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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 $\Bbb R$.

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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 $\mathbb{R}$)

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  • $\begingroup$ Thanks for your answer. I wonder then if $\Bbb Q[[x]]$ embeds into $\Bbb R$. $\endgroup$
    – metaverse
    Jan 18 at 16:40
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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.

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    $\begingroup$ Thanks for your answer. Are you talking about the puiseux series? The field of laurent series doesn't allow fractional exponents. $\endgroup$
    – metaverse
    Jan 19 at 0:10
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    $\begingroup$ @metaverse $(1+nx)^{1/2}=\sum_{k\ge 0} {1/2\choose k} n^k x^k$ is in $\Bbb{Q}[[x]]$ $\endgroup$
    – reuns
    Jan 19 at 6:29
  • $\begingroup$ Oh, this is Newton's generalized binomial theorem, many thanks! $\endgroup$
    – metaverse
    Jan 19 at 6:40

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