Question on the argument proving primary decomposition theorem

48 Views Asked by At

Lang - Algebra p.150, Lemma 7.6

Let $E$ be a torsion module of exponent $p^r(r\geq 1)$ for some prime element $p$. Let $x_1\in E$ be an element of period $p^r$. Let $\bar E = E/(x_1)$.

Let $\bar y\in \bar E$ have period $p^n$ for some $n\geq 1$. Let $y$ be a representative of $\bar y$ in $E$. Then $p^n y\in (x_1)$, and hence

$p^n y= p^s c x_1, c\in R, p$ does not divide $c$ for some $s\leq r$.

Why $s\leq r$? I supposed $s>r$ to lead a contradiction, but I couldn't. How does this inequality hold?

1

There are 1 best solutions below

0
On BEST ANSWER

$p^n \bar{y}=0$ in $E$ means that $p^ny \in (x_1)$. The last equation simply states this fact - general element of $(x_1)$ is of the form $p^s c x_1$ for $s \leq r $ (since $p^r c x_1=c(p^r x_1)=0, p^{r+1} c x_1=pc(p^r x_1)=0, \dots$ and so on - one simply does not neeed to consider exponents $s$ bigger than $r$).