A passage from Bourbaki's Algebre X reads,

"... l'homothetie de rapport $a_1$ dans $\oplus_{i\geq0}I^iM/I^{i+1}M$ est injective,..."

Here $M$ is an $A$-module and $I=(a_1,\ldots,a_n)\subset A$. According to Google Translate, "homothetie" means "homothety" in English, which I have never heard of. Am I looking at something new, or just lost in translation? What does the phrase mean?


To elaborate on Pierre-Yves Gaillard's comment above:

Take the filtration of $M$ by powers of $I$, and form the direct sum as given by taking successive quotients. Each summand is a quotient of submodules of $M$, and is thus still an $A$ module; hence so is $\oplus_{i \geq 0} I^iM/I^{i+1}M$. In particular, multiplication by $a_1$ gives a linear transformation on $\oplus_{i \geq 0} I^iM/I^{i+1}M$.

The word homothety is sometimes used to describe such a linear transformation. See for instance the proof of proposition 1.1.2 in Bruns' and Herzog's "Cohen Macaulay Rings" (Revised Edition) (pg.4).

The upshot of the phrase is that this transformation is injective. Thus if $x \in \oplus_{i \geq 0} I^iM/I^{i+1}M$ and $a_1x=0$ then $x=0$.

In the commutative algebra literature it is sometimes said in this situation that $a_1$ is regular on $\oplus_{i \geq 0} I^iM/I^{i+1}M$, (provided that the homothety is not surjective).

  • $\begingroup$ Thanks for your answer. From the context, it seems that what Bourbaki meant was interpreting $\oplus_{i\geq0}I^iM/I^{i+1}M$ as an $\oplus_{i\geq0}I^i/I^{i+1}$-module, with $a_1$ residing as the image in $I/I^2$, rather than interpreting $\oplus_{i\geq0}I^iM/I^{i+1}M$ as an $A$-module (otherwise multipliying by $a_1$ would be a zero map.) $\endgroup$
    – ashpool
    Aug 26 '11 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.