This exercise is part 3 of some connected exercises. Important for my question will be several things:
$P[x]$ is the set of all polynomials $\sum_{i=0}^{N}a_ix^{i}, N \in \mathbb{N} \cup \{0\}, a_i \in \mathbb{Z}, a_N \neq 0$ for all $N > 0$
$P_+[x]$ is the set of all polynomials $\sum_{i=0}^{N}a_ix^{i}, a_N >0$.
The relation $\sim$ is defined on $P[x] \times P_+[x]$ by:
$$(p,q) \sim (\hat{p}, \hat{q}) \iff p \hat{q} = \hat{p} q$$ It is an equivalence relation. (already shown)
We want the set $$K=\{[(p,q)]_\sim : (p,q) \in P[x] \times P_+[x]\}$$ to become a field, therefore we define addition and multiplication just like in $\mathbb{Q}$. $$(p,q) + (p^{'}, q^{'}) = (pq^{'} + p^{'}q,qq^{'})$$
It truly is a field. (already shown)
Now for the third and final part of the question:
Let $K_+ = \{[(p,q)]_\sim \in K : (p,q) \in P_+ \times P_+\}$. Note the x is missing here from $P_+[x]$. I don't know if this is deliberate and x has been chosen some particular value (which field should i choose from, $\mathbb{R}$?) or if it's just missing.
Show that with this $K_+$, $K$ becomes an ordered field (already shown), which does not have the archimedian property. (not yet shown)
Using the following definition: K has the archimedian property if $\mathbb{N}$ as a subset of K does not have any upper bounds.
So my first question: how are the natural numbers even a subset here? $(\mathbb{N} \times \mathbb{N})?$
I know I have an ordered field, but these things, I think, are still not numbers because x has not been chosen. How can I compare them then? Should I choose this x? Same with the N from above in the sum. As far as I see it, this N can be anything, so there are polynomials of every degree in my field. Or should this N be fixed too?
What I think I need to do is, after being clear on the entries in these fields (which I am not), I need to find an $N \in \mathbb{N}$ for which $\forall x \in K : x \leq N,$ meaning $ N + (-x) \in K_+$. But this N cannot simply be in $\mathbb{N}$ and all of that leaves me confused so I turn to you.
