Htykuut

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

For a, I do:

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

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?

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?

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 $[mathbb Q(root3of2):mathbb Q]=3$ and $[mathbb Q(rootof5):mathbb Q]=2$ (And that you can do with Eisenstein) then it follows that $[mathbb Q(sqrt 5, root3of2):mathbb Q]=6$ 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 $mathbb Q(sqrt 5, root3of2)$.

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 $Bbb Zbigl[frac{1+sqrt5}2bigr]$. 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 $G(X)=gcdotbar g=f(X-sqrt5,)f(X+sqrt5,)$ has its coefficients in $Bbb Q$. Explicitly,
$$
G(X)=bigl((X-sqrt5,)^2-2bigr)bigl(X+sqrt5,)^2-2bigr)=X^4-14X^2+9,,
$$
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 $V(X)=vcdotbar v$, 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.