Background: I came across the following problem.
Let $R$ be a ring with four elements. Show that $R$ is a field if and only if the equation $x^2 + x + 1 = 0$ has a root in $R$.
My solution:
- Suppose $R = \{0,1,a,b\}$ is a field. Then, $R\setminus \{0\}$ is a multiplicative group of order $3$, i.e., $R\setminus \{0\} \cong \Bbb Z/3\Bbb Z$ is a cyclic group. WLOG, suppose $R\setminus \{0\} = \{1,a,b\} = \langle a \rangle$, i.e., $a^3 = 1$. We must have $b = a^{-1} = a^2$. Lastly, $a^3 - 1 = 0$ can be factored as $(a-1)(a^2+a+1) = 0$, implying $a^2 + a + 1 = 0$ as $a \ne 1$.
- Suppose $x^2 + x + 1 = 0$ has a root in $R = \{0,1,a,b\}$. As $1\ne 0$, $0$ cannot be a root of the said equation. If $1$ is a root, then $1 + 1 + 1 = 0$, i.e., $R$ has characteristic $3$. The elements $\{0,1,1+1\}$ form an additive subgroup of $R$, but this is a contradiction since $3$ doesn't divide $4$. WLOG, suppose $a^2 + a + 1 = 0$. Then, $a(-1-a) = 1$, i.e., $a$ is invertible. Clearly, $a^{-1} = b$, so $b$ is also invertible. Thus, $R$ is a field.
Question: I'm wondering if it's possible to generalize the above statement, i.e., if $R$ is a ring with $n$ elements ($n\in \Bbb N$), can we find a polynomial $p_n(x) \in \Bbb Z[x]$ such that $R$ is a field if and only if $p_n(x)$ has a root in $R$?
Thanks for any ideas and suggestions!