Proving Isomorphism of Fields with Four Elements

[Bashyboy]

Homework Statement

Show that any two Fields with four elements are isomorphic.

Homework Equations

The Attempt at a Solution

Let $F= \{0,1,a,b\}$ be a field with four elements. Since $F$ is an additive abelian group, by Lagrange's theorem $1$ must have either an order of $2$ or $4$; i.e., either $1+1=0$ or $1+1+1+1=0$. Since $F$ is a field, every nonzero element is a unit, meaning that the group of units $U(F)$ is a group of $3$ which makes it isomorphic to $\Bbb{Z}_3$. Note that neither $a+b=a$ nor $a+b=b$ can hold. This leaves us with $a+b = 0$ or $a+b=1$...This is where I get stuck. I am trying to show that $1+1 = 0$. If I can rule out $a+b=1$, then I can show $b=-a$, and therefore $(-a)^3 = 1$ or $-1=1$ or $1+1=0$. Another route is to assume that $1+1+1+1=0$ is true, which would imply $F \simeq \Bbb{Z}_4$, and then deduce a contradiction, but I haven't been able to identify the contradiction. Perhaps there is a problem with $F \simeq \Bbb{Z}_4$ and $U(F) \simeq \Bbb{Z}_3$ being simultaneously true...I could use some guidance.

 

[pasmith]

What do you get if you expand (1 + 1)^2?

 

[Bashyboy]

Okay. So $(1+1)^2 = (1+1)(1+1) = 1 + 1 + 1 + 1 = 0$. Since we are working in a field there can be no nonzero divisors which means $1+1=0$.

Would this be another valid way of proving the claim? Suppose that $a+b \neq 0$. This means that $a+b \in U(F) \simeq \Bbb{Z}_3$, which implies that $(a+b)^3 = 1$ or $1 + 3a^2b + 3a b^2 = 0$. Multiplying by $a$ and then $b$ yields the equations $a + 3b + 3a^2 b^2 = 0$ and $b + 3a^2 b^2 + 3a = 0$, and adding the two gives $4a + 4b + 6a^2 b^2 = 0$ or $6a^2 b^2 = 0$, which is a contradiction since neither $a=0$ nor $b=0$. Hence $a + b = 0$ or $b = -a$. Then $(-a)^3 = 1$ implies $-1 = 1$.

Admittedly it is a bit longer, but somewhat interesting, although it may be invalid/unsound.

 

[pasmith]

Okay. So $(1+1)^2 = (1+1)(1+1) = 1 + 1 + 1 + 1 = 0$. Since we are working in a field there can be no nonzero divisors which means $1+1=0$.

Would this be another valid way of proving the claim? Suppose that $a+b \neq 0$. This means that $a+b \in U(F) \simeq \Bbb{Z}_3$, which implies that $(a+b)^3 = 1$ or $1 + 3a^2b + 3a b^2 = 0$. Multiplying by $a$ and then $b$ yields the equations $a + 3b + 3a^2 b^2 = 0$ and $b + 3a^2 b^2 + 3a = 0$, and adding the two gives $4a + 4b + 6a^2 b^2 = 0$ or $6a^2 b^2 = 0$, which is a contradiction since neither $a=0$ nor $b=0$.

How do you justify the implicit assumption that 6 \neq 0?

Hence $a + b = 0$ or $b = -a$. Then $(-a)^3 = 1$ implies $-1 = 1$.

It is in fact true that a + b = 1: it follows from 1 + 1 = 0 and the field axioms.