}\supseteq\mathfrak{a}^{ec}$

rschwieb
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Concerning the second part of the question

Certainly $\mathfrak{a} \subset \mathfrak{a}^{ec}$ because for any $x \in \mathfrak{a}$ we have $ f(x) \in \mathfrak{a}^{e}$ So $f^{-1}(f(x)) \subset \mathfrak{a}^{ec}$ and $x$ is in $f^{-1}(f(x))$. So in particular one of the containments is always true and we only have equality when $\mathfrak{a}^{ec} \subseteq \mathfrak{a}$.

PVAL-inactive
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