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Friday, October 9, 2015

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.

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