How does one prove that : $$\exists !(+,\cdot)|(\mathbb{N}_4,+,\cdot)$$ ,with 0,1 neutral elements of +,., is a field, without using brute force ?
Asked
Active
Viewed 51 times
0
-
1Explain what you write. It's not clear what you mean. – Joshua Tilley Mar 15 '21 at 11:58
-
1First, $(1+1)^2=1+1+1+1=0$ because of Cauchy's theorem, so $1+1=0$, then there's one more element, say $a$ and the forth must be $a+1$. You can check all possible values for $a^2$ and you will find that the only choice that gives a field is $a^2=a+1$. – Berci Mar 15 '21 at 12:10
-
Thanks. I noticed that I understood nothing of Galois theory. I just wrote a code to check all possible operations that satisfy field axioms and was astonished there was only one pair. – QuantumPotatoïd Mar 15 '21 at 18:30