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Pinter's Book of Abstract Algebra chapter 33, problem C2:

Let $p$ be a prime number.

"Explain why $x^p-a$ factors in $\mathbf{F}[x]$ as $x^p-a = p(x)f(x)$, where both factors have degree$\leq 2$.

Let $p = 5$, $a = 1$. $\mathbf{F}$ can be $\mathbb{Q}$ or $\mathbb{R}$, among other fields.

$x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$.

Obviously, the degree of the second factor is greater than 2. In fact, if $p$ is greater than $4$, one or both of $p(x)$ and $f(x)$ have to be greater than 2. Not sure what I'm missing here.

Arturo Magidin
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Alan Ox
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    Your quote makes no sense as written. If both factors have degree at most $2$, then the product has degree at most $4$, so you would have $p\leq 3$. There must be context missing. – Arturo Magidin Aug 02 '21 at 00:11
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    This really doesn't make sense. And writing that "$\mathbf{F}$ can be $\mathbb{Q}$ or $\mathbb{R}$, among other fields." is far too vague. – lulu Aug 02 '21 at 00:17
  • My guess (and it is just a guess) is that you are meant to show that "either $x^p-a$ is irreducible or else it has a root." See, e.g., this question and the question it is linked to as a duplicate. – lulu Aug 02 '21 at 00:29
  • http://www2.math.umd.edu/~jcohen/402/Pinter%20Algebra.pdf FWIW I think question 2 is a continuation of question 1. In it we are assume $\omega$ is a $p$th root of unity in field $F$ and you are assuming $x^p = a$ is not irreducible (and that a root $d$ exists? I'm not really sure.) – fleablood Aug 02 '21 at 00:35
  • @HallaSurvivor: Well, there is actually a lot of missing context. First, the entire sequence of problems assumes that $\mathbf{F}$ contains a primitve $p$th root of unity. This excludes the stated possibility of $\mathbf{F}$ being $\mathbb{Q}$ or $\mathbb{R}$. It is also proceeding under the assumption that $\mathbf{F}$ does not contain a root to $x^p-a$, yet is reducible (end of part 1). However, this is clearly meant to be $\geq 2$ rather than $\leq 2$. – Arturo Magidin Aug 02 '21 at 00:47

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It's a typo. That said, you are incorrect on many things about this problem. You failed to provide critical context, making your post almost unintelligible and impossible to answer. It was only when that context was provided via a link in the comments to the text itself that I was able to puzzle this out.

This is one of a multi-part sequence of exercises meant to show that if $\mathbf{F}$ contains a primitive $p$-th root of unity, then $x^p-a$ either has a root in $\mathbf{F}$ or else is irreducible over $\mathbf{F}$.

The preamble to the set of exercises explicitly states that $\omega\in\mathbf{F}$ is a primitive $p$th root of unity. That means that it is false that "$\mathbf{F}$ can be $\mathbb{Q}$ or $\mathbb{R}$ or anything". In fact, it can be neither $\mathbb{R}$ nor $\mathbb{Q}$ (which contain no primitive $p$-th root of unity for any prime other than $p=2$), nor can it be many finite fields.

Second, the first part of the problem tells you to assume that the field does not contain any roots of $x^p-a$, but that $x^p-a$ is reducible; the goal is to establish a contradiction at the end, so that you will conclude that $x^p-a$ either has a root in $\mathbf{F}$ or else is irreducible over $\mathbf{F}$. Note that this means that you cannot take $a=1$, since that violates the assumption that $x^p-a$ has no roots in $\mathbf{F}$.

This is where the typo comes in. You are expected to explain why you can factor $x^p-a$ into a product $p(x)f(x)$ where both factors have degree at least $2$ ("$\geq 2$" instead of "$\leq 2$").

At which point, you are in the territory of the duplicate indicated above.

Arturo Magidin
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  • thanks - this particular problem hadn't mentioned omega, and I didn't recognize its importance. – Alan Ox Aug 03 '21 at 09:35