Vector space isomorphic fields $\Bbb R(X)\simeq\Bbb R(X^2)$ but $[\Bbb R(X):\Bbb R(X^2)]\neq [\Bbb R(X):\Bbb R(X)]$. Is this correct? I thought the dimension of a vector space remains the same if I replace the field by an isomorphic one? This confuses me to a great extent. Please assist me. Thanks
 A: As vector spaces over ${\mathbb R}$, ${\mathbb R}(X)$ and ${\mathbb R}(X^2)$ are both infinite dimensional. The index of a field extension corresponds to the quotient of dimensions (over the base field ${\mathbb R}$) only if the dimensions are finite.
Or maybe your confusion lies elsewhere. Suppose you have a vector space $V$ over some field $k$ and an isomorphism $\phi \colon k' \to k$ of fields. Then $V$ is not automatically a vector space over $k'$ (as you do not have a scalar multiplication by elements of $k'$), but you can easily make it into one by taking the scalar multiplication to be $\alpha' \cdot v = \phi(\alpha') v$. Let's call this vector space over $k'$ (with the same additive group structure as $V$ but with a different scalar multiplication) $V'$.
Now see what happens in your case. ${\mathbb R}(X)$ is a 2 dimensional vectorspace over ${\mathbb R}(X^2)$. There is a field isomorphism $\phi \colon {\mathbb R}(X) \to {\mathbb R}(X^2)$ given by $\alpha(X) \mapsto \alpha(X^2)$. This results in a new vector space ${\mathbb R}(X)'$ over ${\mathbb R}(X)$ in which the scalar multiplication is defined as $\alpha(X) \cdot g(X) = \alpha(X^2) g(X)$. This vector space ${\mathbb R}(X)'$ (which has the same additive group structure as ${\mathbb R}(X)$, but a different scalar multiplcation) has dimension 2 over ${\mathbb R}(X)$; ${\mathbb R}(X)$ with its normal scalar multiplication has dimension 1 over ${\mathbb R}(X)$, of course.
