# How is the given set a basis for $K$?

I am studying Smith normal form from a lecture note given by our instructor on commutative algebra. Here I find some difficulty in understanding a concept. Here's this $:$

Consider an abelian group $G$ which is generated by two elements $a$ and $b$ which satisfy the relation

$$2a + 4b =0\ \mathrm {and}\ -2a+6b=0.$$

Let $e_1,e_2$ be the standard basis for $\mathbb Z^2$.

Consider the map from $\mathbb Z^2 \longrightarrow G$ given by sending $e_1$ to $a$ and $e_2$ to $b$.

This is an onto group homomorphism.

Let $K$ be the kernel. Let $y_1=2e_1+4e_2$ and $y_2 = -2e_1+6e_2$.

Then $K$ is a free $\mathbb Z$-module which has a basis $\{y_1,y_2 \}$.

This is the statement where I am struggling. My question is " Why is $\{y_1,y_2 \}$ a basis for $K$?" It is clear that $K$ is free since $\mathbb Z$ is a PID and $\mathrm {span}\ \{y_1,y_2 \} \subset K$. How do I prove the other way round? Also since $\mathbb Z$ is a PID so

$2=\mathrm {rank}\ ({\mathrm {span}\ \{y_1,y_2 \}}) \leq \mathrm {rank}\ K \leq \mathrm {rank}\ \mathbb Z^2 = 2 \implies \mathrm {rank}\ K = 2$. Also the set $\{y_1,y_2 \}$ is linearly independent. From here can I conclude that $\{y_1,y_2 \}$ is a basis for $K$? Though this reasoning doesn't hold good for general module over a PID. For instance if we take $\mathbb Z^2$ then as a $\mathbb Z$-module it has rank $2$. Though the linearly independent doubleton set $\{(0,2), (2,0) \}$ fails to be a generating set of $\mathbb Z^2$ and hence not a basis for $\mathbb Z^2$ over $\mathbb Z$ though $\mathbb Z$ is a PID.

• I think the result is not true in general. Take $G$ to be $\mathbb Z_2^2$ .Then let $a=(\bar 1,\bar 0)$ and $b= (\bar 0, \bar 1)$. Clearly $a,b$ generate $G$ and $a,b$ satisfy the given relation. In this case the kernel $K = {2 \mathbb Z}^2 \neq \mathrm {span}\ \{(2,4), (-2,6) \}$. – A.Chattopadhyay Apr 9 '18 at 5:42

EDIT. This answer has been downvoted. I suspect this is because it contains a mistake, but I don't see it, and would be most grateful if somebody could point it out to me.

[Another possible reason for the downvote is this: It seemed (and still seems) clear to me that the intended exercise was not the one stated in the question; that this intended exercise was to assume that the indicated conditions are not only satisfied by the group $G$, but was presented by them; and that this intended exercise is much more interesting than the stated exercise. So the message of the downvoter might be: "You're asked to answer the question as stated, not to speculate about any intended meaning". But I mainly would like to know if there is a mistake, and, if such is the case, what this mistake is. Thank you very much in advance!]

End of the edit.

You should add the assumption that the equalities $$\left\{ \begin{matrix} 2a&+&4b&=0\\ \\ -2a&+&6b&=0 \end{matrix} \right.$$ give a presentation of $G$, because otherwise $G=0$ is a counterexample.

If you do add this assumption, then $G$ is the cokernel of the $\mathbb Z$-linear endomap of $\mathbb Z^2$ attached to the matrix $$A:=\begin{pmatrix} 2&-2\\ 4&6 \end{pmatrix},$$ and $K$ is generated by the columns of $A$. As you said, $K$ is free. Thus it only remains to check that the determinant of $A$ doesn't vanish, which is clear.

(About your comment: Thanks for never denoting $\mathbb Z/(n)$ by $\mathbb Z_n$. By a universal consensus $\mathbb Z_n$ denotes the ring of $n$-adic integers.)

• Do you try to say that $\{G \}$ is the cokernel of the $\mathbb Z$-linear endomap of $\mathbb Z^2$? Because here image is the whole of $G$. Hence the cokernel is singleton. – A.Chattopadhyay Apr 9 '18 at 14:39
• Do you say my comment above doesn't hold good. I am bit confused at this stage.Would you please be more explicit here? – A.Chattopadhyay Apr 9 '18 at 14:45
• @A.Chattopadhyay - I'm saying that (if we make the additional assumption) $G$ is the cokernel of the indicated endomorphism of $\mathbb Z^2$ (note that there is a morphism $\mathbb Z\to G$ and another morphism $\mathbb Z^2\to\mathbb Z^2$). Any proper quotient of $G$ is a counterexample to the original statement; if you replace $\mathbb Z_2$ with $\mathbb Z/(2)$ in the comment to your question, you get such a counterexample. – Pierre-Yves Gaillard Apr 9 '18 at 15:09
• A universal consensus that $Z_n$ denotes the ring of $n$-adic integers? Universal? Really? – Gerry Myerson Apr 15 '18 at 21:36
• @GerryMyerson - My phrasing was probably awkward, but it seems to me that the use of a notation other than $\mathbb Z_n$ for the ring of $n$-adic integers is extremely rare in current mathematics. Don't you think so? The best policy is probably to state explicitly what one means by $\mathbb Z_n$ if one uses this notation (whereas notation like $\mathbb Z/n\mathbb Z$ or $\mathbb Z/(n)$ might be considered self-explanatory). I'd be happy to remove this claim form the post if you find it inappropriate. – Pierre-Yves Gaillard Apr 15 '18 at 22:28