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Is $\mathbb{R}$ isomorphic (as a field) to $\mathbb{R}(x)$? (where $x$ is an indeterminate on $\mathbb{R}$)

Qudit
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Kalhac
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4 Answers4

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Another argument.

In the field $\Bbb{R}$ every element $z$ or its additive inverse is a square of some element of the field. This property is preserved under an isomorphism. But the element $x\in\Bbb{R}(x)$ is not a square of any element. Neither is $-x$. Therefore we can conclude that $\Bbb{R}$ and $\Bbb{R}(x)$ are not isomorphic.

Jyrki Lahtonen
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    I felt that these two answers are sufficiently different to warrant separate posts. The other approach uses properties extrinsic to $\Bbb{R}$ (you get an algebraically closed field as a finite extension) whereas this uses intrinsic properties. In some sense the two approaches may be seen as manifestations of the same underlying fact, but I felt that the intrinsic/extrinsic dichotomy is crucial enough. – Jyrki Lahtonen Nov 19 '17 at 17:52
  • Damn - this appeared while I was typing my (identical) answer... – David C. Ullrich Nov 19 '17 at 17:57
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    A logician would say this answer uses only the first-order theory of fields, while the other does not. – David C. Ullrich Nov 19 '17 at 17:58
  • @DavidC.Ullrich It is obvious to anyone looking that we worked on this independently. FWIW I upvoted your answer. Just keep it here. You make it more explicit (and added the relevant logical category that I would not be able to). – Jyrki Lahtonen Nov 19 '17 at 17:58
  • Right. Thanks but I wasn't worried about being accused of plagiarism, just irked that someone else gave "my" answer. Heh... – David C. Ullrich Nov 19 '17 at 18:00
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No. By adjoining to $\Bbb{R}$ a root of the equation $T^2=-1$ we get an algebraically closed field. If $\Bbb{R}$ and $\Bbb{R}(x)$ were isomorphic, the same would happen by adjoining a root of $T^2=-1$ to $\Bbb{R}(x)$ (observe that any isomorphism maps $-1$ to itself). But $\Bbb{R}(x)[i]\simeq\Bbb{C}(x)$ is not algebraically closed, so this is not the case.

Jyrki Lahtonen
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$\mathbb R$ satisfies the condition $\forall t\exists s(t=s^2\lor -t=s^2)$, but $\mathbb R[x]$ does not (consider $t=x$). So they're not isomorphic (in fact they're not even "elementarily equivalent").

David C. Ullrich
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This question raise from another question: does there exist a field $k$ such that $k$ is isomorphic to $k$(x)? The answer to this question seems to be yes. It is enough to consider a generic field $F$ and let $k = F(x_i : i \in \mathbb{N})$. Then we have the following field isomorphism $$ \varphi:k(x) \rightarrow k \quad \text{such that } \varphi(x) = x_0, \: \varphi(x_i) = x_{i+1},\; \varphi_{|k} = 1_k \text{ (the identity on $k$)}.$$

Thus every field $k$ of the type $F(x_i : i \in I)$ where $I$ is a set of indices at least countable is isomorphic to $k(x)$.

By the last statement we can view $\mathbb{R} = \mathbb{Q}(x_i: i \in I)$ where $\{ x_i : i \in I \}$ is a basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space which must be at least countable. Therefore $\mathbb{R}$ is isomorphic to $\mathbb{R}(x)$.

Kalhac
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    Your argument, indeed, proves that there exists a field $k$ such that $k\simeq k(x)$. But, it doesn't work for the reals. The elements of a $\Bbb{Q}$-basis of $\Bbb{R}$ satisfy certain algebraic equations, and those are not preserved, when you extend the basis and permute them. It is absolutely crucial that the variables $x_i, i\in\Bbb{N}$ are algebraically independent. FWIW I refrain from voting, because the first paragraph describes a useful idea, but it is misapplied to the specific question. – Jyrki Lahtonen Nov 19 '17 at 17:43
  • @JyrkiLahtonen: What about the transcendentals? They certainly are algebraically independent from any algebraic number, and I would expect that you can find a sequence of them that are mutually algebraic independent. – celtschk Nov 19 '17 at 20:38
  • @celtschk: Yes, you can find a transcendence basis $S$ (= a maximal algebraically independent set). But the extension $\Bbb{Q}(S)$ they generate will be a proper subfield of $\Bbb{R}$. For example, it will be missing all the non-rational real algebraic integers. And, it doesn't help if you use $F(S)$ with $F=\overline{\Bbb{Q}}\cap\Bbb{R}$ instead of $\Bbb{Q}(S)$. The field $F(S)$ won't be all of $\Bbb{R}$ either. For example the square roots of the absolute values of elements of $S$ will be missing. – Jyrki Lahtonen Nov 19 '17 at 20:47
  • What we need is a basis of $\mathbb{R}$ (as a $\mathbb{Q}$-vector space) that admit a subset at least countable of mutually algebraic independent elements, in order to permute this subset. Is that correct? – Kalhac Nov 20 '17 at 18:55
  • @Kalhac We have that. You can start by selecting a transcendence basis. You can permute those at will to get automorphisms. But, such permutations won't give rise to automorphisms of $\Bbb{R}$. The group $Aut_{\text{Field}}(\Bbb{R})$ is trivial. The identity is the only field automorphism of $\Bbb{R}$. The explanation is that extending a transcendence basis to a generating set introduces algebraic dependencies, and that kills the possibility to permute. – Jyrki Lahtonen Nov 20 '17 at 19:39
  • If you haven't seen the argument proving that the reals have no non-trivial (field) automorphisms, take a peek at this. I'm sure the argument appeared here earlier, but that's the one I can find fast :-) – Jyrki Lahtonen Nov 20 '17 at 19:42