How to compute the degree of $\mathbb{Q}[\sqrt[3]{2}+\sqrt[5]{2}]$ over $\mathbb{Q}$?

Question

Here is a 2-part problem:

• Find the number of ring homomorphisms $\mathbb Q[\sqrt[3]{2},\sqrt[5]{2}]\to \mathbb C$.
• Prove that the degree of $\mathbb{Q}[\sqrt[3]{2}+\sqrt[5]{2}]$ over $\mathbb Q$ is $15$.

Am I supposed to deduce the second part from the first? I don't see how they are related. Also, I don't know even what fact I should use to solve the first part. Any hints?

Answer 1

Any ring homomorphism $\sigma : \mathbb{Q}[\sqrt[3]{2},\sqrt[5]{2}] \to \mathbb{C}$ has the property that it permutes roots of polynomials. Moreover, any ring homomorphism is completely determined by where it maps $\sqrt[3]{2}$ and $\sqrt[5]{2}$. The minimal polynomial over $\mathbb{Q}$ of $\sqrt[3]{2}$ is $x^3 - 2$ of $\sqrt[5]{2}$ is $x^5 - 2$. So, we have three choices for $\sigma(\sqrt[3]{2})$ and five choices for $\sigma(\sqrt[5]{2})$. So, we have at most $15$ homomorphisms from $\mathbb{Q}[\sqrt[3]{2},\sqrt[5]{2}]$ to $\mathbb{C}$. It can be checked (via a tedious calculation) that each of these $15$ maps are actually homomorphisms.

Since $[\mathbb{Q}(\sqrt[3]{2}): \mathbb{Q}] = 3$ and $[\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5$, and $\gcd(3,5)=1$, we have that $[\mathbb{Q}[\sqrt[3]{2},\sqrt[5]{2}]:\mathbb{Q}] = 3 \cdot 5 = 15$ (prove this!). If we can show that $\mathbb{Q}[\sqrt[3]{2},\sqrt[5]{2}] = \mathbb{Q}[\sqrt[3]{2}+\sqrt[5]{2}]$, then we will have proved the second part. Since $\mathbb{Q}[\sqrt[3]{2}+\sqrt[5]{2}] \subseteq \mathbb{Q}[\sqrt[3]{2},\sqrt[5]{2}]$, it suffices to show that $\sqrt[3]{2},\sqrt[5]{2} \in \mathbb{Q}[\sqrt[3]{2}+\sqrt[5]{2}]$. Finding the multiplicative inverse of 2^1/3 + 2^1/5 is usually helpful in such problems. (see update2 below)

I'll update this answer if I find a simplification of these arguments.

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Update: The only way I know to simplify these calculations is to use the Isomorphism Extension Theorem, or its variants.

Lemma 3.17. (page 33, Field and Galois Theory, Patrick Morandi) Let $\sigma : F \to F'$ be a field isomorphism. Let $f(x) \in F[x]$ be irreducible, let $\alpha$ be a root of $f$ in some extension field $K$ of $F$, and let $\alpha'$ be a root of $\sigma(f)$ in some extension field $K'$ of $F'$. Then there is an isomorphism $\tau : F(\alpha) \to F'(\alpha')$ with $\tau(\alpha) = \tau(\alpha')$ and $\tau |_F = \sigma$.

Let $\operatorname{id} : \mathbb{Q} \to \mathbb{Q}$ be the identity map, $f(x) = x^5 - 2$, $\alpha = \sqrt[5]{2}$ and $\alpha'$ be any root of $f(x)$ in $\mathbb{C}$. Then, the above lemma guarantees that there is a homomorphism $\tau : \mathbb{Q}[\sqrt[5]{2}] \to \mathbb{C}$ that maps $\sqrt[5]{2}$ to any other root of $x^5 - 2$.

Next, fix any such $\tau : \mathbb{Q}[\sqrt[5]{2}] \to \mathbb{Q}[\alpha']$. Let $g(x) = x^3 - 2$. This is irreducible over $\mathbb{Q}[\sqrt[5]{2}]$ as well. Let $\beta = \sqrt[3]{2}$ and $\beta'$ be any root of $g(x)$ in $\mathbb{C}$. Then, the above lemma guarantees that there is a homomorphism $\sigma : \mathbb{Q}[\sqrt[5]{2},\sqrt[3]{2}] \to \mathbb{C}$ that maps $\beta$ to $\beta'$ and also maps $\alpha$ to $\alpha'$.

Thus, we have shown that there are at least $15$ homomorphisms from $\mathbb{Q}[\sqrt[3]{2},\sqrt[5]{2}]$ to $\mathbb{C}$. From the previous discussion, there are no more than $15$, so we have found all the homomorphisms.

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Update2: Finding the inverse gave me no insight. I am truly sorry for the misguiding comment. I will update again if I find a simpler argument to prove the second part.

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Update3: Here is a simplification for the second part. $[\mathbb{Q}[\sqrt[3]{2}+\sqrt[5]{2}]:\mathbb{Q}]$ must divide $15$ by the Tower Law, so it can only be $1$, $3$, $5$ or $15$. Since $1$ is obviously ruled out, we need only show that it is not $3$ or $5$. This amounts to showing that $\sqrt[3]{2}+\sqrt[5]{2}$ is not satisfied by any monic polynomial over $\mathbb{Q}$ of degree $3$ or $5$.

Suppose $f(x) = x^3 + ax^2 + bx + c \in \mathbb{Q}[x]$ has $\sqrt[3]{2}+\sqrt[5]{2}$ as a root. Then, plugging it in and simplifying, we get $$ \begin{align} 0 &= (\sqrt[3]{2}+\sqrt[5]{2})^3 + a(\sqrt[3]{2}+\sqrt[5]{2})^2 + b(\sqrt[3]{2}+\sqrt[5]{2}) + c\\ &= (2+c) + b2^{1/3} + a2^{2/3} + b2^{1/5} + a2^{2/5} + 2^{3/5} + (2a)2^{1/3}2^{1/5} + (3) 2^{1/3} 2^{2/5} + (3) 2^{2/3} 2^{1/5}. \end{align} $$ The RHS is a $\mathbb{Q}$-linear combination of elements which are part of a basis of $\mathbb{Q}[\sqrt[3]{2},\sqrt[5]{2}]$ over $\mathbb{Q}$; in other words, the RHS is a linear combination of a linearly independent set that is equal to zero. Hence, each of the coefficients must be zero, which is false. Hence, there is no polynomial of degree $3$ that $\sqrt[3]{2}+\sqrt[5]{2}$ satisfies.

Next, let $g(x) = x^5 + a x^4 + bx^3 + cx^2 + dx + e \in \mathbb{Q}[x]$ have $\sqrt[3]{2}+\sqrt[5]{2}$ as a root. Then, substituting and simplifying, we get $$ \begin{align} 0 &= (\sqrt[3]{2}+\sqrt[5]{2})^5 + a(\sqrt[3]{2}+\sqrt[5]{2})^4 + b(\sqrt[3]{2}+\sqrt[5]{2})^3 + c(\sqrt[3]{2}+\sqrt[5]{2})^2 + d(\sqrt[3]{2}+\sqrt[5]{2}) + e\\ &= (2+2b+e)+(2a+d)2^{1/3} + (2+c)2^{2/3} + (8a + d) 2^{1/5} + (20+c)2^{2/5} + (b) 2^{3/5} + (a) 2^{4/5}\\ &\qquad \quad {}+{} (10+2c) 2^{1/3}2^{1/5} + (3b) 2^{1/3} 2^{2/5} + (4a) 2^{1/3} 2^{3/5} + (5) 2^{1/3} 2^{4/5} + (3b) 2^{2/3} 2^{1/5}\\ &\qquad \quad {}+{} (6a) 2^{2/3} 2^{2/5} + (10) 2^{2/3} 2^{3/5}. \end{align} $$

Again, the RHS is a $\mathbb{Q}$-linear combination of a $\mathbb{Q}$-linearly independent set in $\mathbb{Q}[\sqrt[3]{2},\sqrt[5]{2}]$, so each of the coefficients must be zero, which is clearly not true. Hence, there is no polynomial of degree $5$ that $\sqrt[3]{2}+\sqrt[5]{2}$ satisfies.

Hence, $[\mathbb{Q}[\sqrt[3]{2}+\sqrt[5]{2}]:\mathbb{Q}] = 15$.

Answer 2

You can prove that $\mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2}) = \mathbb{Q}(\sqrt[3]{2},\sqrt[5]{2})$. In fact this follows by the Primitive Element Theorem, as $\sqrt[3]{2} + \sqrt[5]{2}$ satisfies the condition to be a primitive element of $\mathbb{Q}(\sqrt[3]{2},\sqrt[5]{2})$. Nevertheless here's the explicit construction below:

Let $f(x) = x^3-2$. Now consider $g(x) = f(\sqrt[3]{2} + \sqrt[5]{2}-x) \in \mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2})[x]$. Obviously we have that $\sqrt[5]{2}$ is one of the roots ,while it's not hard to conclude that the other ones are $\sqrt[3]{2} + \sqrt[5]{2} - \zeta_3\sqrt[3]{2}$ and $\sqrt[3]{2} + \sqrt[5]{2} - \zeta_3^2\sqrt[3]{2}$, where $\zeta_3$ is the third root of unity.

On the other side $\sqrt[5]{2}$ satisfies $x^5-2$, whose roots are $\sqrt[5]{2}, \zeta_5\sqrt[5]{2}, \zeta_5^2\sqrt[5]{2}, \zeta_5^3\sqrt[5]{2}, \zeta_5^4\sqrt[5]{2}$. Now the minimal polynomial of $\sqrt[5]{2}$ over $\mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2})$ must divide both $g(x)$ and $x^5-2$ and so their greatest common divisor. But as their only common root is $\sqrt[5]{2}$, we get that it divides $x-\sqrt[5]{2}$ and as the polynomial is linear it must be itself. Hence $\min(\sqrt[5]{2},\mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2})) = x-\sqrt[5]{2}$. This yields that $\sqrt[5]{2} \in \mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2})$. Now it's not hard to conlcude that $\sqrt[3]{2} \in \mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2})$ and therefore $\mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2}) = \mathbb{Q}(\sqrt[3]{2},\sqrt[5]{2})$

As mentioned in the comments from here you can conclude that $\left[\mathbb{Q}(\sqrt[3]{2} + \sqrt[5]{2}):\mathbb{Q}\right]=15 $