I have just started out learning about fields and had tried practicing solving some questions. However, when I encountered this question, I wasn't really sure how to tackle it.
Let $\mathbb{F} = \{0,1,a,b\}$ be a field with four elements.
Prove that $1 + 1 = 0$.
I wanted to show that $1 + 1 = a$, $ 1 + 1 = 1$, and $1 + 1 = b$ would all result in a contradiction. So, I started with supposing that $1+1= a$.
Since I have already proved that $a \cdot b =1$, $a^2=b$, and $b^2=a$, I thought about using at least one of these pieces of information to arrive at a contradiction, but I can't seem to figure how to use them in this case.
In the question, a hint is also given which suggests showing how assuming that $1+1 =a $ will let us make use of $1+a$ and $1+b$ to arrive at $a^2 =0$, which is impossible. But I can't seem to understand how to use it.
Any help would be appreciated.
First, I claim that $1+a=b$.
For if $1+a=0$, then $a=-1$ so $a^2=1$, but you proved $a^2=b$;
if $1+a=1$, then $a=0, $ a contradiction;
and if $1+a=a$, then $1=0$, a contradiction.
A similar argument proves that $1+b=a$.
Adding $1+a=b$ and $1+b=a$ yields $(1+a)+(1+b)=b+a$
or $1+1+a+b=a+b$ (using commutativity of addition),
which implies (subtracting $a+b$ from both sides) that $1+1=0$.
Quod erat demonstrandum.