Irreducible polynomial in extension field

[masterslave]

Homework Statement

Let p\in\mathbb{Q}[x] be an irreducible polynomial. Suppose K is an extension field of \mathbb{Q} that contains a root \alpha of p such that p(\alpha^2)=0. Prove that p splits in K[x].

The Attempt at a Solution

I was thinking contradiction, but if p does not split in K, the only logical conclusion I can come to is that there is an extension field L[x] such that p splits and K[x] \subseteq L[x].

 

[Billy Bob]

I have some thoughts but they are not complete. By some theorem, Q(alpha) is isomorphic to Q[x]/<p(x)>. Use this to show p(x^2)=p(x)q(x) (assume WLOG that p is monic if you want).

Now show p(alpha^4)=0. In fact, alpha^{power of 2} is a root of p. All these powers of alpha can't be distinct. Obtain that alpha is a root of unity.

Here is where my gut tells me you can conclude p is a cyclotomic polynomial, but I don't know enough about them to prove that myself.

 

[VKint]

Nice work, Billy Bob! You're essentially done. Since f is irreducible in \mathbb{Q}[x], f is the minimal polynomial of \alpha. However, \alpha is a root of unity, so the minimal polynomial is, by definition, some cyclotomic polynomial \Phi_k for some k. Thus, f = \Phi_k (without loss of generality, f is monic), so the roots of f are just \alpha, \alpha^2, \alpha^3, \ldots. In particular, f splits in \mathbb{Q}(\alpha)[x] \subseteq K[x].

 

[masterslave]

I see all of it, except the logic you use to conclude that \alpha is a primitive root of unity. My thought is because p(x) is irreducible and \alpha, \alpha^2, \alpha^3, \ldots are all roots of it, then \alpha =\alpha^2 =\alpha^3 =\ldots =1 implying that \alpha is a root of unity.

Any thoughts?

 

[VKint]

I see all of it, except the logic you use to conclude that \alpha is a primitive root of unity.

If \alpha is a primitive s^th root of unity, then \alpha = e^{2 \pi r/s} for some r. We can rewrite this as \alpha = e^{2 \pi t/n}, where \textrm{gcd}(t,n) = 1. In other words, every s^th root of unity is a primitive n^th root of unity for some n, so its minimal polynomial is cyclotomic.

 

[Hurkyl]

An easier way to carry out Billy Bob's plan:

If E is the splitting field of p, then there is an element of Gal(E/Q) that, among other things, sends alpha to alpha^2. Where does it send alpha^2?

 

[Billy Bob]

I see all of it, except the logic you use to conclude that \alpha is a primitive root of unity. My thought is because p(x) is irreducible and \alpha, \alpha^2, \alpha^3, \ldots are all roots of it, then \alpha =\alpha^2 =\alpha^3 =\ldots =1 implying that \alpha is a root of unity.

Any thoughts?

\alpha, \alpha^2, \alpha^4, \alpha^8, \alpha^{16} \ldots are all roots, but maybe not \alpha^3 etc. The point is, there can't be infinitely many roots, so \alpha^i=\alpha^j for some i not equal to j. Now an example will give you an idea for how to handle the general case. If \alpha^8=\alpha^2, then \alpha^8-\alpha^2=0, then \alpha^6-1=0, so alpha is a sixth root of unity.

Thanks VKint and Hurkyl for helping me out.