# Can the field of Laurent series be made into an Archimedean ordered field?

In today's analysis class, my professor introduced the field of formal Laurent series $$\Bbb R((x))$$. He also talked about the dictionary order on $$\Bbb R((x))$$ and why it is not an Archimedean ordered field. One natural question arises:

Can $$\Bbb R((x))$$ be made into an Archimedean ordered field?

My professor answered that there are uncountably orders on $$\Bbb R((x))$$, but none that he knows of make it an Archimedean field. Since we know that every Archimedean field can be embedded into the real number and conversely, every subfield of $$\Bbb R$$ is Archimedean, the question transforms into:

Can $$\Bbb R((x))$$ (as a field) be embedded into $$\Bbb R$$?

The new formulation doesn't seem easier, but it makes the problem a purely algebraic one.

Thanks for nombre's answer, $$\Bbb R((x))$$ cannot be embedded into $$\Bbb R$$ for the simple reason that it contains a copy of $$\Bbb R$$. Now I wonder if $$\Bbb Q((x))$$ embeds into $$\Bbb R$$.

In general no proper extension of $$\mathbb{R}$$ can be embedded into $$\mathbb{R}$$, otherwise their would be two distinct embeddings of $$\mathbb{R}$$ into itself. (see any Archimedean ordered field embeds uniquely into $$\mathbb{R}$$)
• Thanks for your answer. I wonder then if $\Bbb Q[[x]]$ embeds into $\Bbb R$. Jan 18 at 16:40
There is no embedding $$f:\Bbb{Q}[[x]]\to \Bbb{R}$$ neither: take $$n\in \Bbb{Z}$$ such that $$f(1+nx)=1+nf(x)< 0$$ then $$f((1+nx)^{1/2})$$ is purely imaginary.
• @metaverse $(1+nx)^{1/2}=\sum_{k\ge 0} {1/2\choose k} n^k x^k$ is in $\Bbb{Q}[[x]]$ Jan 19 at 6:29