If I was trying to construct a splitting field for $(x^2-3)(x^2-5)$ over $\Bbb Q(\sqrt{2})$.

Then obviuously I would begin by checking if $\sqrt3$ was an element of $\Bbb Q(\sqrt{2})$.

To do this I would assume it is get the equation


and show that this is impossible for the three cases $a=0,b=0,ab=0$

And so adjoin this element to get

$$\Bbb Q(\sqrt{3},\sqrt{2}):=\{a+b\sqrt2+c\sqrt3+\sqrt{6}|a,b,c,d \in \Bbb Q\}$$

But proving that $\sqrt{5}\notin \Bbb Q(\sqrt{3},\sqrt{2})$ in the same manner as above , although rather straightforward, is messy and time consuming. Therefore in an exam setting I would like to avoid this method.

Is there a more elegant method to show that $\sqrt{5}\notin \Bbb Q(\sqrt{3},\sqrt{2})$ ?

  • $\begingroup$ One way is to show that the non-trivial subfields of $\mathbb Q(\sqrt3, \sqrt 2)$ are $\mathbb Q(\sqrt3)$, $\mathbb Q(\sqrt6)$ and $\mathbb Q(\sqrt 2)$. $\endgroup$ – Hw Chu May 20 '18 at 1:25
  • $\begingroup$ @HwChu oh that seems like a nice way , let me see if I understand what you mean though ? the basis for the extension is $\{1,\sqrt{3},\sqrt{2},\sqrt{6}\}$ as these form a basis they are linearly independent from one another and so each one forms a subfield of $\Bbb Q(\sqrt{2},\sqrt{3})$. Then we can show individually by the method I used to show that $\sqrt{3}\notin \Bbb Q \sqrt{2}$, that $\sqrt{5}$ is not an element of any of them ? I feel like perhaps my reasoning for why they constitute subfields is abit vague though , if you feel the same I would love you to elaborate :) $\endgroup$ – excalibirr May 20 '18 at 1:33
  • $\begingroup$ My feeling is that it is not sufficient. For instance, there are some arguments missing to ensure you will not get another field (why is $\mathbb Q(\sqrt2 + \sqrt 3)$ not a proper subfield of $\mathbb Q(\sqrt2, \sqrt 3)$? ). I was thinking of some arguments based on Galois theory ($Gal(\mathbb Q(\sqrt2, \sqrt 3)/\mathbb Q)\cong (\mathbb Z/2\mathbb Z)^2$), but I am not sure you know (or want to use) them $\endgroup$ – Hw Chu May 20 '18 at 1:40
  • $\begingroup$ @HwChu NoI know very little about Galois theory , its an optional course next year. Is there an way to show that these are the only proper subfields of $\Bbb Q(\sqrt{2},\sqrt{3})$ employing just the knowledge one would learn in Rings and fields 1 and 2 ? (I also took a course on group theory if that helps at all as, i know its connect to Galois theory , but I doubt it will ) $\endgroup$ – excalibirr May 20 '18 at 1:45
  • $\begingroup$ @HwChu alternatively do you think in an exam setting that they would expect you to rigourously prove that these are the only proper subfields (If the question only wanted you to find a splitting field )or do you think that it could be taken as a fact and used without proof ? $\endgroup$ – excalibirr May 20 '18 at 1:47

Seems like you do not have a complete toolkit yet, and I guess proving from the definition is probably the only hope. But let me try to make it look less painful.

If $\sqrt 5 \in \mathbb Q[1, \sqrt2, \sqrt3, \sqrt6]$, so is $\sqrt{\alpha}$ for $\alpha \in \{5,10,15,30\}$. Thus, assume

$$ \sqrt\alpha = a + b\sqrt2 + c\sqrt3 + d\sqrt 6, $$

where $a,b,c,d \in \mathbb Q$ and $a \neq 0$. Since $\sqrt\alpha^2 \in \mathbb Q$, by comparing terms we have

$$ \begin{aligned} 2bc + 2ad &= 0;\\ 2ab + 6cd &= 0;\\ 2ac + 4bd &= 0. \end{aligned} $$

From the first equation $d = -\frac{bc}{a}$. Plug into the second, $b(a^2 - 3c^2)=0$. Since $a \neq 0$, $b = 0$, so $d =0$, so $c = 0$, so $\sqrt \alpha \in \mathbb Q$, contradiction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.