# Thread: A infinite dimensional linear space

1. ## A infinite dimensional linear space

Let $B = \left\{ \sqrt{x} \; | \; x = 1\text{ or } x\text{ is prime}\right\},\; S = \left\{ \sum_{k=1}^n a_k b_k \;|\; n\in N, \;a_k \in Q,\; b_k \in B \right\}$
Prove that $S$ is a linear space over $Q$ with a basis $B$

2. Originally Posted by elim
Let $B = \left\{ \sqrt{x} \; | \; x = 1\text{ or } x\text{ is prime}\right\},\; S = \left\{ \sum_{k=1}^n a_k b_k \;|\; n\in N, \;a_k \in Q,\; b_k \in B \right\}$
Prove that $S$ is a linear space over $Q$ with a basis $B$
Where is it you are stuck on this problem?

I believe proving that B is a basis is the hard bit;

Note that B forms a spanning set for your space by it's very definition. Then, $\sqrt{p} \in \mathbb{R}\setminus \mathbb{Q}$ (this is quite a nice proof to do) and so your basis intersects trivially with the rationals. This gives you the fact that your spanning set is linearly independent (why?).

To prove that it is a linear space, you can either hit it with a stick (go through the axioms in a boring way), or you should note that the real numbers form a vector space with the rationals as the field. Here you are just taking a subset of the reals as your basis, and so everything is nice.

3. we need to prove that $1,\sqrt{2},\sqrt{3},\sqrt{5},\cdots$ are linear independent over $\mathbb{Q}$.
Which is much more difficult than prove that $\sqrt{p} \in \mathbb{R} \setminus \mathbb{Q}$

4. Suffices to show any finite subset of B is l.i. Do an induction on the size of the finite subset.

5. Using induction is easier to think of, but it's not that easy for this problem.

Need induction + prove by contradiction + some sort of field expansion...

6. Originally Posted by elim
Using induction is easier to think of, but it's not that easy for this problem.

Need induction + prove by contradiction + some sort of field expansion...
Are you familiar with field automorphisms?

Look at the field $\mathbb{Q}(\sqrt{p_1}, \ldots, \sqrt{p_n})$. Then let $\phi_i$ be the automorphism such that $\sqrt{p_i}\mapsto -\sqrt{p_i}$ and $\sqrt{p_j} \mapsto \sqrt{p_j}$ for $j \neq i$. As this is an Automorphism it preserves everything. So, if these square roots are linearly dependent then what happens if you hit your sum $\alpha_1 \sqrt{p_n} + \ldots + \alpha_n \sqrt{p_n} = 0$ with one of these automorphisms?

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