The set called $GF(6)=\{0,1,2,3,4,5\}$ also has the mathematical operations addition modulo $6$ and multiplication modulo $6$. We want to prove that this is not a field, since it does not match with the law of fields. In which one do we see the mistake?
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1$\begingroup$ Hint: $2\cdot 3 = 6 = 0$. $\endgroup$– lisyarusCommented Oct 20, 2018 at 9:58
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$\begingroup$ why equal to 6? $\endgroup$– M.PapapetrosCommented Oct 20, 2018 at 16:43
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2 Answers
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Hint: In this one:$$(\forall x\in K\setminus\{0\})(\exists y\in K):xy=yx=1.$$
answered Oct 20, 2018 at 10:01
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$\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$– Aloizio Macedo ♦Commented Oct 21, 2018 at 5:50
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For $GF$ to be a field it has to have for every element (except for $0$) an multiplicative inverse. Hence only $1$ and $5$ has one and the others don’t, it’s not an field. For every finite field such as it has an number of elements which is not prime $p$ many or $p^n$ for $n$ is natural, it can’t be an field.
