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Consider the field extension $\mathbb{Q}(3^{1/4},i)$ over $\mathbb{Q}$. Find a basis for this feild extension over $\mathbb{Q}$.

I was thinking I would find a basis for $\mathbb{Q}(i)$ and a basis for $\mathbb{Q}(3^{1/4})$ over $\mathbb{Q}$. Then to take the pairwise product of each element in the basis to get a basis for the whole field extension. So far I've managed to prove that the set $B_{1}=\{1,i\}$ is a basis for $\mathbb{Q}(i)$.

What I'm having trouble with is showing that the set $B_{2} = \{1, 3^{1/4}, 9^{1/4}, 27^{1/4}\}$ is a basis for $\mathbb{Q}(3^{1/4})$.

Once that is show I could conclude that we have the following basis for the whole feild extension.

$$\{1, 3^{1/4}, 9^{1/4}, 27^{1/4}, i , 3^{1/4}i, 9^{1/4}i, 27^{1/4}i\}$$ Thanks in advance.

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Great idea. Now, put $\;w:=\sqrt[4]3\;$ . If there are $\;a,b,c,d\in\Bbb Q\;$ with $\;a+bw+cw^2+dw^3=0\;$ , then it must be that $\;a=b=c=d=0\;$ , otherwise $\;w\;$ is the root of a polynomial of degree$\,<4\;$, which is impossible since $\;w\;$ is a root of the irreducible $\;x^4-3\in\Bbb Q[x]\;$ .

Since $\;[\Bbb Q(w):\Bbb Q]=4\;$, the above proves your set indeed is a basis for $\;\Bbb Q(w)/\Bbb Q\;$ .

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  • $\begingroup$ What do you mean by "Since $[\mathbb{Q}(w): \mathbb{Q}]$". Should this say something like "Since $[\mathbb{Q}(w): \mathbb{Q}]=4$" $\endgroup$ – Justin Stevenson Apr 17 '18 at 18:17
  • $\begingroup$ @JustinStevenson Indeed so. Edited. $\endgroup$ – DonAntonio Apr 17 '18 at 18:32
  • $\begingroup$ Okay. Now how do we know that it's $4$? Isn't that what we would conclude after we've shown that the set really is a basis? Or is there another way to get that? $\endgroup$ – Justin Stevenson Apr 17 '18 at 18:35
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    $\begingroup$ @JustinStevenson Because its minimal polynomial over the rationals is of degree four...But I can't tell what have you studied yet and what you haven't. $\endgroup$ – DonAntonio Apr 17 '18 at 18:38
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By Eisenstein Criteria, the Polynomial $X^4-3$ is irreducible over $\mathbb Q$. Therefore, it is the minimal polynomial of $\sqrt[4]{3}$ over $\mathbb Q$.

From here it is easy to conclude that $B_2$ is a basis (this probably is a result from the textbook/class so may need no proof).

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  • $\begingroup$ We haven't quite learned this technique yet but I have seen it used quite a bit in other posts. i'm assuming we are going to learn it tomorrow. Just to clarify, does finding the minimal polynomial in general and showing its irreducible prove that a set is linearly independent. Then from there I would have to show it spans the feild extension? $\endgroup$ – Justin Stevenson Apr 17 '18 at 18:09
  • $\begingroup$ @JustinStevenson If $P(X)=X^n+a_{n-1}X^{n-1}+...+a_1X+a_0$ is the minimal polynomial of $\theta$ over $K$, it is an easy exercise to prove that $1, \theta, \theta^2,.., \theta^{n-1}$ is a basis for $K(\theta)$ over $K$. Linear independence is straightforward, as a linear combination of those giving zero is a polynomial of degree at most $n-1$ which has $\theta$ as a root, and spanning is easy : if $y \in K(\theta)$ then $y=Q(\theta)$ for some polynomial $Q$. Now, do long division of $Q(X)$ by $P(X)$, plug in $\theta$ and you are done. $\endgroup$ – N. S. Apr 17 '18 at 19:03

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