Splitting field/subfield is isomorphic (1)Let $h$= $\mathbb{Q}$[t]/($t^2-2$). Show that  there exists only one subfield of $\mathbb{R}$ isomorphic to $h$.
(2)Let $h$= $\mathbb{Q}$[t]/($t^3-2$). Show that there exists three(3) subfields of $\mathbb{C}$ isomorphic to $h$.
I am accustomed to doing questions like 'Find the splitting field of the following polynomials' in my graduate algebra worksheet but I came across this question and I am unable to understand how to begin. 
In (1) I know that h produces a field containing  $\mathbb{Q}$ and in which there is a square root of 2. $\mathbb{R}$ contains two distinct roots of 2 namely $\mp$ $\sqrt2$ but what can I do from here and how do I use what I know to find an 'isomorphic' mapping?
Additionally can anyone advise a book to use that uses this type of question notation? I am currently using the dummit and foote only and would like to expand my reading. 
 A: If $\alpha $ is a root of  an irreducible polynomial $p(x)$ over some field  $F $, then $F (\alpha)\cong \dfrac {F [x]}{(p (x))} $(a proof can be found in Dummit and Foote ). 
For the first case,the roots of the polynomial  $t^2-2$ are $\pm\sqrt2$. So $$\Bbb Q(\sqrt2)\cong\mathbb{Q}[t]/(t^2-2)\cong\Bbb Q(-\sqrt2). $$
But note that $\Bbb Q(\sqrt2)=\Bbb Q(-\sqrt2)$(why?). So there is only one field. 
For $\mathbb{Q}[t]/(t^3-2)$, consider the roots of the polynomial  $t^3-2$. The roots are $\sqrt [3]{2},\sqrt [3]{2}\omega, \sqrt [3]{2}\omega^2$, where $\omega$ is a non-trivial cube root of unity. By the same argument used above, we can see that $\Bbb Q(\sqrt [3]{2})\cong\Bbb Q(\sqrt [3]{2}\omega)\cong\Bbb Q(\sqrt [3]{2}\omega^2)$. Clearly $\Bbb Q(\sqrt [3]{2})\neq\Bbb Q(\sqrt [3]{2}\omega)$ and $\Bbb Q(\sqrt [3]{2})\neq\Bbb Q(\sqrt [3]{2}\omega^2)$. You can also show $\Bbb Q(\sqrt [3]{2}\omega)\neq\Bbb Q(\sqrt [3]{2}\omega^2)$ either by a proof by contradiction or by finding out the Galois group of $t^3-2$ over $\Bbb Q  $.
A: Look at the dimension of these fields over the rationals, which is invariant under isomorphisms. Also, if $f:h\rightarrow F$ is an isomorphism of fields, what can you say about $f(t)$? What about its irrreducible polynomial? How many options are there for $f(t)$? Are the different options inside the same field of the given dimension, or do they generate different fields?
