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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.

user26857
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sam10269
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2 Answers2

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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.)

Eric Towers
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  • PID is not required – egreg Mar 23 '18 at 13:22
  • @egreg : I agree. The link I give is not more general than claimed. – Eric Towers Mar 23 '18 at 14:38
  • @egreg Let $R$ be the endomorphism ring of an infinite-dimensional vector space. Then, per this answer of yours (it's a bit funny :p I wasn't looking for this specifically), you have $R \cong R \oplus R$ as left modules. So you can build an exact sequence $0 \to R^2 \to R^1 \to R^0 \to R^0 \to 0$. So you at least need commutative. – Najib Idrissi Mar 23 '18 at 14:44
  • @NajibIdrissi We're working over a domain. – egreg Mar 23 '18 at 14:54
  • @egreg Says who...? OP is working over $\mathbb{Z}$, if you want to state a generalization you need to state hypotheses. You never said that "domain" was one of them, just that "PID" wasn't. Maybe it was clear for you that "domain" is required, but let's say I'm a bit slow and I prefer to give examples to future readers to make sure they don't make the same mistake of thinking the answer could be in full generality. – Najib Idrissi Mar 23 '18 at 15:17
  • @NajibIdrissi The comment was directed to Eric, specifically targeted to “any module over a PID”. The generalization is in my answer. Don't read too much in short comments. – egreg Mar 23 '18 at 15:23
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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?

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).

egreg
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