Char(F)=p; poly NOT irreduc. => must have a root?

Homework Statement

Let F have prime characteristic p and let a be in F. Show that the polynomial f(x)=x^p-a either splits or is irreducible in F[x].

I was given a hit: "what can you say about all of the roots of f in a splitting field?"

The Attempt at a Solution

Suppose f has a root b. That is, b^p=a. Then using Freshman's dream, we have (x-b)^p=x^p+(-b)^p=x^p-a, and so it splits. So, it suffices to show that if that polynomial is NOT irreducible, then it must have a root. That's where I got stuck. The statement seems to be false for field of zero characteristic, so the fact that Char(F)=p should be somehow used.

Dick
Homework Helper
Why don't you use the correct binomial expansion instead of the incorrect Freshman's dream?

Why don't you use the correct binomial expansion instead of the incorrect Freshman's dream?

Because it is indeed correct? In a field F of characteristic p for any two x and y in F we have (x+y)^p=x^p+y^p. This is to see using the mentioned binomial expansion and convincing yourself that p divides every binomial coefficient except the frist and the last. Thus, each term in the expansion will look like p*k*x^r*y^s, where k, r and s are some integers whose values are not important. Since for every x in the field we have px=0, all the terms in the expansion (except the first and the last) dissapear. The freshman's dream is a theorem, not a stupid mistake.

Dick
Homework Helper
Because it is indeed correct? In a field F of characteristic p for any two x and y in F we have (x+y)^p=x^p+y^p. This is to see using the mentioned binomial expansion and convincing yourself that p divides every binomial coefficient except the frist and the last. Thus, each term in the expansion will look like p*k*x^r*y^s, where k, r and s are some integers whose values are not important. Since for every x in the field we have px=0, all the terms in the expansion (except the first and the last) dissapear. The freshman's dream is a theorem, not a stupid mistake.

Ok, so then what's your question? Doesn't that show x^p-a splits if it has a root, since your argument then implies that it has p roots? The Freshman's dream isn't true in characteristic 0.

If it has a root everything is clear. What is not clear is why the fact that x^p-a is NOT irreducible implies that it has a root.

Assume for contradiction that x^p - a = p(x)q(x), where p and q are non-trivial and irreducible. From what you've said, you know what p(x) and q(x) must look like in the splitting field...can you arrive at a contradiction? It will help to assume p is odd (you can take care of char F = 2 separately).

Assume for contradiction that x^p - a = p(x)q(x), where p and q are non-trivial and irreducible. From what you've said, you know what p(x) and q(x) must look like in the splitting field...can you arrive at a contradiction? It will help to assume p is odd (you can take care of char F = 2 separately).

First of all, it is not true that x^p-2 can be represented as a product of only two irreducible polynomials.

How about these arguments. All the roots of x^p-a are p-th roots of a.

Suppose that x^p-2 is not irreducible and factorize in into irreducibles. All of the will be of degree >=2. Consider any of them. Look at the constant term of that factor.

It MUST (BUT WHY?!) look like the p-th root of a to some power, say, k. This tells us that a^{k/p} is in the field. Since p is prime, there is a multiplicative inverse of k, say l, and raising to that power (since field is closed under products) we find that a^{1/p} is in the field. Done.

First of all, it is not true that x^p-2 can be represented as a product of only two irreducible polynomials.

Sorry, I misspoke... but I think my argument still works: write x^p - a as a product of irreducibles. In the splitting field each irreducible is of the form (x-b)^k for some k (where b is the root of x^p - a). Look at the irreducible of lowest degree. This will divide the other factors in the splitting field and therefore in F[x] (this is by uniqueness of Euclidean algorithm since both polys are in F[x]), contradicting that they were irreducible in F[x].

Sorry, I misspoke... but I think my argument still works: write x^p - a as a product of irreducibles. In the splitting field each irreducible is of the form (x-b)^k for some k (where b is the root of x^p - a). Look at the irreducible of lowest degree. This will divide the other factors in the splitting field and therefore in F[x] (this is by uniqueness of Euclidean algorithm since both polys are in F[x]), contradicting that they were irreducible in F[x].

Why are you referring to a root of x^p-a as THE root? How do you know there is only one root? What if they behave like roots in the complex field?

Why are you referring to a root of x^p-a as THE root? How do you know there is only one root? What if they behave like roots in the complex field?

In your first post you observed that if b is a root of x^p - a (i.e. b^p = a), then (x-b)^p = x^p - b^p = x^p - a. So in the splitting field x^p - a breaks down as (x-b)^p which has only one root (with multiplicity p).

Unfortunately I missed the last few weeks of my algebra class during which we discussed fields, so I can't help you directly. You mentioned that the constant term of each irreducible factor must be of the form a^{k/p}. I too do not see why this must be so. Did you read that elsewhere?