# Show that for $0 \leq a < p$, the polynomial $f(x)=x^p-a \in \mathbb{Z}_p[x]$ has a linear factor

Good evening! I need your help. Let $p$ be a prime number and consider the ring $\mathbb{Z}_p = \mathbb Z/p\mathbb Z$.
(i.) Find all divisors of $0$ in $\mathbb{Z}_p$
(ii.) Show that for $0 \leq a$, where $a$ is less than p, the polynomial $f(x)=x^p-a \in \mathbb{Z}_p[x]$ has a linear factor in $\mathbb{Z}_p$.

For (i): Since $p$ is a prime then $\mathbb{Z}_p$ doesn’t contain a zero divisor. For (ii): If I choose $p=2$ and $a=1$, I see that $x^2-1$ has linear factors $x+1$ and $x-1$.May someone help me with a well procedure of proving (ii)?

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For (ii), check out Wikipedia on Fermat's littletheorem. –  user2468 Feb 1 '12 at 18:28

Hint: invoke $\:a\ne 0\ \Rightarrow\ a^{p-1} = 1$ in $\mathbb Z/p\$ (Fermat's little Theorem)

$\rm(i)\$ If $a\ne 0$ then $a^{-1} = a^{p-2}$ thus $ab = 0$ times $a^{-1}$ yields $b = 0$.

$\rm(ii)\$ $f(x) = x^p-a$ has root $x = a,\:$ hence $f(x)$ has factor $x-a$ by the Factor Theorem.

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every $a\in\mathbb{F}_p$ satisfies $a^p=a$ (for instance, $a^{p-1}=1$ since the multiplicative group $\mathbb{F}_p^{\times}$ has order $p-1$). hence $x^p-a=(x-a)^p$ since $$(x-a)^p=\sum_k{p\choose k}x^k(-a)^{p-k}=x^p-a^p+\text{terms divisible by } p$$

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For (i), consider that if for $a,b \in \mathbb{Z}_p$ $ab = 0$ with $a \neq 0$ and $b \neq 0$.

Then $ab = 0 \Rightarrow ab = kp$ for some $k \in \mathbb{Z}$. Thus $p$ divides $ab$. We know that if a prime divides $ab$, it must divide at least one of $a$ or $b$. Consider WLOG if $a$ is divisible by $p$. Then $a = np$ for some $n \in \mathbb{Z}$, so $a = 0 \in \mathbb{Z}_p$ which contradicts $a \neq 0$. Thus there are no zero divisors in $\mathbb{Z}_p$.

For (ii), for all $a \in \mathbb{Z}_p$, $a^p = a$, so $a^p - a$ = 0, and $a$ is a zero of $f(x)$. Then $x - a$ is a linear factor of $f(x)$.

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