On the kernel of a certain module epimorphism $\mathbb{Z}^2 \to \mathbb{Z}/6\mathbb{Z}$

Question

In order the construct a certain projective resolution of $\mathbb Z / 6 \mathbb Z$ I need to find the kernel of the ($\mathbb Z$-) module morphism:

$$\epsilon_0 : \mathbb Z^2 \to \mathbb Z / 6 \mathbb Z, (a,b) \mapsto 2a + 3b +6\mathbb Z$$

that is I need to find the set of all pairs of integers $(a,b)$ such that $2a + 3b\in 6 \mathbb Z$.

How do I do that? Is there a nice theorem about integral domains, divisibility and / or ideals that gives an almost immediate answer?

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I have some embarassingly large gaps in my knowledge, so this question looks easy but I don't know at all how to deal with it.

Answer 1

$2a + 3b\in 6 \mathbb Z \subseteq 3 \mathbb Z$ implies $a \in 3 \mathbb Z$.

$2a + 3b\in 6 \mathbb Z \subseteq 2 \mathbb Z$ implies $b \in 2 \mathbb Z$.

Thus, $2a + 3b\in 6 \mathbb Z $ iff $a \in 3 \mathbb Z$ and $b \in 2 \mathbb Z$ and the kernel is $ 3 \mathbb Z \times 2 \mathbb Z$.

Answer 2

When $b$ is odd $3b$ will also be odd and so adding an even number $2a$ won't make $2a+3b$ a multiple of 6.

When $b$ is even $3b$ will be a multiple of $6$ and hence, for $2a+3b$ to be multiple of $6$ we need $2a$ to be a multiple of $6$ which is the same as $a$ being a multiple of $3$.

So pairs of numbers of the form $(a,b)=(3x, 2y)$ form the kernel precisely.

Answer 3

Merely from looking at it, it’s pretty clear that the kernel should be just the submodule generated by $(3,0)$ and $(0,2)$. But what’s a rigorous argument for that and how can one proceed generally?

Well, there’s the structure theorem which shows that you are looking for a rank two submodule $M$ of $ℤ^2$ with elementary divisors $2$ and $3$ (because $ℤ/6ℤ \cong ℤ/2ℤ × ℤ/3ℤ$).

This means that you can find a $ℤ$-base $x_1, x_2$ of $ℤ^2$ such that the system $2x_1, 3x_2$ forms a basis of $M$. And, whenever you have a system in $M$ of the form $2x_1', 3x_2'$ for some basis $x_1', x_2'$ of $ℤ^2$, then look at $N = 〈2x_1', 3x_2'〉 \subseteq M$. You get $ℤ^2/N \cong ℤ/6ℤ \cong ℤ^2/M$. Now, by the third isomorphism theorem, $$ℤ^2/N \cong \frac{ℤ^2}{M} \cong \frac{ℤ^2/N}{M/N}.$$ As $ℤ^2/N$ is finite, this implies $M = N$.

This shows that it suffices to find a base $x_1, x_2$ of $ℤ^2$ such that $2x_1, 3x_2$ are in $M$. Obviously $x_1 = (0,1)$ and $x_2 = (1,0)$ will do. I think this statement ist true generally for the problem of finding a base for a submodule $M ⊂ R^n$ where $R$ is any principal domain: Just find a base of $R^n$ such that multiplication by the elementary divisors of $M$ gives a system in $M$ – it has to be a base automatically.

(Probably one can easily generalize this argument for the general situation of a principal domain and a submodule in module. I’ll think about that later.)

Answer 4

Hint:

It is well known that, if $a$ and $b$ are coprime, and $\;ua+vb=1$ is a Bézout's relation between $a$ and $b$, the solutions to the linear dipohantine equation $$xa+yb=c\quad (c\in\mathbf Z)$$ is given by $$x=uc+kb,\quad y=vc-ka\quad (k\in\mathbf Z).$$