Show that $a+b+c+d$ is even in the following exact sequence

Question

I'm stuck on the following question:

If $0 \rightarrow \mathbb{Z}^a \rightarrow \mathbb{Z}^b \rightarrow \mathbb{Z}^c \rightarrow \mathbb{Z}^d \rightarrow 0$ is an exact sequence, show that $a+b+c+d$ is even.

I'm sure that I'm missing some obvious trick, any help would be appreciated.

Answer 1

From the definition of exact:

• The zero dimensional image of $0$ is the kernel in the map $\mathbb{Z}^a \rightarrow \mathbb{Z}^b$.
• The $a$ dimensional image of $\mathbb{Z}^a$ is the kernel in the map $\mathbb{Z}^b \rightarrow \mathbb{Z}^c$, so the image has dimension $b-a$.
• Likewise, the image in $\mathbb{Z}^c \rightarrow \mathbb{Z}^d$ has dimension $c-(b-a)$.
• Finally, the $c-(b-a)$ dimensional image in $\mathbb{Z}^c \rightarrow \mathbb{Z}^d$ must be all of $\mathbb{Z}^d$, since the next map has all of $\mathbb{Z}^d$ as kernel.

So we have $d = c -(b-a)$. Adding $a$, $b$, and $c$ to both sides, we have \begin{align*} a+b+c+d &= a+b+c + c - (b - a) \\ &= a+b+c + c - b + a \\ &= 2a + 2c \\ &= 2(a+c) \text{,} \end{align*} so is even.

(If you are concerned about the somewhat cavalier use of the rank-nullity theorem on these free $\mathbb{Z}$-modules (in fact for any module over a PID), see Rank-nullity theorem for free $\mathbb Z$-modules.)

Answer 2

You can split the exact sequence into two short exact sequences \begin{gather} 0\to\mathbb{Z}^a\to\mathbb{Z}^b\to A\to0\\ 0\to A\to\mathbb{Z}^c\to\mathbb{Z}^d\to0 \end{gather} The second sequence tells you that $A\cong\mathbb{Z}^{c-d}$, so the first one gives $a+c-d=b$.

Can you finish?

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Actually this is very simple if you tensor with $\mathbb{Q}$, which is flat so it preserves all exact sequences (and this works for finitely generated free modules over any domain): you get the exact sequence $$ 0\to\mathbb{Q}^a\to\mathbb{Q}^b\to\mathbb{Q}^c\to\mathbb{Q}^d\to0 $$ and the statement becomes an easy one for finite dimensional vector spaces, not even requiring to know that $A$ is free (which it couldn't be over a domain which is not a PID).