Galois Field of order 4 operation with addition

Question

Before posting any answers to my question; just a little note that, I only know the name of this such thing but have no clue what it is, i.e. the Galois Field of order 4.

Also, I have looked at the other posts that might explain it but it doesn't really clear it up for me, because how I go about these things is by showing that it fails one of the axioms of a field by the use of corollaries and theorems.

Suppose that we have a finite-field $F = \{0,1,a,b\}$.

Question: How does $1+1 = 0$? I tried to think through in my head the other options and show that it can't be those other options, i.e. $1+1 \neq 1$ was the easy one.

$1+1 = 1 \implies 1 = 0$ but $1 \neq 0$ thus a contradiction.

I'm not sure how to go about how to show that $1+1 \neq x$?

Thanks.

You may find this helpful. Depends. What other algebraic structures have been covered up to this point? Groups? Rings? — Jyrki Lahtonen, Sep 21 '18 at 18:56

score 1 · Answer 1 · answered Sep 21 '18 at 04:31

1

Under addition, $F$ is a group, and $1$ is a nonzero element. By Lagrange, it either has order $2$, i.e., $1+1=0$, or order $4$, and $F$ is cyclic. In the latter case $F=\{0,1,1+1,1+1+1\}$ where $1+1+1+1=0$. Then consider $2=1+1\ne0$. What is $2\times 2$ in $F$?

answered Sep 21 '18 at 04:31

Angina Seng

158,341

This is too complex of an answer as where I am at in my math knowledge. Sorry can't understand it. – javacoder Sep 21 '18 at 04:33

Galois Field of order 4 operation with addition

1 Answers1