About the inductive step on the Structure Theorem on Principal Domains

Question

Im looking for a proof on the following statement Let R be a principal domain and M be an R-module of p-torsion and finitely generated ($p\in R$ prime). Let $x_1 \in M$ with $Ann_R(M)=Ann_R(x_1)$, and consider the quotient $N=M/R.x_1$. If $N=\langle y_2 \rangle \oplus ... \oplus \langle y_n \rangle$ then there are $x_2,...,x_n \in M$ with $\pi (x_i)=y_i$ and $Ann_R(x_i)=Ann(y_i)$ for $2 \le i \le n$ (being $\pi:M \rightarrow N$ the proyection)

Answer 1

$\newcommand{ann}[1]{\operatorname{Ann}_{R}({#1})}$

What follows is a possible approach. I must admit that the proof is rather verbose, with some comments here and there in order to clarify some steps, as the argument is long and not as elegant as one may like.

Lemma: let $R$ be a principal domain and $N$ a finitely generated torsion $R$-module such that $$ \ann{N} = p^mR $$ with $p$ irreducible and $m \in \mathbb{N}$. Suppose that $x_1$ is so that $\ann{x_1} = \ann{N}$ and $$ N/Rx_1 \simeq Ry_2 \oplus \dots \oplus Ry_n. $$ Then, there exist $x_2, \dots, x_n \in N$ such that $\ann{x_i} = \ann{y_i}$ and $\pi(x_i) = y_i$ for all $i$, with $\pi: N \to N/Rx_1$ the canonical projection. Moreover, $$ N \simeq Rx_1 \oplus \dots \oplus Rx_n. \tag{$\star$} $$

Proof. Let's divide the argument into two steps. First, we will see that for any $y \in N/Rx_1$, there exists a 'lift' $z \in \pi^{-1}(y)$ with $\ann{z} = \ann{y}$. Finally, we will conclude that the 'lifts' of $x_2, \dots, x_n$ along with $x_1$ sum the whole module $N$, and that the sum is direct.

Let $y \in N/Rx_1$. Note that since $\ann{N} = p^mR$, for any $[x'] \in N/Rx_1$ we have that

$$ p^m[x'] = [p^mx'] = [0] = 0, $$

and therefore, $p^mR \subset \ann{N/Rx_1}$. The ring $R$ is principal, and the by the former inclusion, we know that $\ann{N/Rx_1}$ divides $p^mR$, so $\ann{N/Rx_1} = p^kR$ with $k \leq m$. More so, since $y$ is an element of $N/Rx_1$, the ring elements that annihilate $N/Rx_1$ will in particular annihilate $y$. Thus, we get that $$ \ann{N/Rx_1} \subset \ann{y} $$ and by the exact same argument, we get $\ann{y} = p^rR$ with $r\leq k \leq m$.

In particular we have that $p^ry = 0$ and so the elements of the class of $y$ are in the same class of $0$ modulo $Rx_1$. Conretely, $y$ is an element of the quotient, so it is of the form $y = [z]$ for some $z \in N$. Since we see that $p^ry = [p^rz] = 0 = [0]$, necessarily $p^rz \sim 0$ which means that $p^rz -0 \in Rx_1$, that is, $y = \pi(z)$ and $p^rz = ax_1$ for some $a \in R$. For future convinience, let's write $a = p^sc$ with $p \not | c$ and $s \geq 0$, that is, $p^s$ is the greatest power of $p$ in the factorization of $a$, with $s$ eventually zero if $p \not | a$.

From now on, we will separate in two cases, depending on wether $s \geq m$ or not. Once again, recall our objective: we must see that there exists $z \in \pi^{-1}(y)$ has the same annihilator as $y$. So far, we only know that there exists $z \in \pi^{-1}(y)$ with $$ p^rz = p^scx_1. \tag{1} $$

• Case 1 ($s \geq m$): this one is straight forward, since $$ p^rz \stackrel{(1)}{=} p^scx_1 \stackrel{(s \geq m)}{=} p^m(p^{s-m}cx_1) = 0 $$ because $p^m$ annihilates $N$, then $p^rR \subset \ann{z}$ and so by a previous argument, $\ann{z} = p^iR$ with $i \leq r$. Hence, $p^iz = 0$ and so $$ p^iy = p^i\pi(z) = \pi(p^iz) = \pi(0) = 0. $$ This also says that $p^i \in \ann{y} = p^rR$ so $p^r | p^i$ and thus $r \leq i$. Since both inequalities holds, necessarily we get $i = r$ and $\ann{z} = \ann{y} = p^rR$.

• Case 2 ($s < m$): sadly, this one is even more technical. We will see that in this case the annihilator of $z$ is not the same one as $y$, but nevertheless we can select an appropriate replacement which is still a 'lift'. Once again $p^m$ annihilates the whole module, so $$ 0 = p^mcx_1 = p^{m-s}(p^scx_1) \stackrel{(1)}{=} p^{m-s}p^rz = p^{m-s+r}z. \tag{2} $$ Equation $(2)$ proves that the annihilator of $p^scx_1$ contains $p^{m-s}R$, so $\ann{p^scx_1} = p^jR$ with $j \leq m-s$. Since $$ p^jp^scx_1 = p^{j+s}cx_1 = 0, $$ then $p^{j+s}c \in \ann{x_1} = p^mR$. We know that $p \not | c$ by construction: it has then to be that $p^m | p^{j+s}$ and so $m \leq j+s$ or equivalently, $j \geq m-s$. Since both inequalities hold, we see that $j = m-s$. Finally, this says that $\ann{z} = p^{m-s+r}R$. In effect, from $(2)$ we have seen that $p^{m-s+r} \in \ann{z}$ and $\ann{z} = p^iR$ with $i \leq m-s+r$. On the other hand, from the last paragraph of case one, we know it always holds that $r \leq i$ because $\ann{z} \subset \ann{y}$. Thus, $$ 0 = p^iz = p^{i-r}p^rz = p^{i-r}p^scx_1. $$ We already know that $\ann{p^scx_1} = p^{m-s}R$, so $m-s \leq i-r$ and thus $m-s+r \leq i$ which proves the other inequality. Hence $\ann{z} = p^{m-s+r}R$ which is not exactly what we wanted, but it can be fixed: since $\ann{N} \subset \ann{z}$, then by the same argument as always it is $m -s +r \leq m$ and $r \leq s$. Thus, $p^{s-r}$ is well defined. We now claim that the element $z' := z - p^{s-r}cx_1$ verifies $\pi(z') = y$ and $\ann{z'} = p^rR = \ann{y}$. From the fact that $\ker \pi = Rx_1$, we conclude that $$ \pi(z') = \pi(z) - p^{s-r}c\pi(x_1) = \pi(z) - 0 = \pi(z) = y. $$ Finally, by a direct calculation, $$ p^rz' = p^r - p^scx_1 = p^rz - p^rz = 0 $$ so if $\ann{z'} = p^lR$, then $l \leq r$ and we always have $r \leq l$ because $\ann{z'} \subset \ann{y}$ since $\pi(z') = y$.

We have thus proven that any $y \in N/Rx_1$ has a 'lift' $z \in N$ with $\pi(z) = y$ and $\ann{z}= \ann{y}$.

Now, for each $2 \leq i \leq n$, let $x_i \in N$ be so that $\pi(x_i) = y_i$ and $\ann{x_i} = \ann{y_i}$, and let's see that $N = Rx_1 \oplus \dots \oplus Rx_n$.

If $x \in N$, since $\{y_i\}_{i = 2}^n$ generate $N/Rx_1$ there exist $a_2, \dots, a_n \in R$ with $$ \pi(x) = \sum_{i = 2}^na_iy_i = \sum_{i = 2}^na_i\pi(x_i). $$ Thus, $\pi(x - \sum_{i = 2}^na_ix_i) = 0$ and therefore $x - \sum_{i = 2}^na_ix_i \in Rx_1$, which means that there exists $a_1 \in R$ with $x - \sum_{i = 2}^na_ix_i = a_1x_1$. Equivalently, $$ x = x - \sum_{i = 2}^na_ix_i + \sum_{i = 2}^na_ix_i = \sum_{i = 1}^na_ix_i. $$ and so $N = \sum_{i=1}^nRx_i$ because $x$ was arbitrary. To conclude, let's see that the sum is direct. If $b_1x_1 + \dots b_nx_n = 0$, then applying $\pi$ we get $$ 0 = b_2y_2 + \dots b_ny_n $$ and since $N/Rx_1$ is the direct sum of $y_2, \dots, y_n$, then $b_iy_i = 0$ if $i > 1$. Consequently $b_i \in \ann{y_i} = \ann{x_i}$ and so $b_ix_i = 0$ for $i > 1$, which also proves that $$ 0 = b_1x_1 + \dots b_nx_n = b_1x_1 + (b_2x_2 + \dots b_nx_n) = b_1x_1 $$ and so the sum is direct, which concludes the proof. $\square$