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I have the following homework question, which I've managed to do the forward implication of, but not the other direction:

Let $\mathbb{K}$ be a field, $ a\in\mathbb{K}$ and $p$ be a prime. Show that the polynomial $x^p - a$ is irreducible over $\mathbb{K}$ iff $a$ is not a $p$-th power of an element in $\mathbb{K}$.

My initial attacks involved attempting to show that $\mathbb{K}[x]/<x^p - a>$ wasn't a field, or beginning by assuming that the polynomial is not irreducible, but neither of these seemed to get me very far. Crucially, I couldn't seem to make use of the (presumably important) fact that $p$ is a prime.

Any hints please?

lokodiz
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    Work over a splitting field. The polynomial breaks up into a product of linear factors. Show no product of a proper subset of the linear factors could have its coefficients all in $K$. (Hint: look at the next-to-leading coefficient of a such a product.) – KCd Apr 04 '13 at 22:01
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    By the way, the real meat in this problem is the reverse direction. I once had a student say he had done "half" the proof of the division algorithm (that $a = bq+r$ for unique $q$ and $r$...), and his "half" was the case that $b|a$. Writing that you "managed" to do the forward implication suggests some real work was needed, when all it requires is that an irreducible polynomial of degree above 1 has no roots in the field. The two directions of the problem are far from being of equal difficulty. – KCd Apr 04 '13 at 22:07
  • I haven't had chance to look at the problem again, but shall do tomorrow, but I appreciate the hint and hope that it turns out to be useful. Also, "managed" wasn't meant to imply that I had to work particularly hard; it was basically a one-liner. On a related note, when marking, my lecturer assigns equal weighting for each direction of an iff proof (at least for homework problems), which is absurd, as this example shows. – lokodiz Apr 06 '13 at 19:02
  • Linked. – Alex Ravsky Jan 08 '19 at 06:16

1 Answers1

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One direction is straightforward:

$$a=k^p\;,\;\;k\in\Bbb K\implies x^p-a=x^p-k^p=(x-k)(x^{p-1}+kx^{p-2}+\ldots+k^{p-2}x+k^{p-1})$$

Suppose now $\,x^p-a\,$ is reducible and assume $\,a\,$ is not a $\;p-$th power of an element in $\,\Bbb K\,$ . Suppose $\,w\,$ is a root of the polynomial in some extension of $\,\Bbb K\,$ , then;

$$[\Bbb K(w):\Bbb K]= r<p\;\;(\text{why ?})$$

But then denoting by $\,N\,$ the norm of $\,\Bbb K(w)/\Bbb K\,$ , we get

$$a^r=N(a)=N(w^p)=\left(N(w)\right)^p\in\Bbb K$$

and since trivially $\;(r,p)=1\;$ , we get that $\,a\,$ itself must be a $\,p-$th power in $\,\Bbb K\,$ .

Note: The norm above is nothing but the product of all the conjugates of $\,a=w^p\,$ in $\,\Bbb K(w)\,$ , and it always yields an element in the base field.

DonAntonio
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    I seemed to understand this, having filled in some of the details. In particular, $[\mathbb{K}(\omega) : \mathbb{K}] =$ deg$(f) < p$, where $f$ is the minimal polynomial of $\omega$, which has degree $< p$ since $x^p-a$ is not irreducible (i.e. where $f$ is a proper factor of $x^p - a$). I also altered one line, since we hadn't covered field norms: Let $\omega_2, ... , \omega_j$ be the other roots of $f$ in \mathbb{K}(\omega). Then $$ a^r = \omega^p...\omega_{j}^p = (\omega ... \omega_j)^p $$ which is in $\mathbb{K}$ since $\omega ... \omega_j$ is the coefficient of 1 in f. – lokodiz Apr 07 '13 at 10:19
  • Your message got cut out, but I guess instead norm you can use the automorphisms of the Galois groups and their product...it's exactly the same. :) – DonAntonio Apr 07 '13 at 10:22
  • Sorry, I keep forgetting that enter submits a comment. Does what I've written now seem correct? And we haven't covered Galois groups yet (we're starting them next week!) – lokodiz Apr 07 '13 at 10:26
  • Ok, great: use the conjugates of $,a,$ (i.e., all the roots of its minimal polynomial) instead: it all is the same! – DonAntonio Apr 07 '13 at 10:33