# Showing that an element is not in a field extension

1. May 2, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Let $K = \mathbb{Q} (1, a_1, a_2, \dots, a_n)$, which is the smallest field containing $\mathbb{Q}$ and $a_1, a_2, \dots, a_n$, where each $a_i$ is the square root of a rational
number. Show that the cube root of 2 is not an element of K.

2. Relevant equations

3. The attempt at a solution
I need some pointers. I am thinking about making an argument with degrees. That is, looking at the degree of $K$ over $\mathbb{Q}$ and seeing how that relates to the degree of $\mathbb{Q}(2^{1/3})$ over $\mathbb{Q}$, such as whether the latter is a divisor of the former.

2. May 2, 2017

### Staff: Mentor

Sounds good. Alternatively, by foot so to say, you could prove it by contradiction. Assume $\sqrt[3]{2} \in \mathbb{Q}(a_1, \ldots ,a_n)$. This means, there is a polynomial $f(x_1,\ldots , x_n) \in \mathbb{Q}[x_1,\ldots , x_n]$ with $f(a_1,\ldots , a_n)^3=2$. Now use the fact, that all powers of the $a_i$ are either $0$ or $1$, even in $f^3$, and deduce the contradiction.

3. May 2, 2017

### Mr Davis 97

So just to be clear, is it correct to say that $[\mathbb{Q}(a_1, \dots ,a_n) : \mathbb{Q}] = [\mathbb{Q}(a_1, \dots ,a_n) : \mathbb{Q}(a_1, \dots ,a_{n-1})] \cdots [\mathbb{Q}(a_1) : \mathbb{Q}] = 2^n$, while $[\mathbb{Q}(2^{1/3}) : \mathbb{Q}] = 3$, and since the latter does not divide the former, $2^{1/3} \not\in \mathbb{Q}(a_1, \dots ,a_n)$?

Note that $[\mathbb{Q}(a_1, \dots ,a_n) : \mathbb{Q}]$ denotes the degree of $\mathbb{Q}(a_1, \dots ,a_n)$ over $\mathbb{Q}$

4. May 2, 2017

### Staff: Mentor

Yes. Only a minor thought: If some $\sqrt{a_i}$ happen to be a rational itself or already contained in other $\mathbb{Q}(a_i)$, which you haven't excluded, then the degree is only $2^m < 2^n$. It doesn't change the argument, but it is a possibility.

Btw.: This is basically the reason, why one cannot construct the third of an angle only by compass and ruler.

5. May 2, 2017

### Mr Davis 97

Also, one last thing. Why must $[\mathbb{Q}(2^{1/3}) : \mathbb{Q}]$ divide $[\mathbb{Q}(a_1, \dots ,a_n) : \mathbb{Q}]$ in order for $2^{1/3} \in \mathbb{Q}(a_1, \dots ,a_n)$? I want to make sure I have my reasons right

6. May 2, 2017

### Staff: Mentor

For a tower of algebraic extensions $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt[3]{2}) \subseteq \mathbb{Q}(a_1,\ldots ,a_n)$ we have
$$2^m = [\,\mathbb{Q}(a_1,\ldots ,a_n)\, : \, \mathbb{Q}\,] = [\,\mathbb{Q}(a_1,\ldots ,a_n)\, : \,\mathbb{Q}(\sqrt[3]{2})\,]\,\cdot \,[\,\mathbb{Q}(\sqrt[3]{2})\, : \,\mathbb{Q}\,]\\ = [\,\mathbb{Q}(a_1,\ldots ,a_n)\, : \,\mathbb{Q}(\sqrt[3]{2})\,]\,\cdot \,3$$
(https://en.wikipedia.org/wiki/Degree_of_a_field_extension#The_multiplicativity_formula_for_degrees)
which is impossible. If you may not or don't want to use this, then you'll have to do the work (I think):

An induction should work and is somehow clearer to work with, than to deal with all $a_i$ in one step.
Just write $\sqrt[3]{2} = r_0 + r_1a_1$ and show that this is impossible. All powers of $a_1$ greater than $1$ can be reduced, because $a_1^2=r\in \mathbb{Q}$.