Ciro Santilli $$Sponsor Ciro$$ 中国独裁统治 China Dictatorship 新疆改造中心、六四事件、法轮功、郝海东、709大抓捕、2015巴拿马文件 邓家贵、低端人口、西藏骚乱
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# GF(4)

| nosplit | ↑ parent "Finite field" | words: 375
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Ciro Santilli tried to add this example to Wikipedia, but revert, so here we are.
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This is a good first example of a field of a finite field of non-prime order, this one is a prime power order instead.
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, so one way to represent the elements of the field will be the to use the 4 polynomials of degree 1 over GF(2):
• 0X + 0
• 0X + 1
• 1X + 0
• 1X + 1
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Note that we refer in this definition to anther field, but that is fine, because we only refer to fields of prime order such as GF(2), because we are dealing with prime powers only. And we have already defined fields of prime order easily previously with modular arithmetic.
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Over GF(2), there is only one irreducible polynomial of degree 2: $$X2+X+1 (164)$$
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Addition is defined element-wise with modular arithmetic modulo 2 as defined over GF(2), e.g.: $$(1X+0)+(1X+1)=(1+1)X+(0+1)=0X+1 (165)$$
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Multiplication is done modulo , which ensures that the result is also of degree 1.
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For example first we do a regular multiplication: $$(1X+0)×(1X+1)=(1×1)X2+(1×1)X+(0×1)X+(0×1)=1X2+1X+0 (166)$$
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Without modulo, that would not be one of the elements of the field anymore due to the !
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So we take the modulo, we note that: $$1X2+1X+0=1(X2+X+1)+(0X+1) (167)$$ and by the definition of modulo: $$(1X2+1X+0)mod(X2+X+1)=(0X+1) (168)$$ which is the final result of the multiplication.
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TODO show how taking a reducible polynomial for modulo fails. Presumably it is for a similar reason to why things fail for the prime case.
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