Applied Discrete Structures

First the field must contain the additive and multiplicative identities, 0 and 1, so, without loss of generality, we can assume that the field we are looking for is of the form $F=\{0,1,a,b\}\text{.}$ Since there are only two nonisomorphic groups of order 4, we have only two choices for the group table for $[F;+]\text{.}$ If the additive group is isomorphic to ${\mathbb{Z}}_4$ then two of the nonzero elements of $F$ would not be their own additive inverse (as are 1 and 3 in ${\mathbb{Z}}_4$). Let’s assume $\beta \in F$ is one of those elements and $\beta +\beta =\gamma \ne 0\text{.}$ An isomorphism between the additive groups $F$ and ${\mathbb{Z}}_4$ would require that $\gamma$ in $F$ correspond with 2 in ${\mathbb{Z}}_4\text{.}$ We could continue our argument and infer that $\gamma \cdot \gamma =0\text{,}$ producing a zero divisor, which we need to avoid if $F$ is to be a field. We leave the remainder of the argument to the reader. We can thus complete the addition table so that $[F;+]$ is isomorphic to ${\mathbb{Z}}_2{}^2\text{:}$

Next, since 1 is the unity of $F\text{,}$ the partial multiplication table must look like:

Hence, to complete the table, we have only four entries to find, and, since $F$ must be commutative, this reduces our task to filling in three entries. Next, each nonzero element of $F$ must have a unique multiplicative inverse. The inverse of $a$ must be either $a$ itself or $b\text{.}$ If $a^{-1}=a\text{,}$ then $b^{-1}=b\text{.}$ (Why?) But $a^{-1}=a\Rightarrow a\cdot a=1\text{.}$ And if $a\cdot a=1\text{,}$ then $a\cdot b$ is equal to $a$ or $b\text{.}$ In either case, by the cancellation law, we obtain $a=1$ or $b=1\text{,}$ which is impossible. Therefore we are forced to conclude that $a^{-1}=b$ and $b^{-1}=a\text{.}$ To determine the final two products of the table, simply note that, $a\cdot a\ne a$ because the equation $x^2=x$ has only two solutions, 0 and 1 in any field. We also know that $a\cdot a$ cannot be 1 because $a$ doesn’t invert itself and cannot be 0 because $a$ can’t be a zero divisor. This leaves us with one possible conclusion, that $a\cdot a=b$ and similarly $b\cdot b=a\text{.}$ Hence, our multiplication table for $F$ is: