# translation from French

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?

• Dilation? See en.wikipedia.org/wiki/Homothetic_transformation
– Did
Aug 26 '11 at 0:31
• If $M$ is an $A$-module and $a$ en element of $A$, then the map $M\to M$, $m\mapsto am$ is the "homothety" of "ratio" $a$ on $M$. Aug 26 '11 at 0:31

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 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).
• 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.) Aug 26 '11 at 12:54