I have read questions related to $\mathbb{Z}$-module and exact sequence of it. I have a question about generalising this result (or finding counter example)
Let $R$ be an integral domain and $0\rightarrow M_1\rightarrow M\rightarrow M_2\rightarrow 0$ be an exact sequence of finitely generated $R$-module. Then, is $rank(M)=rank(M_1)+rank(M_2)$ true?
(Definition of rank: $rank(M)$ denotes the cardinality of the largest linearly independent subset of $M$)
I have read statements online and I found that "flat modules" satisfy this result (though I have not learnt tahings about maps and tensor product so I cannot effectively understand all proofs), but there are no general result about integral domains. Hence, I would like to ask this question here and seek for help.
Thank you for your kind help!
(I have stuck in this for half a week and I am not making any progress. The current state of my work is that:
I have (kind of?) proved that $Rank(M)\geq Rank(M_1)+Rank(M_2)$ by putting their linear independent sets together and prove that it is still linearly independent.
I have proved that the statement is true if $Rank(M)-Rank(M_1)=0$ or $1$. The proof of $0$ is by torsion module, and the proof of $1$ is proved by hardcoding (but it cannot be generalised to $2$ or larger. )
Let $K$ be the fraction field of $R$. $K$ is flat over $R$ (this is a general fact about localizations).
One can show that $Rank(M)=\operatorname{dim}_K(K \otimes_R M)$
So if we have an exact sequence of $R$-modules
$0\rightarrow M_1\rightarrow M\rightarrow M_2\rightarrow 0$
We can tensor with $K$, using that $K$ is flat over $R$ and get an exact sequence of vector spaces.
$0\rightarrow K \otimes_R M_1\rightarrow K \otimes_R M\rightarrow K \otimes_R M_2\rightarrow 0$
Now the result follows from the usual rank-nullity for vector spaces and the above relationship between rank and dimension over $K$.