# 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.

## 2 Answers

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 $$\sqrt {2},\sqrt {2}\omega, \sqrt {2}\omega^2$$, where $$\omega$$ is a non-trivial cube root of unity. By the same argument used above, we can see that $$\Bbb Q(\sqrt {2})\cong\Bbb Q(\sqrt {2}\omega)\cong\Bbb Q(\sqrt {2}\omega^2)$$. Clearly $$\Bbb Q(\sqrt {2})\neq\Bbb Q(\sqrt {2}\omega)$$ and $$\Bbb Q(\sqrt {2})\neq\Bbb Q(\sqrt {2}\omega^2)$$. You can also show $$\Bbb Q(\sqrt {2}\omega)\neq\Bbb Q(\sqrt {2}\omega^2)$$ either by a proof by contradiction or by finding out the Galois group of $$t^3-2$$ over $$\Bbb Q$$.

• We don't need the machinery of the Galois group: If a field $F$ contains $a:=\sqrt2\omega$ and $b:=\sqrt2\omega^2$, then it contains their sum. Since the irreducible polynomial of those numbers along with $\sqrt2$ is $t^3-2$, we have by Vieta's relations that $a+b+\sqrt2=0$, so $a+b=-\sqrt2$ and $\sqrt2\in F$. – Jose Brox Mar 28 at 8:49
• Btw, in my opinion in this case it would have been better to give hints for the OP than the full answer! – Jose Brox Mar 28 at 8:50
• @JoseBrox Yeah, hints would have been sufficient. I just got carried away. BTW, thanks for the hint;) – Thomas Shelby Mar 28 at 11:53

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?