UTM Ideals Varieties and Algorithm - Chapter 1 Section 1 Exercise 3

Problem:

Solution:

(Part a)

The closure, associativity, identity properties are obvious. The only thing unclear is why such an operation has multiplicative inverse.

Note that for a fixed $a \neq 0$, if $x \neq y$, then $ax \neq ay$ because otherwise $a(x - y) = 0$ and that is impossible if both of them are non-zero.

When $x$ varies from 1 to $p$, there are $p$ different $ax$ values ranging from 1 to $p$. By the pigeonhole principle one of them must be 1.

(Part b)

Thanks for the hint for using the Lagrange's theorem, we know that the subgroup generated by $a$ have an order $s$ which is a factor of $p - 1$. Now we can write $a^{p-1} = a^{ks} = (a^s)^k = 1^k = 1$.

For myself in the future, I need to explain more why $a^s = 1$. Consider the following sequence that represent the cyclic subgroup.

$a^0, a^1 \cdots a^{s-1}$.

Now it is obvious why $a^s = 1$

(Part c)

Again, thanks for the hint. It is obvious now if $a = 0, a^p = 0 = a$. If $a \neq 0$, part b applies and just multiply both side by $a$.

(Part d)

What else would that be? $a^p - a$, that has to be 0.