Extension Fields¶
Definition¶
Let \(i = \sqrt{-1}\).
Extension fields function by introducing a new element and defining its relationship with existing elements in the field (in this case, \(i^2 + 1 = 0\)) (also referred to as a Quadratic field).
Extensions of Finite Fields¶
Standard operations apply, with the exception of division:
\(a / b = (a \cdot b^{p^{2}-2}) \pmod {p}\)
\(x^{p^2 - 1} \equiv 1 \pmod p\) => \(p^2 - 1\) is the **order of the multiplicative group in the field **
Examples¶
Consider extending the prime field modulo 7 with \(i\):
- \((2 + 3i) + (4 + 2i) = 6 + 5i\)
- \((5 + 2i) + 3 = 1 + 2i\)
- \((6 + 2i) \cdot 2 = 5 + 4i\)
- \(4i \cdot (2 + i) = 3 + i\)
However, the result of \((2 + 3i) / (4 + 2i)\) may be a bit challenging to understand. Here’s the breakdown: we first decompose the product into \(4i \cdot 2 + 4i \cdot i\), yielding \(8i - 4\). Since we’re working in modulo 7 arithmetic, this simplifies to \(i + 3\).