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