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(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.

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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 $.

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    $\begingroup$ We don't need the machinery of the Galois group: If a field $F$ contains $a:=\sqrt[3]2\omega$ and $b:=\sqrt[3]2\omega^2$, then it contains their sum. Since the irreducible polynomial of those numbers along with $\sqrt[3]2$ is $t^3-2$, we have by Vieta's relations that $a+b+\sqrt[3]2=0$, so $a+b=-\sqrt[3]2$ and $\sqrt[3]2\in F$. $\endgroup$ – Jose Brox Mar 28 at 8:49
  • $\begingroup$ Btw, in my opinion in this case it would have been better to give hints for the OP than the full answer! $\endgroup$ – Jose Brox Mar 28 at 8:50
  • $\begingroup$ @JoseBrox Yeah, hints would have been sufficient. I just got carried away. BTW, thanks for the hint;) $\endgroup$ – Thomas Shelby Mar 28 at 11:53
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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?

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