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

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?

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?

• Hint: $b-\sqrt{5}=\sqrt[3]2$ – Empy2 Dec 2 '18 at 2:54
• Eisenstein's criterion is merely a sufficient condition for irreducibility. Failing to find such a $p$ in no way guarantees reducibility! – bounceback Dec 2 '18 at 2:56
• For $b$, start with $x=b\implies x-\sqrt5=\sqrt[3]2$. Now cube both sides and you’ll only have one term with a square root on the left. Rearrange and you can square both sides to get rid of the rest of the roots. – Clayton Dec 2 '18 at 2:57
• I'd guess that looking at this question might help: Minimal polynomial of $\sqrt[3]{2} + \sqrt{3}$. (Solving this problem with $\sqrt3$ instead of $\sqrt5$ is probably not that much different.) – Martin Sleziak Dec 2 '18 at 5:29

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 $$\mathbb Q(\sqrt 5, \root3\of2)$$.

• Thanks, so if $\{1,\sqrt{2},\sqrt{5},\sqrt{10}\}$ are linearly independent, it means that $Q[\sqrt{2}+\sqrt{5}/Q]$ has degree at least 4. Is there a theorem that forces a polynomial of order less than or equal to the degree of the field extension to be irreducible? Is this obvious since if the polynomial factors, then the factors will have the same root so it would reduce the basis of the field extension, leading to a contradiction? – Mike Dec 2 '18 at 3:09
• There's a little transition you've gotta make to say that the extension of the sum is the same as the double extension (it is at most the double extension, but in this case it is true and you can show that $\sqrt 2,\sqrt 5$ are in the sum extension pretty easily. Another thing you have to use here to show the polynomial is irreducible is that the degree of the minimal polynomial is the degree of the field extension for a simple extension. – NL1992 Dec 2 '18 at 3:24
• Right, I can show that the double extension is the same as the sum extension. I see you mention the minimal polynomial at the end, is there a connection between minimal polynomial and irreducible polynomial here? Are they the same polynomial in certain cases? Does the degree of the minimal polynomial equaling the degree of the field extension tell us that the degree of the irreducible polynomial has to equal the degree of the field extension? – Mike Dec 2 '18 at 4:26
• The minimal polynomial of $\alpha$ is the polynomial with least degree (over our original field) that takes $\alpha$ as a root. It is an easy exercise to show this polynomial is irreducible. – NL1992 Dec 2 '18 at 15:13
• I'm sorry, I got a little lost. Can you please clarify how $\{1,\sqrt{2},\sqrt{5},\sqrt{10}\}$ being linearly independent implies that $p(X)$ is irreducible? Also, for b I cannot see how your discussion on order can actually lead to the computation of an irreducible polynomial for b. – Mike Dec 3 '18 at 0:54

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.