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I wanted to find the degree of extension from $\mathbb{Q}$ to $\mathbb{Q}(i, \sqrt[4]2)$. The minimal polynomial of $\sqrt[4]2$ over $\mathbb{Q}$ is $f(x)=x^4-2$. Now the roots of $f(x)$ are $\{\sqrt[4]2, -\sqrt[4]2, \sqrt[4]2i, -\sqrt[4]2i\}$. So we can observe $\frac{\sqrt[4]2i}{\sqrt[4]2}=i$ so $i \in \mathbb{Q}(\sqrt[4]2)$ so we can already see the degree of the extension from $\mathbb{Q}$ to $\mathbb{Q(i, \sqrt[4]2)}$ is 4. But also there is another approach to this problem:

Since we know $[\mathbb{Q}(\sqrt[4]2):\mathbb{Q}]$=4, then we can say $\mathbb{Q}(\sqrt[4]2)=\{a+b\sqrt[4]2+c\sqrt[4]2^2+d\sqrt[4]2^3 | a,b,c,d \in\mathbb{Q}\}$. Now I am not sure if my understanding is correct from here or not, but I think we should be able to express $i$ as an element of this linear combination. I have tried a few variations and I cannot seem to show this is possible. There is no negative under the radical sign. Is there something wrong with this second approach or can we express $i$ as a linear combination of the basis?

Steven Stadnicki
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Josh
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    All elements of $\mathbb{Q}(\sqrt[4]{2})$ are real, so $i$ can't be a member of it. It's correct to say that $\mathbb{Q}[x]/\langle f(x)\rangle\equiv \mathbb{Q}(\sqrt[4]{2})$, but that doesn't imply that all roots of $f()$ are members of the extension. – Steven Stadnicki Feb 21 '21 at 19:29
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    "Now the roots of $f(x)$ are ${\sqrt[4]2, -\sqrt[4]2, \sqrt[4]2i, -\sqrt[4]2i}$." This is where your misunderstanding lies, and here is how to fix your thinking: when you say "the roots of $f$ are...", you should stop and ask yourself "the roots of $f$ in which field?" For example, $f$ has no roots in $\mathbb{Q},$ that's the whole point; but $f$ does have roots in other fields. The roots depend not just on $f$ but also on the field being considered. Your confusion stems from a habit of thinking of everything as living in $\mathbb{C}$, which is not the case. Lose this habit. – Will R Feb 21 '21 at 19:36

2 Answers2

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I don't understand how did you get to $i\in\mathbb{Q}(\sqrt[4]{2})$. It's not true that all the roots of the minimal polynomial of $\sqrt[4]{2}$ must belong to this field. Actually, in this example it is clearly false. By definition, $\mathbb{Q}(\sqrt[4]{2})$ is the intersection of all subfields of $\mathbb{C}$ which contain the element $\sqrt[4]{2}$. In particular, $\mathbb{R}$ is such a field, so it is one of the fields in the intersection. This means $\mathbb{Q}(\sqrt[4]{2})$ is a subfield of $\mathbb{R}$, and so $i$ can't be an element of this field. This tells us that we must have $[\mathbb{Q}({i,\sqrt[4]{2})}:\mathbb{Q}(\sqrt[4]{2})]=2$, and so by the tower rule the extension degree you are looking for is $8$.

Mark
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  • I was wrongly under the impression that the roots must belong to this field. In this case, I believe to find the appropriate minimal polynomial, it would not be correct to simply multiply $x^4-2$ and $x^2+1$ since the degree would be 6. Is there anything that can be said about these two minimal polynomials that can be informative for finding the minimal polynomial of $\mathbb{Q}(i, \sqrt4)$ over $\mathbb{Q}$? If needed, I can post another question about this. – Josh Feb 21 '21 at 19:40
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    @Josh There is no such thing as the minimal polynomial of a field. Every specific algebraic element has a minimal polynomial, not the whole field. It's just that specifically, when $\alpha$ is algebraic over $F$ then $[F(\alpha):F]$ is the degree of the minimal polynomial of $\alpha$. But here our field extension $\mathbb{Q}(i,\sqrt[4]{2})/\mathbb{Q}$ is generated by $2$ elements, so in order to find the extension degree we use the tower rule. – Mark Feb 21 '21 at 19:44
  • I see your point. Then maybe an appropriate question would be to find the minimal polynomial of $i+\sqrt[4]2$. I think $i+\sqrt[4]2 \in \mathbb{Q}(i, \sqrt[4]2)$ and $i+\sqrt[4]2$ is not in $\mathbb{Q}(\sqrt[4]2)$ or $\mathbb{Q}(i)$. Would it be fair to say that the minimal polynomials of $\mathbb{Q}(\sqrt[4]2)$ and $\mathbb{Q}(i)$ are not giving any information about the minimal polynomial of $i+\sqrt[4]2$? – Josh Feb 21 '21 at 20:00
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    Yes, that's how it is. In general it is difficult to find the minimal polynomial of a sum or a product. Actually, from the definition itself it is even far from obvious that the algebraic elements over $F$ form a field - so this fact is usually proved using equivalent definitions. – Mark Feb 21 '21 at 20:18
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    Thank you for clarifying my misunderstanding and providing additional information! – Josh Feb 21 '21 at 20:21
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$i \in \Bbb Q(\sqrt[4]{2})$ is not correct. Indeed, $\Bbb Q(\sqrt[4]{2})$ is completely contained in $\Bbb R$.
Note that this extension is not a normal extension, it does not contain all the roots of the minimal polynomial. This is why your analysis fails.

Since $i \notin \Bbb Q(\sqrt[4]{2})$, you can conclude that $[\Bbb Q(\sqrt[4]{2})(i):\Bbb Q(\sqrt[4]{2})] \ge 2.$ On the other hand, the polynomial $x^2 + 1 \in \Bbb Q(\sqrt[4]{2})[x]$ shows that the degree is exactly $2$.

Since the degrees multiply, we see that the extension $\Bbb Q(\sqrt[4]{2})(i) = \Bbb Q(\sqrt[4]{2}, i) \supset \Bbb Q$ is of degree $\boxed{8}$.

Aryaman Maithani
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