Isomorphism of lattices

Question

This is an exercise that appears in some sheet of exercises in toric geometry. Take two lattices, $N=\mathbb{Z}^{3}$ and $N'=\operatorname{Span}_\mathbb{Z}\{(-1,-1,1), (-1,2,1), (2,-1,1)\}$. The exercise ask to show that the group $N/N'$ is isomorphic to $\mathbb{Z}/3\mathbb{Z}$. I tried to do this constructing an homomorphism beetween $N$ and $\mathbb{Z}/3\mathbb{Z}$ whith kernel $N'$ and using the first homomorphism theorem, but I didn't reach the result. Then I found the relation, between the three generators of $N'$, say $u_{1},u_{2},u_{3}$, $u_{1}+u_{2}+u_{3}=(0,0,3)$, and so I tried to show that $u_{1}$ and $u_{2}$ are part of a basis of $\mathbb{Z}^3$. Doing so, I wanted to use $N/N'=(u_{1}\mathbb{Z}\oplus u_{2}\mathbb{Z}\oplus e_{3}\mathbb{Z})/(u_{1}\mathbb{Z}\oplus u_{2}\mathbb{Z} \oplus 3e_{3}\mathbb{Z})\simeq\mathbb{Z}/3\mathbb{Z}$. But the point here is that $u_1$ and $u_2$ are not part of a basis of $\mathbb{Z}^{3}$, their wedge is proportional to $3$. Have you got any other idea? I'm sure that the result is right, because it's in an exercise. Thank you.

Answer 1

For an explicit homomorphism try $$(a, b, c) \mapsto b + c$$ This is surjective and has $N'$ in the kernel. All that's left is for you to prove that $N'$ is the kernel, which it must be for the statement to be correct.

Edit: So spin is right, the vector $(1, 0, 0)$ is in the kernel of this homomorphism but $(a, 0, 0)$ is in $N'$ if and only if $3$ divides $a$. So $N/N'$ is indeed $(\mathbb Z/3)^2$ and an explicit homomorphism giving that is, for example, $(a, b, c) \mapsto (a + c, b + c)$.

Answer 2

For all integer $a,b,c$ $$ (3a,3b,3c)=(c-a-b)(-1,-1,1)+(b+c)(-1,2,1)+(a+c)(2,-1,1), $$ so $N'\supset 3\mathbb{Z}^3$. Therefore $N/N'\cong M/M'$ where $M=\mathbb{Z}^3/3\mathbb{Z}^3\cong\mathbb{Z}_3^3$ and $M'$ is the image of $N'$ in $M'$. But the images of all vectors $(-1,-1,1), (-1,2,1), (2,-1,1)$ coincide in $M$. Hence $N/N'\cong \mathbb{Z}_3^2$.

Answer 3

The relations matrix for $N'$ is

$$\begin{pmatrix}-1 & -1 & 1 \\ -1 & 2 & 1 \\ 2 & - 1 & 1 \end{pmatrix}$$

with row and column reduction we see that this matrix has Smith normal form

$$\begin{pmatrix}1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}$$

in $\mathbb{Z}$. Hence $\mathbb{Z}^3 /N' \cong \mathbb{Z} / (1) \oplus \mathbb{Z} / (3) \oplus \mathbb{Z} / (3) = \mathbb{Z} / 3\mathbb{Z} \oplus \mathbb{Z} / 3 \mathbb{Z}$.

For details on why this works, and why it also works for any finitely generated module over a PID, see the Basic Algebra I by Jacobson. Besides the isomorphism, you also get a nice algorithm for finding generators that express a finitely generated module over a PID as a direct sum of cyclic submodules.