Linear independence over $\Bbb Z$ to prove an abstract algebra problem

Question

Show that there exists a subgroup of $\Bbb R$ that is isomorphic to $\Bbb ℤ \times \Bbb ℤ$.

I am instructed to consider "linear independece over $\Bbb Z$" in this problem, but I have no idea what this means? What I thought was that I needed to find a homomorphism $\eta: \Bbb Z \times \Bbb Z \to \Bbb R$ for which $\ker(\eta)=0$ so that $\Bbb Z \times \Bbb Z /\ker(\eta) \cong \Bbb Z \times \Bbb Z \cong \text{Im}(\eta)$.

Apparently this kind of homomorphism can be fonud using the "linear independece over $\Bbb Z$". Could I have a explanation on what does this mean in practice?

Answer 1

Well lets look at the structure of homomorphisms (they are in a sense just like linear transformations)

(see that: $<a, 0> = \sum_{n=1}^{a} <1, 0> $)

$$ (<a,b>) = (<a,0>) + (<0,b>) = (\sum_{n=1}^{a}<1,0> )+ (\sum_{n=1}^{b}<0,1> ) $$ $$ =\sum_{n=1}^{a} (<1,0>) + \sum_{n=1}^{b} (<0,1>)$$ $$ = a*(<1,0>) + b*(<0,1>)$$

lets say (<1,0>) = $c_1$ and (<0,1>) = $c_2$ ($c_1,c_2 \in ℝ$)

So we know $(<a,b>) = a c_1 + b c_2$ (Notice this as a linear combination -- you may even want to try and describe this as a matrix)

Now for such a homomorphism to have a kernel (well one that is not the trivial group) $$ac_1 + bc_2 = 0$$ $$ac_1 = -bc_2$$ $$c_1 = \frac{-b}{a}c_2$$ this shows us that $c_1,c_2$ can be made from rational multiples of each other (they are in a sense linearly dependent)

Now if we pick $c_1, c_2$ such that they are not rational multiples of each other (in that sense linearly independent) then no kernel will exist.

This is quite easy as an irrational number ($c_1$) can not be a rational multiple of a rational number ($c_2$)

so if $c_1 \not \in \mathbb{Q}, c_2 \in \mathbb{Q}$

then $(<a,b>) = ac_1 + bc_2$ ($a, b \in \mathbb{Z}$) will have $\ker() = \{0\}$

This idea of linear independence and vector spaces with the real numbers (and any other field for that matter) is a core idea of field extensions and very worth looking into.

A good next step is to generalize for $:\mathbb{Z}^{n}→\mathbb{R}$, which leads to a very important idea about a certain higher idea of numbers being linearly dependent (Think the difference between $\sqrt{2}$ and $\pi$).

I hope I was able to help some.

Answer 2

Linear independence over $\mathbb{Z}$ is really the same as linear independence over $\mathbb{Q}$, at least in this context: let $\alpha,\beta\in \mathbb{R}$, and suppose $m\alpha + n\beta = 0$ for some integers $m,n$ implies $m=n=0$, meaning we have linear independence over $\mathbb{Z}$. Suppose next $x\alpha + y\beta = 0$ for some rationals $x,y$: then we may multiply by a common denominator $d$ such that $dx,dy$ are integers, and obtain $dx \alpha + dy \beta = 0$, which implies $dx=dy=0$ by the assumption of $\mathbb{Z}$-linear independence, and therefore $x=y=0$. (The converse is trivial.)

So the question becomes: what does it mean for the images $\eta((1,0))$ and $\eta((0,1))$ to be $\mathbb{Q}$-linearly independent? Hint: it has something to do with irrationality.

Answer 3

Observe that e.g. the real numbers $1$ and $\sqrt{2}$ are linearly independent over $\mathbb{Z}$, in the sense that there don't exist $c_1,c_2 \in \mathbb{Z}$, not both zero, such that $c_1 \cdot 1 + c_2 \cdot \sqrt{2} = 0$ (because $\sqrt{2}$ is irrational). This is a standard usage of the term 'linearly independent', even though of course you can't have a vector space over $\mathbb{Z}$, but the analogy with linear independence in vector spaces is obvious.

Anyway, this precisely means that the homomorphism $(m,n) \mapsto m + n\sqrt{2}$ has zero kernel, which is what you need.