# Irreducible polynomial in extension field

• #1
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## 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 $$K[x] \subseteq L[x]$$.

## Answers and Replies

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• #2
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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.

• #3
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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 $$\mathbb{Q}(\alpha)[x] \subseteq K[x]$$.

• #4
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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?

• #5
139
12
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.

• #6
Hurkyl
Staff Emeritus
Gold Member
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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?

• #7
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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.

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