1
$\begingroup$

I came across this Problem in Terms of my exam preparation:

a.) Let N $\subset \mathbb{Z}^3$ be the submodule generated by the set {(2,4,1),(2,-1,1)}. Find a Basis {$f_1,f_2,f_3$} for $\mathbb{Z}^3$, and Elements $d_1,d_2,d_3 \in \mathbb{Z}$ such that the non-Zero Elements of the set {$d_1f_1,d_2f_2,d_3f_3$} form a Basis for N and $d_1 |d_2 |d_3$.

b.) Write $\mathbb{Z}^3/N$ as a direct sum of non-trivial cyclic $\mathbb{Z}$-modules.

c.) Why is there no $u \in \mathbb{Z}^3$ such that {(2,4,1),(2,-1,1),u} is a Basis of $\mathbb{Z}^3$?

I think that those Questions are related to the structure Theorem for finitely generated modules and from a similar Problem I think I was able to solve b.): The smith normal form of \begin{pmatrix} 2 & 4&1\\ 2&-1&1\end{pmatrix} is given by \begin{pmatrix} 1 & 0&0\\ 0&5&0\end{pmatrix} and therefore

$\mathbb{Z}^3/N$ is isomorphic to $\mathbb{Z}/5\mathbb{Z} \oplus \mathbb{Z}$. Is this correct? Unfortunately I fail to see WHY this solution works. Can anyone explain why this proofs that they are isomorphic (or maybe even give an explicit isomorphism?)

And then the second Question, how do I solve part a.)? I guess if I solved a.) I canuse that to find an answer to c.)

Thanks a lot for any help.

$\endgroup$
0
$\begingroup$

That is right, that is the Smith normal form, and determines the structure of the quotient group, just as you state.

To solve (a) you have to do a bit more. If $A$ is your original matrix and $S$ is the Smith normal form, then you have to find matrices $U$ and $V$ with integer entries and determinant $\pm1$ with $S=UAV$. In effect this involves keeping track of the elementary row/column operations done during the SNF computation.

Then $A=U^{-1}SV^{-1}$. Then $N$ is the row space of $A$ and is the same as the row space of $UA=SV^{-1}$. If $f_1$, $f_2$ and $f_3$ are the rows of $V^{-1}$, then they are a basis of $\Bbb Z^3$. Then $f_1$ and $5f_2$ are a basis of the row space of $SV^{-1}$, that is $N$.

$\endgroup$
  • $\begingroup$ Hello Lord Shark the Unknown. Thanks a lot for your answer. I did what you said, and it worked perfectly. I have the solution and was able to check that it is right. $f_1 = (2,4,1)$ and $f_2= (0,1,0)$. However, I still can't really see why this worked. Would you care explain a little more? I tried to compare to the homomorphisms in the answer @cat pointed out, math.stackexchange.com/questions/1441046/…, is A transposed the homomorphism g? Thanks a lot. $\endgroup$ – math_dealer Jun 7 '18 at 7:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.