This is a well known lemma that consistently appears in textbooks, either as a statement without proof, or as an exercise (see for example pp. 146 of Hatcher)

If $0 \stackrel{id}{\to} A \stackrel{f}{\to} B \stackrel{g}{\to} C\stackrel{h}{\to} 0$ is a short exact sequence of finitely generated abelian groups, then $\operatorname{rank} B = \operatorname{rank} A + \operatorname{rank} C$.

I've been trying to prove this unsuccessfully.

What do we know? $f$ is injective, $g$ is surjective, $\mathrm{Im} f = \mathrm{ker} g$, $\mathrm{Im} g = \mathrm{ker} h$, $C\simeq B/A$

So I start with a maximally linearly independent subset $\{ a_\alpha \}$ of $A$ such that the sum (with only finite non-zero entries) $$\sum n_\alpha a_\alpha=0$$ for $n_\alpha \in \mathbb{Z}$, implies that $n_\alpha=0$.

Where to go from here is a puzzle? Any hints would be appreciated


Pick a maximal linearly independent subset $\{c_{\beta}\}$ of $C$. Now push the $a_{\alpha}$ to $B$ using $f$, and for each $c_{\beta}$ pick $c'_{\beta}\in B$ such that $g(c'_{\beta}) = c_{\beta}$.

Now suppose that you have a finite linear combination of the $a_{\alpha}$ and the $c'_{\beta}$ that is equal to $0$, $$n_{\alpha_1}f(a_{\alpha_1}) + \cdots + n_{\alpha_k}f(a_{\alpha_k}) + m_{\beta_1}c'_{\beta_1} + \cdots + m_{\beta_{\ell}}c'_{\beta_{\ell}} = 0.$$ Use $g$ to get a conclusion about the $m_{\beta_j}$; then use $f$ to get a conclusion about the $n_{\alpha_i}$. This will give you that $\mathrm{rank}(B)\geq \mathrm{rank}(A)+\mathrm{rank}(C)$.

Do you also need help with the converse inequality?

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  • $\begingroup$ @Arturo - thank you, I will have a look and see if I can work it out from here $\endgroup$ – Juan S Feb 28 '11 at 5:32
  • $\begingroup$ @Arturo - are we taking a finite linear combination of all the $a_\alpha$'s and $c'_\beta$'s? (i.e in your formula the rank of $A$ is $k$ and the rank of $C$ is t? $\endgroup$ – Juan S Feb 28 '11 at 6:20
  • $\begingroup$ @Qwirk: Linear combinations, by definition, are finite (or have only finitely many nonzero coefficients). So I'm not assuming that the ranks are finite, but you only consider finite linear combinations in any case, just as in linear algebra. $\endgroup$ – Arturo Magidin Feb 28 '11 at 14:05
  • $\begingroup$ @Arturo - sorry, maybe my question wasn't clear. Say that $A$ has the maximum independent subset $a_{\alpha 1}, \cdots a_{\alpha_k}$, then when we form the linear combination do we use $a_{\alpha 1}, \cdots a_{\alpha_k}$, or $a_{\alpha 1}, \cdots a_{\alpha_i}$ for some $i \le k$? (obviously we need the $c'_{\beta}$ terms as well) $\endgroup$ – Juan S Feb 28 '11 at 20:22
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    $\begingroup$ @Fahad: The standard way is to tensor with $\mathbb{Q}$: the argument above shows that the tensored sequence is exact, and using the fact that the tensor product distributes over the direct sum shows that the rank of a finitely generated abelian group $A$ is equal to the dimension of the $\mathbb{Q}$-vector space $A\otimes\mathbb{Q}$. Alternatively, show that no subgroup of a group of rank $k$ can have rank strictly larger than $k$, and show that a linearly independent subset of $B$ of size greater than rank A + rank C would lead to a lin. ind. subset of A of size larger than rank A. $\endgroup$ – Arturo Magidin May 23 '11 at 19:13

If you're not familiar with tensor products, then learning them for this problem is overkill. But if you are familiar with them, then I think the following solution is nice. $\newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}}$

For a finitely generated abelian group $A$, note that $\mathrm{rank}(A) = \dim_\Q(A\otimes_\Z \Q)$. One can show (basically following Arturo's answer) that given a short exact sequence of abelian groups $0\to A\to B\to C\to 0$, the sequence obtained by tensoring with $\Q$ is also exact: $$0\to A\otimes_\Z \Q \to B\otimes_\Z \Q \to C\otimes_\Z \Q \to 0$$

Now the statement about ranks boils down to an easy statement about how dimensions of vector spaces add in a short exact sequence.

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    $\begingroup$ thanks. I knew there was an answer involving tensor products, but I am not really comfortable with them yet, so I will accept Arturo's answer $\endgroup$ – Juan S Feb 28 '11 at 5:32
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    $\begingroup$ @Qwirk: In this case, it is all fairly straightforward because you can "clear denominators" and get an expression involving only integer coefficients. $\endgroup$ – Arturo Magidin Feb 28 '11 at 5:34

This holds for integral domains $R$ and finitely generated $R$-modules. If we put $A = M_1, B = M_2,$ and $C = M_3$, then the given short exact sequence is isomorphic to the following short exact sequence: $$ 0 \to M_1 \to M_2 \to M_2/M_1 \to 0. $$ If $r_1 = \operatorname{rank}M_1$, $r_2 = \operatorname{rank} M_2$, and $r_3 = \operatorname{rank} M_3 = \operatorname{rank} M_2/M_1$, then let $u_1,\dots,u_{r_1}$ be linearly independent in $M_1$ and $\bar v_1,\dots,\bar v_{r_3}$ be linearly independent in $M_2/M_1$. Then, if $$ a_1u_1 + \dotsb + a_{r_1}u_{r_1} + b_1v_1 + \dotsb + b_{r_3}v_{r_3} = 0, $$ in $M$, then reducing this equation modulo $M_1$ gives $$ b_1\bar v_1 + \dotsb + b_{r_3}\bar v_{r_3} = \bar 0\quad\text{in $M_2/M_1$}, $$ so $b_1 = \dots = b_{r_3} = 0$ by linear independence of these elements, and thus $a_1 = \dotsb = a_{r_1} = 0$ by linear independence of the $u_i$. Hence $\operatorname{rank} M_2 \geqslant r_1 + r_3$.

Suppose that $w_1,\dots,w_s$ are linearly independent in $M_2$ with $s > r_1 + r_3$. Now (exercise) there is a free submodule $N/M_1\subset M_2/M_1$ with $R^{r_3}\cong N/M_1$ such that $(M_2/M_1)\big/(N/M_1)\cong M_2/N$ is a torsion module.

Since $N/M_1$ is free of rank $r_3$, there is $0\leqslant n\leqslant r_3$ such that $n$ of the $\bar w_1,\dots,\bar w_s$ are nonzero in $M_2/M_1$ and belong to $N/M_1$. After reindexing, assume that $\bar w_1,\dots,\bar w_{s-n}$ are either zero in $M_2/M_1$ or belong to $(M_2/M_1)\smallsetminus(N/M_1)$. Since $0\leqslant n\leqslant r_3$, we have $s-n\geqslant s-r_3 > r_1$. Removing the elements in the list that are equal to $\bar 0$ in $M_2/M_1$ and reindexing, we obtain a new list $\bar w_1,\dots,\bar w_j$ with $0\leqslant j \leqslant s-n$ and such that each $\bar w_j$ is nonzero in $M_2/M_1$, and even nonzero in $T:=(M_2/M_1)\big/(N/M_1)$. Since $T$ is a torsion module, for each $j$, there is an $x_j\in R\smallsetminus\{0\}$ such that $x_j\bar w_j = \bar 0 \bmod N/M_1$ in $T$. Thus, we end up with $s-n > r_1$ elements: $$x_1\bar w_1=\dots=x_j\bar w_j=\bar w_{j+1}=\dots=\bar w_{j+(s-n-j)}=\bar w_{s-n}=\bar 0\quad\text{in $N/M_1$}. $$

Thus, we have $x_1w_1,\dots,x_jw_j,w_{j+1},\dots,w_{s-n}$ belong to $M_1$, and it is easy to see that these elements are linearly independent in $M_1$ since $R$ is an integral domain and the elements $w_1,\dots,w_j,w_{j+1},\dots,w_{s-n}$ are linearly independent by assumption. But the rank of $M_1$ is $r_1 < s-n$, so this is a contradiction. Consequently, $\operatorname{rank} M_2 \leqslant r_1 + r_3$, so $$ \operatorname{rank} M_2 = r_1 + r_3 = \operatorname{rank} M_1 + \operatorname{rank} M_3, $$ as desired.

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