5
$\begingroup$

Suppose $d_1d_2=n$ and let $0 \to d_1\mathbb Z_n \overset {i} \to \mathbb Z_n \stackrel {d_2\cdot} \to d_2\mathbb Z_n\to 0$ be a short exact sequence. Show that sequence splits iff $\gcd(d_1,d_2)=1$.

Suppose s.e.s. split then $\mathbb Z_n=d_1\mathbb Z_n +d_2\mathbb Z_n$, hence $1=d_1a+d_2b$ hence $\gcd=1$. How to prove the converse?

$\endgroup$
1
  • $\begingroup$ Calling integers $d_i$ in the context of homological algebra is not a good idea... It took me some time to understand what was going on in your question. $\endgroup$ Commented Feb 18, 2016 at 8:39

2 Answers 2

2
$\begingroup$

If $(d_1,d_2)=1$, show that every $a \in \mathbb Z_n$ can be written as $a=xd_1+yd_2$ where $x,y\in \mathbb Z_n$ are unique modulo $d_2$ and $d_1$ respectively.

Then define $f:\mathbb Z_n \to d_1\mathbb Z_n$ by $f(xd_1+yd_2)=xd_1$. This is well defined since $x$ is unique modulo $d_2$ and $d_1d_2=n=0$. Show that is a group homomorphism and that $fi=\text{Id}_{d_1\mathbb Z_n}$.

$\endgroup$
0
$\begingroup$

Or, alternatively to Nitrogen's: you are looking for a section of the multiplication by $d_2$ map. Therefore you are looking for an element $x$ such that $$d_1x \equiv 0\pmod n$$ (because $d_2$ has order $d_1$) and such that $$d_2x \equiv d_2 \pmod n.$$ Dividing through, this is equivalent to solving $x \equiv 0 \pmod {d_2}$ and $x \equiv 1 \pmod {d_1}$, which is possible if $d_1$ and $d_2$ are coprime. In any event, $x = b d_2$, where $b$ is as above (i.e., so your section is $k\,d_2 \mapsto k\,x$, for $k \in \mathbb Z$).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .