Compute the irreducible polynomials over Q for $a=\sqrt{2}+\sqrt{5}$ and $b=\sqrt[3]{2}+\sqrt{5}$

Question

Compute the irreducible polynomials over Q for $a=\sqrt{2}+\sqrt{5}$ and $b=\sqrt[3]{2}+\sqrt{5}$

For a, I do:

$$a=\sqrt{2}+\sqrt{5} \Rightarrow a^2-7=2\sqrt{10} \Rightarrow a^4-14a^2+9=0$$

So $p(X)=X^4-14X^2+9 \in Q[X]$ is a candidate. Mathematica says this is irreducible.

How do I know this is irreducible, I can't find a prime that divides both 14 and 9, so Eisenstein's criterion (if I'm using it correctly) should imply this is reducible, what am I doing wrong?

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For b, I haven't been able to use the method above to find any polynomials, let alone check if they are reducible, should I be algebraically smarter via the method above, or is there another way to approach this?

Answer 1

Eisenstein's criterion doesn't imply that something is reducible, it can only imply something is irreducible.

For (a) try and show that the set $\{1,\sqrt 2,\sqrt 5,\sqrt 10\}$ is linearly independent over $\mathbb Q$.

For (b) if we find that $[\mathbb Q(\root3\of2):\mathbb Q]=3$ and $[\mathbb Q(\root\of5):\mathbb Q]=2$ (And that you can do with Eisenstein) then it follows that $[\mathbb Q(\sqrt 5, \root3\of2):\mathbb Q]=6$ from basic field theory and some factorisation arguments. Then it is enough to show that the required extension is not of orders $2,3$, as it is contained in $\mathbb Q(\sqrt 5, \root3\of2)$.

Answer 2

Here’s a method that I like for showing irreducibility in cases like this, even though like all such, it doesn’t always apply (or work!). It involves some arithmetic of other fields than $\Bbb Q$, facts that you may not know, but look at this:

Look at your extensions as extensions of $K=\Bbb Q(\sqrt5\,)$, whose ring of integers is $\Bbb Z\bigl[\frac{1+\sqrt5}2\bigr]$. You need to know at the outset that this ring is a unique factorization domain, and that $2$ is still a prime element there.

Now, just as $f(X)=X^2-2$ is irreducible over $\Bbb Q$, it’s also irreducible over $K$, by Eisenstein. Over $K$, $g(X)=f(X-\sqrt5\,)$ is still irreducible, and in $K[X]$, you see that $G(X)=g\cdot\bar g=f(X-\sqrt5\,)f(X+\sqrt5\,)$ has its coefficients in $\Bbb Q$. Explicitly, $$ G(X)=\bigl((X-\sqrt5\,)^2-2\bigr)\bigl(X+\sqrt5\,)^2-2\bigr)=X^4-14X^2+9\,, $$ just the polynomial you got. Now comes the magic: you have factored $G$ as the product of two $K$-irreducible (quadratic) polynomials, and this is the only nontrivial factorization that there is for $G$ in $K[X]$. Suppose that $G$ had a $\Bbb Q$-factorization. Then this would be a $K$ factorization as well. But we know that $G$ has a $K$-factorization, and it’s plainly not a $\Bbb Q$-factorization. Thus there is no $\Bbb Q$ factorization (other than the trivial one), and therefore $G$ is $\Bbb Q$-irreducible.

Now, for $\sqrt[3]2+\sqrt5$, you do the same thing, forming $u(X)=X^3-2$, irreducible over both $\Bbb Q$ and $K$, then $v(X)=u(X-\sqrt5\,)$, also irreducible over $K$. Then your $\Bbb Q$-polynomial is $V(X)=v\cdot\bar v$, same as before, I’ll let you compute it out, but even without calculating the (sextic) product, you see that you have expressed $V$, a $\Bbb Q$-polynomial with $\sqrt[3]2+\sqrt5$ for one of its roots, as a product of two $K$-irreducibles. Once more, this is the only nontrivial factorization of $V(X)$ possible over $K$. Therefore, by the same magic as above, $V$ is $\Bbb Q$-irreducible.