Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)?

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

3 Answers 3

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.

share|improve this answer

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 $$ (x-a)^p=\sum_k{p\choose k}x^k(-a)^{p-k}=x^p-a^p+\text{terms divisible by } p $$

share|improve this answer

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 $f(x)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.