0

Assume that $R$ is and integral domain.

If $M$ is an $R$-module that contains a submodule $N$ free of rank $n$, such that $M/N$ is a torsion $R$-module, then $M$ is of rank $n$.

My reasoning: That $M/N$ is a torsion-module implies that for each $y_i \in M \setminus N$ there exists a non-zero $r_i \in R$ s such that

$$r_iy_i = r^{i}_1x_1+ \ldots + r^{i}_nx_n$$ where $x_1,\ldots,x_n$ is a basis for $N$.

Now, let´s choose arbitrary $y_1,\ldots,y_{n+1} \in M \setminus N$.

We know that since $$N \cong R^n$$ then $$\{e_1,\ldots,e_n\}$$ is a basis for $N$.

We can then write $$r_iy_i = a_{1i}e_1 + \ldots + a_{ni} e_n$$

where $r_i$ is choosen as the annihilator to $\bar{y_i} \in M/N$.

Now,we get a matrix $A$ of dimension $$(n+1) \times (n+1)$$ such that the last row is constantly $0$, where the columns $A_i$ are made up of the coefficients for $r_iy_i$ in the basis $\{e_1,\ldots,e_n\}$.

By computing the determinant from the last row:s elements as coefficients for the minors, we see that $$\operatorname{det}(A) = 0$$ and it follows from lemma that the columns are $R$-linearly dependent.

I.e. there exists atleast two columns $A_i,A_j$ for $i \neq j$ such that $rA_i = A_j$ for some $r \in R$ (I think).

Since $$rA_i = r(a_{1i}e_i+ \ldots + a_{ni}e_n) = r(r_iy_i) = a_{1j} e_1+ \ldots + a_{nj} e_n = A_j = r_jy_j$$ it follows that $$(rr_i)y_i-r_jy_j= = 0$$ but I think by assumption $r_i$ and $r_j$ are non-zero, and since $R$ is an integral domain, both $rr_i$ and $r_j$ are non-zero, so that we find that $y_i,y_j$ are $R$-linearly dependent.

Now, this extends to elements in $N$ ofc, since $N$ is free.

Is this argument correct? I feel I could possibly make it better; for example, by not dividing it into the case $M\setminus{N}$ and $N$, and also my argument for why $r_i$ and $r_j$ are non-zero.

Ben123
  • 952
  • 5
  • 10
  • Are you sure that this is true? Take for instance $R = \mathbb Z$, $M = \mathbb Z^n \times \mathbb Z/2\mathbb Z$ and $N = \mathbb Z^n \times {0}$. – Marktmeister Oct 21 '23 at 14:16
  • What is your definition of rank? What's your definition of a torsion module? What's your definition of free? – Arthur Oct 21 '23 at 14:17
  • I added "free of rank $n$" for the submodule $N$. But yes, I believe it is true. Also, I changed "$M$ free of rank $n$" to just "$M$ of rank $n$". – Ben123 Oct 21 '23 at 14:21
  • @Marktmeister $M$ is of rank $n$ in that example, right? I don´t see that this is a counterexample. – Ben123 Oct 21 '23 at 14:30

1 Answers1

1

One fact that is often helpful for reasoning about rank is that the rank of $M$ is thr dimension of $K \otimes_R M$ over $K$, where $K$ is the fraction field of $R$. Using this, one can reduce some questions about rank to linear algebra.

For example, if $0\to M' \to M \to M''\to 0$ is exact, then tensoring with $K$, which is flat over $R$, we obtain $$0\to K\otimes_R M' \to K \otimes_R M \to K \otimes_R M'' \to 0$$ Which implies $\operatorname{rank}(M)=\operatorname{rank}(M')+\operatorname{rank}(M'')$.

This implies the statement in question, even without the freeness assumption, because torsion modules have rank zero.

Lukas Heger
  • 20,801
  • Are you here using that the exact sequence splits, to say $$\operatorname{rank}(M) = \operatorname{rank}(M')+\operatorname{rank}(M'')$$?

    I suppose in my particular case you would take

    $$0 \to \operatorname{ker}(\pi) \to M \twoheadrightarrow M/N \to 0$$ where $$\pi:M \to M/N$$ is the projection onto the quotient?

    Can you explain why the sequence splits? I.e. could you for example show me that there exists a rectraction or a section in the tensored SES?

    Also, I´d really like some feedback if there is something wrong with my argument, before trying to think more broadly.

     – Ben123 Oct 21 '23 at 14:47
  • The sequence does not need to split: I tensor with $K$, get a new SES and then compare dimensions. – Lukas Heger Oct 21 '23 at 14:52
  • I mean, over a field any SES splits, but we don't need this: rank-nullity is enough. – Lukas Heger Oct 21 '23 at 14:53
  • Tensoring like that is exact because $K$ is flat over $R$. So that $$K \otimes_{R}-$$ is exact iff $K$ is flat. But how do you mean that you compare dimensions afterwards? – Ben123 Oct 21 '23 at 15:00
  • Yeah, I don´t have time to go into this more now, but please leave an answer if you want. Ofc. – Ben123 Oct 21 '23 at 15:02
  • @Ben123 if $0 \to V' \to V \to V'' \to 0$ is a short exact sequence of vector spaces over a field, then $\mathrm{dim}(V)=\mathrm{dim}(V')+\mathrm{dim}(V'')$, this is essentially the rank-nullity theorem from linear algebra. If we combine this with $\mathrm{dim}_K(K\otimes_R M)=\mathrm{rank}_R(M)$, the argument follows – Lukas Heger Oct 21 '23 at 18:42
  • Oh, I see, I did not know about this, I think. :) – Ben123 Oct 21 '23 at 18:50