first time taking abstract algebra stuck on this question and have no idea how to go about it. Show that $x^{p}+a$ is reducible in $\mathbb{Z_{p}}[x]$: for each $a$ in $\mathbb{Z_{p}}$, where $p$ is a prime number. Any help is greatly appreciated.
Show $x^{p}+a$ is reducible in $\mathbb{Z}_p[x]$
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abstract-algebra
irreducible-polynomials
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1Use Fermat's Little Theorem to show that the polynomial has a root. – 2017-02-19
1 Answers
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.Little tricks are useful to know. Here's one to remember.
So we have $x^p + a \mod p$. Since we are working $\mod p$, and $p$ is a prime, I want you to note that $\binom p k = \frac{p!}{(p-k)!k!}$ is divisible by $p$ for all $ 0 < k < p$. This is because $p$ will be in the numerator in $p!$, but in the denominator, $k!$ and $(n-k)!$ are products of numbers smaller than $p$, so cannot cancel out the $p$ from the numerator as $p$ is a prime, so none of these numbers can divide $p$.
Here's the trick: \begin{split} x^p + a & \equiv x^p + a^p \mod p \quad \text{ (because $a^p \equiv a \mod p$)} \\ & \equiv x^p + \sum_{k=1}^{n-1} \binom pk x^{k}a^{p-k} + a^p \mod p \quad \text{(by our logic above)} \\ & \equiv \sum_{k=0}^n \binom pk x^k a^{p-k} \mod p \quad \text{(Check this yourself!)} \\ & \equiv (x+a)^p \mod p \quad \text{(Ordinary binomial expansion)} \end{split}
But then, we have factored $x^p + a$ as $(x+a)^p$ $\mod p$.