<Home>Qualifying Exams><Algebra>Fall 2016 - Problem 6

Fall 2016 - Problem 6

Galois theory

Let $F$ be a field of characteristic $p > 0$ . Prove that for every $a \in F$ , the polynomial $x^p - a$ is either irreducible or splits into a product of linear factors.

Solution.

Notice that in $F\p{a^{1/p}}$ , $f = x^p - a$ splits via the Frobenius homomorphism:

f = \p{x - a^{1/p}}^p.

If $a^{1/p} \in F$ , then $f$ splits as above. Otherwise, suppose $a^{1/p} \notin F$ , let $g$ be the minimal polynomial of $a^{1/p}$ over $F$ with $k = \deg{g} < p$ , and write $f = gh$ . Since $F\p{a^{1/p}}$ is a UFD, it follows that $f = \p{x - a^{1/p}}^k$ . But the coefficient of $x^{k-1}$ is $\p{-1}^k ka^{k/p}$ , which means $a^{k/p} \in F$ . Since $p$ is prime, there exist $b, c \in \Z$ such that $bp + ck = 1$ , which gives

F \ni \p{a^{k/p}}^c \cdot a^b = a^{\p{1 - bp}/p} \cdot a^b = a^{1/p},

a contradiction. Hence, $\deg{g} = p$ , i.e., $f = g$ and $f$ is irreducible.