# Tag Info

### Is there a maximal ordered field? What about $\mathbb R$?

Let me give you two seemingly contradictory answers: There is no Maximal Ordered Field We can prove that there is no maximal ordered field. Based on your post, I think some of this will be new to you, ...
Accepted

1 vote

### Example of a complete, non-archimedean ordered field

The set of functions $$\mathbb{R}((\mathbb{Q})):=\{f:\mathbb{Q}\to \mathbb{R}\ |\ supp(f)\mbox{ is well-ordered}\},$$ where $supp(f):=\{x\in \mathbb{Q}\ | \ f(x)\neq 0\}$, is a field under the ...
1 vote

Upon further review, your $27$ linear polynomials reduce to three. First off, the $1$ is not under any root signs, so you can just call it $1$ without multiplying in any roots of unity. That leaves $1+... 1 vote ### Trace on Finite fields We have $$Tr_{q^r|q}(\alpha)=\alpha+\alpha^q...+\alpha^{q^{r-1}}\in \mathbb{F}_q.$$ If$\alpha=\beta^q-\beta$for some$\beta\in \mathbb{F}_{q^r}$, then$\$Tr_{q^r|q}(\alpha)=Tr_{q^r|q}(\beta)^q-Tr_{q^...

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