# The degree of $\sqrt{2} + \sqrt{5}$ over $\mathbb Q$

I know that the degree is at most $6$, since $\sqrt{2} + \sqrt[3]{5} \in \mathbb Q(\sqrt{2}, \sqrt[3]{5})$, which has degree $6$ over $\mathbb Q$. I’m trying to construct a polynomial with root $\sqrt{2} + \sqrt[3]{5}$ and coefficients in $\mathbb Q$ of degree $6$, and then show that it is irreducible over $\mathbb Q$.

I managed to find that it is a root of the polynomial $x^6 – 6x^4 – 10x^3 +12x^2 – 60x +17$. This is where I run in to some problems. I don’t know how to show that this is irreducible over $\mathbb Q$ (or $\mathbb Z$). The only criteria we have learned is Eisenstein’s criteria, which clearly does not apply here. How else can I show that the degree of this number over $\mathbb Q$ is $6$?

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Here’s how you can show irreducibility of your polynomial, using the same method I used in the argument I gave in the reference that @BillDubuque mentioned:

First form $x^3-5$, the minimal polynomial for $\root3\of5$ over $\Bbb Q$. Now think of it as a polynomial over the principal ideal domain $\Bbb Z[\sqrt2\,]$. It happens that $5$ is still prime there, so Eisenstein still applies, and this is the minimal polynomial for $\root3\of5$ over the extension. Consequently $(x-\sqrt2)^3-5$ is still irreducible over $\Bbb Z[\sqrt2\,]$, and it’s the minimal polynomial for $\sqrt2+\root3\of5$. Let’s call this polynomial $f(x)$. It’s actually $x^3-3\sqrt2x^2+6x-5+2\sqrt2$. Now take the conjugate polynomial, call it $\bar f$, which you get by replacing $\sqrt2$ by $-\sqrt2$ in $f$ wherever it appears. Finally, multiply, and get $f\bar f$, which turns out to be exactly the sextic polynomial you calculated.

This clearly is a $\Bbb Q$-polynomial that has $\sqrt2+\root3\of5$ as a root. But it is also a $\Bbb Z[\sqrt2\,]$-polynomial, and we know its factorization into irreducibles there, namely $f\bar f$. And by uniqueness of factorization there, this is the only possible factorization with coefficients in $\Bbb Z[\sqrt2\,]$. But a $\Bbb Q$-factorization of your polynomial would also be a $\Bbb Z[\sqrt2\,]$-factorization, and we already have one such, and it’s unique. So there is no factorization of your polynomial over $\Bbb Z$ or $\Bbb Q$.

1. Let $V$ be the 6-dimensional vector space consisting of all numbers of the form $$a+b2^{1/2} + c5^{1/3} + d2^{1/2}5^{1/3} + e5^{2/3} + f2^{1/2}5^{2/3}$$ where $a,b,c,d,e,f\in \Bbb Q$

2. Let $w = \sqrt2+\sqrt[3]5$. The numbers $w^0, w^1, w^2, w^3, w^4, w^5$ are vectors from $V$. Since $V$ is 6-dimensional, these 6 vectors either span $V$ or else any them can be expressed as a linear combination of the others. Your undergraduate linear algebra class taught you how to check which it is.

3. If your $6$th-degree polynomial were reducible, it would have some factor $P$ of degree $d$ less than $6$; say $P(x) = x^d + Q(x)$. Since $P(w) = 0$, you have $$w^d = -Q(w)$$ where $Q$ has degree at most $d-1$. This expresses $x^d$ in terms of $x^0, x^1,\ldots, x^{d-1}$, contradicting your finding in step 2.