Notation: For an ID (integral domain) $R$ and $R$-module $C$, denote the torsion submodule: $C_t$. A unital ring $R$ has $\textbf{invariant basis number}$ [IBN] If every basis of a given free $R$-module has the same cardinality. If $R$ is a PID, $M$ is an $R$-module, and $m\in M$, then we denote the order ideal: $\mathcal{O}_m:= \{s\in R\mid sm = 0\}$, which equals $(r)$ for some $r\in R$.
I would like verification of my proof of:
$\textbf{Theorem.}$ Let $R$ be a PID, let $F$ and $E$ be free $R$-modules of finite rank, and let $M$ and $N$ be torsion $R$-modules s.t. $A:=F\bigoplus M$ and $B:=E\bigoplus N$ are f.g. and $A\cong B$. Then we have:
- $A_t = 0\bigoplus M$ and $B_t = 0\bigoplus N$;
- $\frac{F\bigoplus M}{0\bigoplus M}\cong \frac{E\bigoplus N}{0\bigoplus N}$.
Hence, $\text{rank F= rank E}$.
(I would also like to ask if this "Theorem" is correct, I am not sure if it is, the statement came from me). Attempted proof:
$\textbf{Proof of 1.} $ Let $n:=\text{rankF}$, so let $X = \{x_1,...,x_n\}$ be a basis of $F$. If $(f,m)\in F\bigoplus M$ has a nonzero order ideal, lets say that $\mathcal{O}_{(f,m)} = (r)$. So $r\neq 0_R$ and $r(f,m) = 0_A$, while $f = r_1x_1+...+r_nx_n$ for some $r_1,...,r_n\in R$.Then $0_A = (0,0)= (rr_1x_1+...+rr_nx_n,rm)$ so that every $r_i = 0$ by linear independence of $X$ (since $R$ has no zero divisors and $r\neq 0$) and $rm = 0$. Hence $(f,m) = (0,m)\in 0\bigoplus M$. So $A_t\subset 0\bigoplus M$. Conversely, if $(0,w)\in 0\bigoplus M$, write $\mathcal{O}_w = (s)$, then $s\neq 0$ (since $w\in M$ and $M$ is a torsion module). So $s(0,w) = (0,sw) = (0,0)$ so that $(0,w)\in A_t$.
$\textbf{Proof of 2.} $ Let $\phi:F\bigoplus M\xrightarrow{\sim}E\bigoplus N$ be an isomorphism. Well $\phi(0\bigoplus M)\subset 0\bigoplus N$ (if $(0,m)\in 0\bigoplus M$, say $\mathcal{O}_m =(r)$, where $r\neq 0_R$, then $r\phi(0,m) = \phi(r(0,m)) = \phi(0_A) = 0_B$, so that $\phi(0,m)\in B_t = 0\bigoplus N$ (1)). So $\phi$ induces an $R$-module homomorphism $\overline{\phi}:\frac{F\bigoplus M}{0\bigoplus M}\to \frac{E\bigoplus N}{0\bigoplus N}$, but $\overline{\phi}$ is an isomorphism since $\phi$ is an isomorphism and $\phi(0\bigoplus M)= 0\bigoplus N$ ($\phi(0\bigoplus M)\supset 0\bigoplus N$ by an almost identical argument to (1)).
The last statement follows immediately from the fact that $F\cong \frac{F\bigoplus M}{0\bigoplus M}\cong \frac{E\bigoplus N}{0\bigoplus N}\cong E$ and two free modules over a ring $S$ with IBN are iso (as $S$-modules) iff they have the same rank.