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Tuesday, October 20, 2015

UTM Ideals Varieties and Algorithm - Chapter 1 Section 1 Exercise 4

Problem:


Solution:

Step 1: The set $ \mathbf{F} - \{0\} $ together with multiplication is a group - all axioms can be trivially verified by virtue that $ \mathbf{F} $ is a finite field.

Step 2: The set $ a^p $, $ \forall p \in \mathbf{Z} $, $ a \neq 0 \in \mathbf F $ is the cyclic subgroup generated by $ a $. As such, the subgroup has an order $ r \mid q-1 $. Therefore $ a^{q-1} = (a^r)^k = 1^k = 1 $. $a^r = 1 $ is because it is a cyclic subgroup.

Step 3: For $ a \ne 0 $, the above show that $ a^{q} = a(a^{q-1}) = a $, therefore $ a^q - a = 0$. It is obvious that $ a^q - a = 0 $ for $ a = 0 $ as well.

Therefore we have just proved that $ a^q - a = 0 $ for all $ a \in \mathbf{F} $.

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