Ring $A$ with $4$ elements

Question

We have a ring $A$ with order $4$. Show that $A$ is a field if and only if the equation $x^{2}+x+1=0$ has a root in $A$. I minded checking all 11 rings with order 4, but i am sure there is a better method.

Answer 1

Let $R = \{0,1,a,b\}$ be the ring in question. Suppose $R$ is in fact a field. Then $a$ has to have order 3 with inverse $b$, since it is the only other element available. Thus $b = a^2$, and $R = \{0,1,a,a^2\}$.

Notice that $R$ has characteristic 2. The see this, consider the sequence $1,1+1,1+1+1,...$ and see when it starts repeating.

Now consider what the expression $a+1$ is equal to in $R$. We cannot have $a+1 \in \{1,a\}$ because $1$ and $a$ are not zero. Thus $a+1 \in \{0,a^2\}$

Since we are in characteristic 2, $a+1 = 0$ would imply $a=-1=1$ which is a contradiction, so $a+1 = a^2 = -a^2 \implies a^2 + a + 1 =0$.

Likewise, if we assume $R = \{0,1,a,b\}$ has an element (wlog a) such that $a^2+a+1 = 0$ then $a(a+1) = -1$, which is invertible. By a similar argument as before, its inverse must be $b$ so all elements are invertible. Thus $R$ is a field.