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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.

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    $\begingroup$ $\alpha$ 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). $\endgroup$ – Jyrki Lahtonen Mar 5 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 at 18:42
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    $\begingroup$ 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 at 18:47
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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.

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