Prove $2^{1/2}+3^{1/3}$ is irrational using Galois theory.

Question

So, I want to prove that $2^{1/2}+3^{1/3}$ is irrational, and I need to prove it using Galois theory.

To start, let's forget about the sum and deal with the individual numbers and $F_1 = \mathbb{Q}(2^{1/2})$ and $F_2 = \mathbb{Q}(3^{1/3})$. Both $2^{1/2}$ and $3^{1/3}$ are clearly irrational with easily determined minimal polynomials over $\mathbb{Q}$, namely $f_1(x) = x^2 -2$ and $f_2(x) = x^3 -3$. Also, $[F_1:\mathbb{Q}] = 2$ and $[F_2:\mathbb{Q}] = 3$. Let $F_3 = \mathbb{Q}(2^{1/2},3^{1/3})$, then $[F_3:\mathbb{Q}] = 6$ since the $f_1$ is irreducible over $F_2$ and $f_2$ is irreducible over $F_1.$ Since $2^{1/2}+3^{1/3} \in F_3$, take $\sigma \in {\rm Gal}(F_3/\mathbb{Q})$ and consider $\sigma(2^{1/2}+3^{1/3})$. Then $\sigma(2^{1/2}+3^{1/3})$ will lie in $\mathbb{Q}$ if and only if $2^{1/2}+3^{1/3}$ lies in $\mathbb{Q}$. So $\sigma(2^{1/2}+3^{1/3}) = \sigma(2^{1/2})+\sigma(3^{1/3})$, and neither one of those images will lie in $\mathbb{Q}$.

Here I feel like I'm back at square one with showing that the sum of two irrationals is again irrational, but I don't really know how to proceed. I don't even know if my approach is a good one, so any help with regards to new approaches, mistakes I've made or how to carry on from where I am will be much appreciated.

Answer 1

The minimal polynomial of $\alpha=\sqrt{2}$ over $\mathbb{Q}$ is $p(x)=x^2-2$ and the minimal polynomial of $\beta=\sqrt[3]{3}$ is $q(y)=y^3-3$. Assuming that $\alpha+\beta=\frac{u}{v}\in\mathbb{Q}$, $\beta=\frac{u}{v}-\alpha$ is a root of: $$ r(x)=v^2\cdot p\left(\frac{u}{v}-x\right) \in\mathbb{Q}[x] $$ but since $\partial r=2$, that contradicts $[\mathbb{Q}(\beta):\mathbb{Q}]=3$.

Answer 2

I know I'm answering my own question without really knowing if it's correct, but I wanted some feedback on my solution.

Assume that $2^{1/2}+3^{1/3}$ is rational, then for all $\sigma \in {\rm Gal}(F_3/\mathbb{Q})$ we would have $\sigma(2^{1/2}+3^{1/3}) = 2^{1/2}+3^{1/3}$ since they fix $\mathbb{Q}$. Take for instance $$\sigma: 2^{1/2} \rightarrow -2^{1/2}, \: 3^{1/3} \rightarrow 3^{1/3}$$ then $\sigma$ doesn't fix $2^{1/2}+3^{1/3}$ so it can't be rational.

Answer 3

Let $ K = \mathbb{Q}(2^{1/2}, 3^{1/3}) $ and let $ L $ be the normal closure of $ K/\mathbb{Q} $. Define the field trace by

$$ T(x) = \textrm{Tr}_{L/\mathbb{Q}}(x) = \sum_{\sigma \in \textrm{Gal}(L/\mathbb{Q})} \sigma(x) $$

We will need the following properties of the trace:

• $ T(a+b) = T(a) + T(b) $, which is obvious.
• $ T(qa) = q T(a) $ for $ q \in \mathbb{Q} $. This follows because the Galois group fixes $ \mathbb{Q} $.
• $ T(1) = [L : \mathbb{Q}] $, and hence $ T(q) = q [L : \mathbb{Q}] $ for any rational $ q $. This implies that for rational $ q $, $ T(q) = 0 $ iff $q = 0 $.
• Let $ \alpha \in L $ be such that the sum of its $ \mathbb{Q}$-conjugates is $ 0 $. Then, $ T(\alpha) = 0 $. This follows because for any conjugate $ \beta $, the automorphism $ \mathbb{Q}(\alpha) \to \mathbb{Q}(\beta) $ extends to a $ \mathbb{Q} $-automorphism of $ L $, therefore $ \alpha $ has full orbit under the action of the Galois group.

To finish the argument, we first note that $ \alpha = 2^{1/2} + 3^{1/3} \neq 0 $. On the other hand, the field trace yields $ T(2^{1/2} + 3^{1/3}) = T(2^{1/2}) + T(3^{1/3}) = 0 + 0 = 0 $. Since $ \alpha \neq 0 $ but $ T(\alpha) = 0 $, $ \alpha $ cannot be rational.

Answer 4

If you want to do it by Galois theory, take any (finite) Galois extension $E/\mathbb{Q}$ which contains both $2^{1/2}$ and $3^{1/3}$. Write $G$ for its Galois group. Now $\alpha = 2^{1/2}+3^{1/3}$ is rational iff $\sigma(\alpha)=\alpha$ for all $\sigma \in G$. However, the set $\{ \sigma(2^{1/2}) \}_{\sigma \in G}$ has two elements, whereas the set $\{ \sigma(3^{1/3}) \}_{\sigma \in G}$ has three elements, since $2^{1/2}$ is of degree $2$ over $\mathbb{Q}$ and $3^{1/3}$ is of degree $3$. It follows that there are $\sigma_1,\sigma_2 \in G$ with $\sigma_1(2^{1/2})=\sigma_2(2^{1/2})$ but $\sigma_1(3^{1/3})\neq \sigma_2(3^{1/3})$. It then immediately follows that $\sigma_1(\alpha)\neq \sigma_2(\alpha)$, implying $\alpha$ is non-rational. $\square$

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But actually no Galois theory is needed, just basic facts about algebraic fields. The number $\alpha=2^{1/2}+3^{1/3}$ is an algebraic integer (i.e. it is a zero of a monic polynomial with coefficients in $\mathbb{Z}$), it being the sum of two algebraic integers; therefore if it is rational, it must be an integer. However we have $2 < \alpha < 3$, contradiction. $\square$

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[To prove $\alpha < 3$ without a calculator, just show $2^{1/2}<3/2$ and $3^{1/3}<3/2$, by squaring and cubing the respective inequalities.]