I have been told that $\mathbb{Q}(\sqrt[4]{3})$ as a vector space over $\mathbb{Q}$ has dimension $4$. If $\alpha = \sqrt[4]{3}$, then I am guessing a basis is $1, \alpha, \alpha^2, \alpha^3$. I can see that these elements span all of $\mathbb{Q}(\alpha)$, but I am not sure how to show that they are linearly independent.

Now, I think I can show that any two of the elements are linearly independent, but I know this doesn't imply that all four are linearly independent.

  • 2
    $\begingroup$ $\alpha$ is a root of $p(x)=x^4-3$. Have you covered enough theory to show that $p(x)$ is irreducible over $\Bbb{Q}$? If so, then you are basically done, because a linear dependency relation here would mean that $\alpha$ is also a zero of a lower degree polynomial (and an irreducible polynomial with $\alpha$ as a zero is easily seen to be the lowest degree one). $\endgroup$ Mar 5 '19 at 18:39
  • $\begingroup$ @JyrkiLahtonen: I kinda see this ahead, but we aren't there yet. Is there a more direct way to do it? $\endgroup$
    – John Doe
    Mar 5 '19 at 18:42
  • 1
    $\begingroup$ I think any proof that $[\Bbb{Q}(\sqrt[4]{3}):\Bbb{Q}]=4$ will directly or indirectly involve proving that $X^4-3$ is the minimal polynomial of $\sqrt[4]{3}$ over $\Bbb{Q}$, and hence proving that $X^4-3$ is irreducible over $\Bbb{Q}$. $\endgroup$
    – Servaes
    Mar 5 '19 at 18:47

If the set $\{1,\alpha,\alpha^2,\alpha^3\}$ is linearly dependent over $\Bbb{Q}$, then there exist $x_0,x_1,x_2,x_3\in\Bbb{Q}$ not all zero such that $$x_0+x_1\alpha+x_2\alpha^2+x_3\alpha^3=0.$$ This means $\alpha$ is a root of the polynomial $f:=x_3X^3+x_2X^2+x_1X+x_0\in\Bbb{Q}[X]$. Because $\alpha$ is also a root of the polynomial $g:=X^4-3\in\Bbb{Q}[X]$, it is also a root of $\gcd(f,g)$. Now it suffices to show that $X^4-3$ is irreducible in $\Bbb{Q}[X]$ to reach a 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.